- The supplied derivative tells how the tangent slope changes with \(t\), but the question asks how it changes with \(x\). - Use the given \(x\)-rate to convert between those two kinds of change. - Check the sign: a negative numerator divided by a positive \(x\)-rate stays negative.

1. Convert differentiation with respect to \(t\) into differentiation with respect to \(x\): \(\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}\). 2. Thus, \(\frac{d^2y}{dx^2}=\frac{-6}{2}=-3\).

\(-3\)

- First decide whether \(x\) is increasing or decreasing as the parameter changes. - Then determine whether the tangent slope is increasing or decreasing with the parameter. - Concavity describes how tangent slope changes as \(x\) increases.

1. Panel a) shows that \(x(t)\) is increasing, so \(\frac{dx}{dt}>0\). 2. Panel b) shows that the tangent slope \(m(t)\) is decreasing, so \(\frac{dm}{dt}<0\). 3. Thus \(\frac{d^{2}y}{dx^{2}}=\frac{dm/dt}{dx/dt}<0\), and the curve is concave down.

Concave down at \(t=0\).

- Compare the tangent slopes just to the left and right of the point. - How does the slope change as you move to the right?

1. The curve bends upward around the marked point, and its slope increases as \(x\) increases. 2. It is concave up there.

Concave up.

- Increasing in parameter order is not necessarily increasing from left to right. - What happens to the sign when a positive rate is divided by a negative rate?

1. Increasing tangent slope as \(t\) increases means \(\frac{d}{dt}\left(\frac{dy}{dx}\right)>0\). 2. Since \(\frac{dx}{dt}<0\), dividing the first rate by the second gives \(\frac{d^2y}{dx^2}<0\).

Negative.

- Follow the curve from left to right through P. - Compare the tangent line's tilt just before and just after P.

1. The curve bends downward, so the tangent changes from steeper upward slopes on the left to lower slopes on the right. 2. Its tangent slopes decrease.

The tangent slopes decrease.

- Distinguish the tangent slope from its rate of change with respect to the parameter. - How does the horizontal coordinate rate enter when converting that change to a rate with respect to x?

1. The derivative of the tangent slope with respect to \(t\) is \(0\). 2. Dividing by \(\frac{dx}{dt}=2\) gives \(\frac{d^2y}{dx^2}=0\).

\(0\).

- First determine how the tangent slope changes with the parameter. - The requested derivative measures that slope change with respect to x, not with respect to \(t\). - Use all four entries in the table and keep the stages distinct.

1. Differentiate the first-derivative ratio with respect to \(t\): \(\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{4(2)-3(-1)}{2^2}=\frac{11}{4}\). 2. Divide this rate by \(\frac{dx}{dt}=2\). 3. Therefore, \(\frac{d^2y}{dx^2}=\frac{11}{8}\).

\(\frac{11}{8}\)

- Distinguish the rate of slope change with respect to \(t\) from the rate with respect to x. - Relate the two rates using the horizontal component rate. - Preserve the sign information when solving.

1. The second derivative satisfies \(\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}\). 2. Substitution gives \(3\frac{dx}{dt}=-6\). 3. Therefore, \(\frac{dx}{dt}=-2\).

\(\frac{dx}{dt}=-2\)

- The table gives how tangent slope changes with the parameter, not directly with x. - Account for the sign of the horizontal component rate at each row. - Classify concavity from the sign of the converted rate.

1. At \(t=-1\), \(\frac{d^2y}{dx^2}=\frac{6}{-2}=-3\), so the curve is concave down. 2. At \(t=0\), \(\frac{d^2y}{dx^2}=\frac{-3}{3}=-1\), so the curve is concave down. 3. At \(t=2\), \(\frac{d^2y}{dx^2}=\frac{-4}{-1}=4\), so the curve is concave up.

\(t=-1\): concave down \(t=0\): concave down \(t=2\): concave up

- The problem already provides the first derivative as a function of the parameter. - Differentiate that expression, then account for how x changes with the parameter. - Evaluate only after forming the second derivative.

1. Differentiate the given tangent slope with respect to \(t\): \(\frac{d}{dt}\left(\frac{dy}{dx}\right)=3t^2-1\). 2. The horizontal component rate is \(\frac{dx}{dt}=2t\). 3. Therefore, \(\frac{d^2y}{dx^2}=\frac{3t^2-1}{2t}\). 4. At \(t=1\), the value is \(1\).

\(1\)

- Watch how the tangent slopes change from left to right. - What might happen to the tangent slope between two measurements?

1. The tangent slope becomes larger as \(x\) increases. 2. These samples suggest concave up because their slopes rise. 3. Three values do not show how the slope changes at every point between them, so they do not prove concavity throughout the interval.

The data suggest concave up, but do not prove concavity everywhere between the sampled points.

- Check which variable the second derivative measures change against. - What does a negative horizontal rate say about the direction of travel?

1. The slope's rate with respect to the parameter is \(\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{d^2y}{dx^2}\frac{dx}{dt}\). 2. Substituting gives \(4(-3)=-12\). 3. The slope falls as \(t\) increases because \(x\) is decreasing, but the slope rises as \(x\) increases; that is concave up.

\(\frac{d}{dt}\left(\frac{dy}{dx}\right)=-12\). The curve is concave up because the tangent slope increases with \(x\), even though \(x\) falls as \(t\) increases.

- Start with the two component derivatives, not with the second derivatives. - A rate of change of slope with respect to \(t\) is not yet a rate with respect to \(x\).

1. Differentiate to get \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=2t\), hence \(\frac{dy}{dx}=t\). 2. Differentiate the slope with respect to \(t\) and divide by \(2\): \(\frac{d^2y}{dx^2}=\frac{1}{2}\).

\(\frac{1}{2}\).

- The graph shows how the tangent slope changes with \(t\), not with \(x\). - How quickly is \(x\) changing as \(t\) increases?

1. The slope graph passes through \((0,2)\) and \((2,0)\), so \(\frac{ds}{dt}=\frac{0-2}{2-0}=-1\). 2. From \(x(t)=2-t\), \(\frac{dx}{dt}=-1\). 3. Thus \(\frac{d^2y}{dx^2}=\frac{ds/dt}{dx/dt}=\frac{-1}{-1}=1\).

\(\frac{d^2y}{dx^2}=1\).

- Look at how the curve bends separately on the two sides of the origin. - The curve need not have the same concavity everywhere.

1. On the left of the origin near P, slopes fall as \(x\) increases, giving concave down. 2. On the right near Q, slopes rise as \(x\) increases, giving concave up.

No. P: concave down; Q: concave up.

- Differentiate the quotient \(\frac{y'}{x'}\) before finding the second derivative with respect to \(x\). - Check how many factors of the horizontal rate end up in the denominator.

1. The tangent slope is \(\frac{y'}{x'}\). Its derivative with respect to \(t\) is \(\frac{x'y''-y'x''}{(x')^2}\). 2. Divide again by \(x'\): \(\frac{d^2y}{dx^2}=\frac{2\cdot4-3\cdot1}{2^3}=\frac{5}{8}\).

\(\frac{5}{8}\).

- Locate the parameter value for P, then find how the coordinates change there. - Compare the signs of those rates with the drawn arrow. - To find concavity, differentiate the Cartesian slope with respect to \(x\), not just \(t\).

1. At P, \(x=1\) gives \(t=-1\). The component rates are \(\frac{dx}{dt}=-1\) and \(\frac{dy}{dt}=2t=-2\), so increasing \(t\) moves left and down. 2. The arrow points right and up, so it shows the opposite direction. 3. Since \(\frac{dy}{dx}=\frac{2t}{-1}=-2t\), the Cartesian second derivative is \(\frac{d^2y}{dx^2}=\frac{-2}{-1}=2>0\).

No; the arrow points against increasing \(t\). The second derivative is \(2>0\), so the curve is concave up.

- Which variable did Mira differentiate with respect to? - The requested second derivative measures slope change per unit of \(x\), not \(t\).

1. The result \(6\) is the change of slope per unit of \(t\). 2. Divide by the horizontal coordinate rate: \(\frac{d^2y}{dx^2}=\frac{6}{3}=2\).

\(\frac{d^2y}{dx^2}=2\).

- The second derivative is the derivative of the first derivative. - What is the derivative of a constant?

1. The tangent slope is constant, so its rate of change is zero. 2. Therefore the Cartesian second derivative is \(0\).

\(0\); a constant tangent slope has zero derivative.

- Start from the tangent slope as a function of the parameter. - Ask how a rate of change with respect to \(t\) can be converted into a rate of change with respect to \(x\). - Check that the horizontal component rate is nonzero at the parameter value where you evaluate the result.

1. The first derivative is \(\frac{dy}{dx}=\frac{3t^{2}-2}{2t}=\frac{3t}{2}-\frac{1}{t}\). 2. Differentiate with respect to \(t\): \(\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{3}{2}+\frac{1}{t^{2}}\). 3. Divide by \(\frac{dx}{dt}=2t\): \(\frac{d^{2}y}{dx^{2}}=\frac{3t^{2}+2}{4t^{3}}\). 4. At \(t=1\), the value is \(\frac{5}{4}\).

\(\frac{d^{2}y}{dx^{2}}=\frac{5}{4}\)

- Find the first and second derivatives of each coordinate with respect to \(t\). - The second derivative with respect to \(x\) is not simply the second derivative of \(y\) with respect to \(t\). - Verify that the horizontal component rate is nonzero.

1. At \(t=0\), the coordinate rates are \(x'=2\), \(x''=1\), \(y'=1\), and \(y''=3\), with primes denoting derivatives with respect to \(t\). 2. For a parametric curve, \(\frac{d^2y}{dx^2}=\frac{y''x'-y'x''}{(x')^3}\) when \(x'\ne0\). 3. Substitution gives \(\frac{3\cdot2-1\cdot1}{2^3}=\frac{5}{8}\).

\(\frac{d^2y}{dx^2}=\frac{5}{8}\)

- Find and simplify the tangent slope before trying to evaluate a second derivative. - Keep track of which variable each derivative is taken with respect to. - Before evaluating, verify that the horizontal component rate is nonzero at \(t=\frac{\pi}{2}\).

1. The first derivative is \(\frac{dy}{dx}=-\frac{\sin(t)}{1+\cos(t)}\). 2. Differentiate with respect to \(t\): \(\frac{d}{dt}\left(\frac{dy}{dx}\right)=-\frac{1}{1+\cos(t)}\). 3. Divide by \(\frac{dx}{dt}=1+\cos(t)\): \(\frac{d^{2}y}{dx^{2}}=-\frac{1}{(1+\cos(t))^{2}}\). 4. At \(t=\frac{\pi}{2}\), the value is \(-1\).

\(\frac{d^{2}y}{dx^{2}}=-1\)

- Keep the first derivative in a form that makes its change with the parameter manageable. - A derivative of the slope with respect to \(t\) is not yet the requested second derivative with respect to \(x\). - Check the parameter value against any denominator restrictions before evaluating.

1. The first derivative is \(\frac{dy}{dx}=\frac{2t}{3(t^{2} - 1)}\). 2. Differentiate with respect to \(t\): \(\frac{d}{dt}\left(\frac{dy}{dx}\right)=-\frac{2(t^{2} + 1)}{3(t^{2} - 1)^{2}}\). 3. Divide by \(\frac{dx}{dt}=3(t^{2} - 1)\): \(\frac{d^{2}y}{dx^{2}}=-\frac{2(t^{2} + 1)}{9(t^{2} - 1)^{3}}\). 4. At \(t=2\), the value is \(-\frac{10}{243}\).

\(\frac{d^{2}y}{dx^{2}}=-\frac{10}{243}\)

- The logarithm requires a positive parameter, which includes the specified value. - Find the first and second rates of each coordinate with respect to \(t\). - The requested second derivative measures how \(\frac{dy}{dx}\) changes with \(x\), not just with \(t\).

1. At \(t=1\), the component derivatives are \(x'=1+\frac{1}{t}=2\), \(x''=-\frac{1}{t^{2}}=-1\), \(y'=2t=2\), and \(y''=2\). 2. Use \(\frac{d^{2}y}{dx^{2}}=\frac{y''x'-y'x''}{(x')^{3}}\). 3. Substituting gives \(\frac{2\cdot2-2(-1)}{2^{3}}=\frac{3}{4}\).

\(\frac{d^{2}y}{dx^{2}}=\frac{3}{4}\)

- Watch the signs when differentiating the reciprocal terms. - Find the four coordinate-derivative values at the given parameter before substituting. - Check that the horizontal rate is nonzero.

1. At \(t=2\), \(x'=1-\frac{1}{t^2}=\frac{3}{4}\) and \(x''=\frac{2}{t^3}=\frac{1}{4}\). 2. Also, \(y'=2t+\frac{1}{t^2}=\frac{17}{4}\) and \(y''=2-\frac{2}{t^3}=\frac{7}{4}\). 3. The Cartesian second derivative is \(\frac{d^2y}{dx^2}=\frac{y''x'-y'x''}{(x')^3}=\frac{\frac{7}{4}\cdot\frac{3}{4}-\frac{17}{4}\cdot\frac{1}{4}}{\left(\frac{3}{4}\right)^3}=\frac{16}{27}\).

\(\frac{d^2y}{dx^2}=\frac{16}{27}\)

- Determine which component rate controls whether differentiation with respect to \(x\) is valid. - Keep the parameter form rather than trying to solve either coordinate equation for \(t\). - After combining the component derivatives, inspect the denominator for possible zeros.

1. The parameter derivatives are \(x'=1+e^t\), \(x''=e^t\), \(y'=3t^2+e^t\), and \(y''=6t+e^t\). 2. Therefore, \(\frac{d^2y}{dx^2}=\frac{(6t+e^t)(1+e^t)-(3t^2+e^t)e^t}{(1+e^t)^3}\). 3. Simplifying gives \(\frac{6t+e^t(1+6t-3t^2)}{(1+e^t)^3}\). 4. Since \(1+e^t>0\) for every real \(t\), the expression is defined for all real \(t\).

\(\frac{d^2y}{dx^2}=\frac{6t+e^t(1+6t-3t^2)}{(1+e^t)^3}\); defined for all real \(t\).

- The horizontal rate, not merely the trigonometric denominator after simplification, determines where the x-based derivative can fail. - Simplify only after you have distinguished parameter-rates from x-rates. - Solve the resulting horizontal-rate condition over all real parameter values.

1. The first component rates are \(x'=1+\cos(t)\) and \(y'=-\sin(t)\). 2. Differentiating the tangent slope with respect to the parameter and converting to an x-rate gives \(\frac{d^2y}{dx^2}=-\frac{1}{(1+\cos(t))^2}\) wherever \(1+\cos(t)\ne0\). 3. Therefore, the expression is undefined when \(\cos(t)=-1\), or \(t=(2k+1)\pi\), \(k\in\mathbb Z\).

\(\frac{d^2y}{dx^2}=-\frac{1}{(1+\cos(t))^2}\), undefined at \(t=(2k+1)\pi\), where \(k\in\mathbb Z\).

- Differentiate the polynomial coordinates and simplify the first derivative as a rational function. - Apply the parametric second-derivative formula without canceling factors across invalid parameter values. - Identify where \(\frac{dx}{dt}\) vanishes and check the resulting denominator.

1. The first derivative is \(\frac{dy}{dx}=\frac{2t - 1}{3t^{2}}\). 2. Differentiating with respect to \(t\) and dividing by \(\frac{dx}{dt}=3t^{2}\) gives \(\frac{d^{2}y}{dx^{2}}=\frac{2(1-t)}{9t^{5}}\). 3. The expression is undefined at \(t=0\).

\(\frac{d^{2}y}{dx^{2}}=\frac{2(1-t)}{9t^{5}}\), undefined at \(t=0\).

- Compute the first parametric derivative before forming the second derivative. - Evaluate the second derivative at the specified parameter without rounding. - Use a positive value to identify the local concavity.

1. The first derivative is \(\frac{dy}{dx}=\frac{3t^{2}}{2(t + 1)}\). 2. Differentiate with respect to \(t\) and divide by \(\frac{dx}{dt}=2(t+1)\) to obtain \(\frac{d^{2}y}{dx^{2}}=\frac{3t(t+2)}{4(t+1)^{3}}\). 3. At \(t=1\), the value is \(\frac{9}{32}>0\), so the curve is concave up.

The curve is concave up; \(\frac{d^{2}y}{dx^{2}}=\frac{9}{32}\).

- Check the logarithm's parameter domain before evaluating any rates. - Concavity depends on both coordinates' first and second parameter-rates. - Use the sign of the x-based second derivative, not the sign of \(y''(t)\) alone.

1. At \(t=0\), \(x'=\frac32\), \(x''=-\frac14\), \(y'=-1\), and \(y''=3\). 2. Thus \(\frac{d^2y}{dx^2}=\frac{3(\frac32)-(-1)(-\frac14)}{(\frac32)^3}=\frac{34}{27}\). 3. Since \(\frac{34}{27}>0\), the curve is concave up.

The curve is concave up; \(\frac{d^2y}{dx^2}=\frac{34}{27}\).

- Form the second derivative from the same rational expression used for the tangent slope. - Check that \(\frac{dx}{dt}\) is nonzero at the specified parameter. - Use the sign of the exact second derivative to classify concavity.

1. The first derivative is \(\frac{dy}{dx}=\frac{2t}{3(t^{2} - 1)}\). 2. Differentiating with respect to \(t\) and dividing by \(\frac{dx}{dt}=3(t^{2}-1)\) gives \(\frac{d^{2}y}{dx^{2}}=-\frac{2(t^{2}+1)}{9(t^{2}-1)^{3}}\). 3. At \(t=2\), the value is \(-\frac{10}{243}<0\), so the curve is concave down.

The curve is concave down; \(\frac{d^{2}y}{dx^{2}}=-\frac{10}{243}\).

- Differentiate the trigonometric coordinate functions before forming \(\frac{dy}{dx}\). - Apply the parametric second-derivative formula at \(t=0\). - A negative result indicates which way the graph bends locally.

1. The first derivative is \(\frac{dy}{dx}=-\frac{\sin(t)}{2+\cos(t)}\). 2. Differentiating with respect to \(t\) and dividing by \(\frac{dx}{dt}=2+\cos(t)\) gives \(\frac{d^{2}y}{dx^{2}}=-\frac{2\cos(t)+1}{(2+\cos(t))^{3}}\). 3. At \(t=0\), the value is \(-\frac{1}{9}<0\), so the curve is concave down.

The curve is concave down; \(\frac{d^{2}y}{dx^{2}}=-\frac{1}{9}\).

- Respect the logarithm's domain before doing any derivative work. - Concavity requires information from both coordinates, including their second parameter-rates. - Decide concavity only after obtaining the sign of \(\frac{d^{2}y}{dx^{2}}\).

1. At \(t=2\), \(x'=1+\frac{1}{2}=\frac{3}{2}\) and \(x''=-\frac{1}{4}\). 2. Also, \(y'=1-\frac{1}{4}=\frac{3}{4}\) and \(y''=\frac{1}{4}\). 3. Therefore, \(\frac{d^{2}y}{dx^{2}}=\frac{\frac{1}{4}\cdot\frac{3}{2}-\frac{3}{4}(-\frac{1}{4})}{(\frac{3}{2})^{3}}=\frac{1}{6}\). 4. Since \(\frac{1}{6}>0\), the curve is concave up.

The curve is concave up; \(\frac{d^{2}y}{dx^{2}}=\frac{1}{6}\).

- Differentiate the product in the y-component before forming the first parametric derivative. - The exponential factor cancels once in \(\frac{dy}{dx}\), but it matters again when converting the slope's t-rate to an x-rate. - For concavity, separate the sign-changing factor from factors that are always positive.

1. Since \(\frac{dx}{dt}=e^t\) and \(\frac{dy}{dt}=e^t(t^2-2t+1)\), the first derivative is \(\frac{dy}{dx}=(t-1)^2\). 2. Therefore, \(\frac{d}{dt}\left(\frac{dy}{dx}\right)=2(t-1)\). 3. Dividing by \(\frac{dx}{dt}=e^t\) gives \(\frac{d^2y}{dx^2}=2(t-1)e^{-t}\). 4. Because \(e^{-t}>0\), the second derivative changes sign at \(t=1\). The corresponding point is \((e,2e)\).

\(\frac{d}{dt}\left(\frac{dy}{dx}\right)=2(t-1)\) \(\frac{d^2y}{dx^2}=2(t-1)e^{-t}\) Concavity changes at \(t=1\), at \((e,2e)\).

- Find \(\frac{d^{2}y}{dx^{2}}\) in terms of the parameter before substituting either value. - Evaluate at both parameter values, keeping track of the signs. - What does a positive second derivative say about concavity?

1. The second derivative is \(\frac{d^{2}y}{dx^{2}}=\frac{3(t^{2}+1)}{4t^{3}}\). 2. At \(t=1\), it is \(\frac{3}{2}>0\), so the curve is concave up. 3. At \(t=2\), it is \(\frac{15}{32}>0\), so the curve is also concave up.

At \(t=1\), \(\frac{d^{2}y}{dx^{2}}=\frac{3}{2}\), concave up. At \(t=2\), \(\frac{d^{2}y}{dx^{2}}=\frac{15}{32}\), concave up.

- Evaluate the same parametric second-derivative structure at both parameter values. - Do not infer concavity from \(y''(t)\) alone. - Compare the signs of the two final x-based values.

1. At \(t=0\), \(x'=2\), \(x''=1\), \(y'=0\), and \(y''=-2\). Thus \(\frac{d^2y}{dx^2}=-\frac{1}{2}\), so the curve is concave down. 2. At \(t=1\), \(x'=1+e\), \(x''=e\), \(y'=1\), and \(y''=4\). 3. Hence \(\frac{d^2y}{dx^2}=\frac{4(1+e)-e}{(1+e)^3}=\frac{4+3e}{(1+e)^3}>0\), so the curve is concave up.

At \(t=0\), \(\frac{d^2y}{dx^2}=-\frac{1}{2}\), so the curve is concave down. At \(t=1\), \(\frac{d^2y}{dx^2}=\frac{4+3e}{(1+e)^3}\), so the curve is concave up.

- Simplify the common second-derivative expression before evaluating either point. - Substitute the two parameter values separately and retain exact fractions. - Compare the signs first, then the magnitudes if a finer comparison is requested.

1. The second derivative is \(\frac{d^{2}y}{dx^{2}}=\frac{4t^{2}(t^{2}+3)}{9(t^{2}+1)^{3}}\). 2. At \(t=\frac{1}{2}\), \(\frac{d^{2}y}{dx^{2}}=\frac{208}{1125}>0\), so the curve is concave up. 3. At \(t=1\), \(\frac{d^{2}y}{dx^{2}}=\frac{2}{9}>0\), so the curve is also concave up.

At \(t=\frac{1}{2}\), \(\frac{d^{2}y}{dx^{2}}=\frac{208}{1125}\), and the curve is concave up. At \(t=1\), \(\frac{d^{2}y}{dx^{2}}=\frac{2}{9}\), and the curve is also concave up.

- Evaluate the curve coordinates at \(t=2\) to locate the point of tangency. - Use the first derivative for the tangent-line slope and point-slope form. - Compute the second derivative independently and interpret its sign.

1. The point is \((3, 6)\), and the tangent slope is \(\frac{11}{4}\). 2. The tangent line is \(y-6=\frac{11}{4}(x-3)\). 3. The second derivative at the point is \(\frac{13}{32}>0\), so the curve is concave up.

a) \(y-6=\frac{11}{4}(x-3)\) b) Concave up; \(\frac{d^{2}y}{dx^{2}}=\frac{13}{32}\)

- Find the exact point on the curve at \(t=\frac{\pi}{2}\). - Use the component derivatives to build the tangent line. - Apply the parametric second-derivative formula separately to determine concavity.

1. The point is \(\left(0, 1 + \frac{\pi}{2}\right)\), and the tangent slope is \(-1\). 2. The tangent line is \(y-\left(1 + \frac{\pi}{2}\right)=-x\). 3. The second derivative at the point is \(-1<0\), so the curve is concave down.

a) \(y-\left(1 + \frac{\pi}{2}\right)=-x\) b) Concave down; \(\frac{d^{2}y}{dx^{2}}=-1\)

- Use the same parameter value to determine the point and the two first component rates. - The tangent and concavity questions use related but different derivative information. - For concavity, keep track of how the tangent slope changes relative to horizontal motion.

1. At \(t=0\), the point is \((1, 0)\). The component rates are \(x'=2\) and \(y'=1\), so the tangent slope is \(\frac{1}{2}\). 2. The tangent line is \(y=\frac{1}{2}(x-1)\). 3. At \(t=0\), \(x''=1\) and \(y''=2\). Thus \(\frac{d^2y}{dx^2}=\frac{2\cdot2-1\cdot1}{2^3}=\frac{3}{8}\). 4. Since \(\frac{3}{8}>0\), the curve is concave up.

a) \(y=\frac{1}{2}(x-1)\) b) Concave up; \(\frac{d^2y}{dx^2}=\frac{3}{8}\)

- Evaluate both reciprocal coordinate expressions at \(t=2\) to locate the point. - Form the tangent slope from the first component derivatives before writing the line. - Compute the second derivative independently and classify its sign.

1. The point is \(\left(\frac{5}{2}, \frac{3}{2}\right)\), and the tangent slope is \(\frac{5}{3}\). 2. The tangent line is \(y-\frac{3}{2}=\frac{5}{3}\left(x-\frac{5}{2}\right)\). 3. The second derivative at the point is \(-\frac{32}{27}<0\), so the curve is concave down.

a) \(y-\frac{3}{2}=\frac{5}{3}\left(x-\frac{5}{2}\right)\) b) Concave down; \(\frac{d^{2}y}{dx^{2}}=-\frac{32}{27}\)

- Convert the given rate of slope change into a rate with respect to x. - Determine which factors can change sign. - A concavity change requires opposite signs on the two sides of a parameter value.

1. The second derivative is \(\frac{d^2y}{dx^2}=\frac{t-1}{t^2+1}\). 2. The denominator is positive for every real \(t\), so the sign is determined by \(t-1\). 3. The curve is concave down for \(t<1\) and concave up for \(t>1\). 4. The concavity changes at \(t=1\).

Concave down for \(t<1\); concave up for \(t>1\); concavity changes at \(t=1\).

- Work through the second derivative separately for each parameterization. - Account for the horizontal component rate after differentiating the tangent slope. - Compare geometric concavity with the direction in which a curve is traced.

1. For \(\mathbf{r}_1\), \(\frac{dy}{dx}=2t\), so \(\frac{d^2y}{dx^2}=2\). 2. For \(\mathbf{r}_2\), \(\frac{dy}{dx}=-2u\). Differentiating with respect to \(u\) and dividing by \(\frac{dx}{du}=-1\) gives \(\frac{d^2y}{dx^2}=2\). 3. Both parameterizations give the same second derivative because they describe the same geometric curve, even though they trace it in opposite directions.

Both parameterizations give \(\frac{d^2y}{dx^2}=2\).

- Label the independent variable attached to each derivative before comparing the two quantities. - At \(t=0\), find how quickly x itself changes with the parameter. - A rate per unit parameter and a rate per unit horizontal distance need not be equal.

1. The tangent slope is \(\frac{dy}{dx}=\frac{2t}{1+e^t}\). 2. Differentiating this slope with respect to \(t\) gives \(1\) at \(t=0\). 3. At the same parameter, \(\frac{dx}{dt}=2\), so \(\frac{d^2y}{dx^2}=\frac{1}{2}\). 4. The values differ because the first is change of slope per unit \(t\), while the second is change of slope per unit \(x\).

\(\frac{d}{dt}\left(\frac{dy}{dx}\right)=1\) and \(\frac{d^2y}{dx^2}=\frac{1}{2}\) at \(t=0\). They differ because \(\frac{dx}{dt}=2\).

- Determine where horizontal motion vanishes before interpreting the second derivative formula. - The vertical-tangent claim requires checking both component rates, not only \(\frac{dx}{dt}\). - Keep the expression in parameter form; eliminating the parameter is unnecessary.

1. The component rates are \(x'=3t^2-1\), \(x''=6t\), \(y'=2t+1\), and \(y''=2\). 2. Therefore, \(\frac{d^2y}{dx^2}=\frac{2(3t^2-1)-(2t+1)(6t)}{(3t^2-1)^3}=-\frac{2(3t^2+3t+1)}{(3t^2-1)^3}\). 3. At \(t=\frac{1}{\sqrt{3}}\), \(x'=0\) while \(y'=1+\frac{2}{\sqrt{3}}\ne0\), so the tangent is vertical. 4. Because differentiation with respect to \(x\) requires division by the horizontal rate, the displayed second derivative is not defined there.

\(\frac{d^2y}{dx^2}=-\frac{2(3t^2+3t+1)}{(3t^2-1)^3}\) for \(t\ne\pm\frac{1}{\sqrt{3}}\); it is undefined at \(t=\frac{1}{\sqrt{3}}\) because \(\frac{dx}{dt}=0\) there while \(\frac{dy}{dt}\ne0\).

- The horizontal rate is never zero, so the first parametric derivative is defined for every real \(t\). - Use the quotient rule on the first derivative before dividing by the horizontal rate again. - For zeros of the second derivative, focus on the numerator after confirming the denominator cannot vanish.

1. The component rates are \(\frac{dx}{dt}=3t^2+1\) and \(\frac{dy}{dt}=2t\), so \(\frac{dy}{dx}=\frac{2t}{3t^2+1}\). 2. Differentiating with respect to \(t\) gives \(\frac{d}{dt}(\frac{dy}{dx})=\frac{2(1-3t^2)}{(3t^2+1)^2}\). 3. Divide by \(\frac{dx}{dt}=3t^2+1\): \(\frac{d^2y}{dx^2}=\frac{2(1-3t^2)}{(3t^2+1)^3}\). 4. The second derivative is zero at \(t=\pm\frac{1}{\sqrt3}\). At either value, \(y=\frac13\) and \(x=t(t^2+1)=\frac43t\). 5. The points are \(\left(-\frac{4}{3\sqrt3},\frac13\right)\) and \(\left(\frac{4}{3\sqrt3},\frac13\right)\).

\(\frac{dy}{dx}=\frac{2t}{3t^2+1}\) \(\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{2(1-3t^2)}{(3t^2+1)^2}\) \(\frac{d^2y}{dx^2}=\frac{2(1-3t^2)}{(3t^2+1)^3}\) Points: \(\left(-\frac{4}{3\sqrt3},\frac13\right)\) and \(\left(\frac{4}{3\sqrt3},\frac13\right)\)

- Express the requested component acceleration as an unknown. - Relate the given second derivative to the four component-rate quantities. - Solve the resulting linear equation.

1. Let \(\frac{d^2y}{dt^2}=k\). 2. The second derivative with respect to x is \(\frac{2k-1(3)}{2^3}\). 3. Setting \(\frac{2k-3}{8}=-\frac{1}{8}\) gives \(2k-3=-1\). 4. Therefore, \(k=1\).

\(\frac{d^2y}{dt^2}=1\)

- Do not infer an inflection from a zero second derivative alone. - Track the positive denominator separately from the factors that can become zero. - Compare the sign of the final expression for small positive and negative parameter values.

1. The first derivative is \(\frac{dy}{dx}=\frac{4t^3}{3t^2+1}\). 2. Differentiating with respect to \(t\) gives \(\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{12t^2(t^2+1)}{(3t^2+1)^2}\). 3. Dividing by \(\frac{dx}{dt}=3t^2+1\) gives \(\frac{d^2y}{dx^2}=\frac{12t^2(t^2+1)}{(3t^2+1)^3}\). 4. The expression is zero at \(t=0\) but positive for every nearby nonzero \(t\), so the concavity does not change and the claim is false.

\(\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{12t^2(t^2+1)}{(3t^2+1)^2}\) \(\frac{d^2y}{dx^2}=\frac{12t^2(t^2+1)}{(3t^2+1)^3}\) The claim is incorrect because the curve is concave up on both sides of \(t=0\).

- The horizontal rate is positive everywhere, so a horizontal tangent is controlled by the vertical component rate. - Differentiate the quotient for the tangent slope before converting that t-rate into an x-rate. - Use the sign of the final second derivative at each tangent point.

1. Since \(\frac{dx}{dt}=3t^2+1>0\), horizontal tangents occur where \(\frac{dy}{dt}=3t^2-3=0\), so \(t=\pm1\). 2. The points are \((-2,2)\) and \((2,-2)\). 3. The first derivative is \(\frac{3t^2-3}{3t^2+1}\). Its t-derivative is \(\frac{24t}{(3t^2+1)^2}\), so \(\frac{d^2y}{dx^2}=\frac{24t}{(3t^2+1)^3}\). 4. At \(t=-1\), the t-derivative of slope is \(-\frac{3}{2}\) and the second derivative is \(-\frac{3}{8}\), so the curve is concave down. 5. At \(t=1\), the corresponding values are \(\frac{3}{2}\) and \(\frac{3}{8}\), so the curve is concave up.

At \(t=-1\): point \((-2,2)\), \(\frac{d}{dt}\left(\frac{dy}{dx}\right)=-\frac{3}{2}\), \(\frac{d^2y}{dx^2}=-\frac{3}{8}\), concave down. At \(t=1\): point \((2,-2)\), \(\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{3}{2}\), \(\frac{d^2y}{dx^2}=\frac{3}{8}\), concave up.

- Interpret the second derivative as the rate at which tangent slope changes with x. - Evaluate both derivatives at the known parameter value. - Apply the small x-change to estimate the slope change.

1. At \(t=1\), the tangent slope is \(\frac{dy}{dx}=0\). 2. The second derivative is \(\frac{3(t^2+1)}{4t^3}\), so at \(t=1\), \(\frac{d^2y}{dx^2}=\frac{3}{2}\). 3. For a small change \(\Delta x=0.02\), the tangent slope changes by approximately \(\frac{3}{2}(0.02)=0.03\). 4. The estimated new tangent slope is \(0.03\).

Approximately \(0.03\)

- Interpret the two vectors as first- and second-component derivative data. - Account for the fact that the horizontal component rate is negative. - Use the sign of the final result to classify concavity.

1. The component values are \(x'=-2\), \(y'=3\), \(x''=4\), and \(y''=1\). 2. The second derivative is \(\frac{y''x'-y'x''}{(x')^3}=\frac{1(-2)-3(4)}{(-2)^3}=\frac{7}{4}\). 3. Because the second derivative is positive, the curve is concave up at the point.

\(\frac{d^2y}{dx^2}=\frac{7}{4}\); the curve is concave up.

- Find the ellipse’s tangent slope as a function of the parameter. - Differentiate that slope before converting from parameter change to x-change. - Evaluate the sign at the specified angle.

1. The first derivative is \(\frac{dy}{dx}=-\frac{3}{2}\cot(t)\). 2. Differentiating with respect to \(t\) and dividing by \(\frac{dx}{dt}=-2\sin(t)\) gives \(\frac{d^2y}{dx^2}=-\frac{3}{4\sin^3(t)}\). 3. At \(t=\frac{\pi}{6}\), the second derivative is \(-6\). 4. Therefore, the ellipse is concave down at that point.

\(\frac{d^2y}{dx^2}=-6\); concave down.

- Compute the original tangent-slope function before applying the geometric scale factors. - Horizontal and vertical stretching affect the first derivative by different factors. - The final conversion to \(d^2y/dx^2\) must use the stretched horizontal rate as well.

1. For the original curve, \(\frac{dy}{dx}=\frac{2t}{3t^2+1}\), so \(\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{2(1-3t^2)}{(3t^2+1)^2}\). 2. At \(t=1\), the t-derivative of slope is \(-\frac{1}{4}\). Dividing by \(\frac{dx}{dt}=4\) gives \(\frac{d^2y}{dx^2}=-\frac{1}{16}\). 3. A horizontal stretch by \(3\) and vertical stretch by \(2\) multiplies the first derivative by \(\frac{2}{3}\). Thus, at \(t=1\), the stretched t-derivative of slope is \(-\frac{1}{6}\). 4. The stretched horizontal rate is \(12\), so the stretched second derivative is \(-\frac{1}{72}\). 5. The stretched value is \(\frac{2}{9}\) of the original second derivative.

Original at \(t=1\): \(\frac{d}{dt}\left(\frac{dy}{dx}\right)=-\frac{1}{4}\), \(\frac{d^2y}{dx^2}=-\frac{1}{16}\) Stretched at \(t=1\): \(\frac{d}{dt}\left(\frac{dy}{dx}\right)=-\frac{1}{6}\), \(\frac{d^2y}{dx^2}=-\frac{1}{72}\) The stretched second derivative is \(\frac{2}{9}\) of the original.

- Find the tangent slope as a function of the cycloid parameter. - Convert its parameter-rate change into change with respect to x. - Evaluate only after simplifying at the top of the arch.

1. The tangent slope is \(\frac{dy}{dx}=\frac{\sin(t)}{1-\cos(t)}\). 2. Differentiating the slope with respect to \(t\) and dividing by \(\frac{dx}{dt}=1-\cos(t)\) gives \(\frac{d^2y}{dx^2}=-\frac{1}{(1-\cos(t))^2}\). 3. At \(t=\pi\), \(1-\cos(\pi)=2\), so \(\frac{d^2y}{dx^2}=-\frac{1}{4}\). 4. The curve is concave down at the top of the arch.

\(\frac{d^2y}{dx^2}=-\frac{1}{4}\); concave down.

- Separate the sign analysis from the algebraic derivation of the parametric second derivative. - On the stated interval, determine which factors can and cannot change sign. - An inflection conclusion requires both the point and a genuine concavity change across it.

1. The parametric second derivative is \(\frac{d^2y}{dx^2}=\frac{3t\left(2+e^t(2-t)\right)}{(1+e^t)^3}\). 2. On \([-1,1]\), the factor \(2+e^t(2-t)\) and the denominator are positive. 3. Therefore, the second derivative is negative for \(t<0\) and positive for \(t>0\): the curve is concave down, then concave up. 4. At \(t=0\), the point is \((0,0)\). Because concavity changes there, the origin is an inflection point.

The curve is concave down for \(-1<t<0\) and concave up for \(0<t<1\). The origin is an inflection point.

- Compute the second derivative in parameter form. - Use the positive-parameter restriction when solving the resulting equation. - Convert the valid parameter value to a point on the curve.

1. The first derivative is \(\frac{dy}{dx}=\frac{3t^2+1}{2t}\). 2. The second derivative is \(\frac{3t^2-1}{4t^3}\). 3. Setting this equal to \(\frac{1}{2}\) gives \(2t^3-3t^2+1=0\), or \((t-1)^2(2t+1)=0\). 4. The positive solution is \(t=1\), which gives the point \((2,2)\).

\((2,2)\)

- Estimate how the tangent slope changes as \(t\) changes. - The second derivative describes slope change with respect to \(x\), not \(t\). - What does the sign of the estimate suggest, and what can an estimate not guarantee?

1. Estimate the rate at which the slope changes with \(t\): \(\frac{3.7-4.1}{2.1-1.9}=-2\). 2. Divide by \(\frac{dx}{dt}=5\): \(\frac{d^{2}y}{dx^{2}}\approx-\frac{2}{5}\). 3. The negative estimate suggests that the curve is concave down near \(t=2\); these two data points alone do not prove its exact concavity at \(t=2\).

\(\frac{d^{2}y}{dx^{2}}\approx-\frac{2}{5}\). It appears concave down near \(t=2\).

- Treat \(a\) as a constant when finding the parameter derivatives. - Evaluate the component-rate data at \(t=0\) before imposing the required second derivative. - The target value supplies an equation for \(a\); its sign also determines concavity.

1. At \(t=0\), \(x'=2\), \(x''=1\), \(y'=0\), and \(y''=2a\). 2. Therefore, \(\frac{d^2y}{dx^2}=\frac{(2a)(2)-0\cdot1}{2^3}=\frac{a}{2}\). 3. Setting \(\frac{a}{2}=1\) gives \(a=2\). 4. The prescribed second derivative is positive, so the curve is concave up at \(t=0\).

\(a=2\); the curve is concave up at \(t=0\).

- Differentiate the product in y carefully; its trigonometric combination is chosen to simplify. - After the first derivative simplifies, remember that the second derivative still divides by \(dx/dt=e^t\). - The exponential factor never changes sign, so the concavity changes are controlled by the sine factor.

1. Differentiating the y-component gives \(\frac{dy}{dt}=e^t\cos(2t)\), while \(\frac{dx}{dt}=e^t\). Therefore, \(\frac{dy}{dx}=\cos(2t)\). 2. Thus, \(\frac{d}{dt}\left(\frac{dy}{dx}\right)=-2\sin(2t)\). 3. Dividing by \(\frac{dx}{dt}=e^t\) gives \(\frac{d^2y}{dx^2}=-2e^{-t}\sin(2t)\). 4. In the interior \((0,\pi)\), the only zero at which the sign changes is \(t=\frac{\pi}{2}\). 5. The corresponding point is \(\left(e^{\pi/2},-\frac{e^{\pi/2}}{5}\right)\).

\(\frac{dy}{dx}=\cos(2t)\) \(\frac{d}{dt}\left(\frac{dy}{dx}\right)=-2\sin(2t)\) \(\frac{d^2y}{dx^2}=-2e^{-t}\sin(2t)\) Inflection point: \(\left(e^{\pi/2},-\frac{e^{\pi/2}}{5}\right)\) at \(t=\frac{\pi}{2}\)

- Form \(\frac{dy}{dx}\) as a function of \(t\), then differentiate it with respect to \(t\). - The parametric second-derivative formula divides by \(\frac{dx}{dt}\), so check where that quantity is zero. - Eliminate \(t\) to distinguish failure of the parametric formula from the behavior of the Cartesian endpoint.

1. For \(t\ne0\), \(\frac{dy}{dx}=\frac{4t^{3}}{2t}=2t^{2}\). 2. Differentiate the slope with respect to \(t\), then divide by \(\frac{dx}{dt}=2t\): \(\frac{d^{2}y}{dx^{2}}=\frac{4t}{2t}=2\) for \(t\ne0\). 3. At \(t=0\), \(\frac{dx}{dt}=0\), so the parametric second-derivative formula cannot be applied. 4. Eliminating the parameter gives \(y=x^{2}+1\) with \(x\ge0\). Thus the Cartesian branch has right-hand second derivative \(2\) at the endpoint \(x=0\), although no two-sided derivative is available there because the branch has no points with \(x<0\).

For \(t\ne0\), \(\frac{d^{2}y}{dx^{2}}=2\). The parametric formula is undefined at \(t=0\) because \(\frac{dx}{dt}=0\). The Cartesian branch \(y=x^{2}+1\), \(x\ge0\), has right-hand second derivative \(2\) at its endpoint.

- Differentiate both coordinates and form \(\frac{dy}{dx}\) before applying the parametric second-derivative formula. - On the stated interval, \(t\) is positive; determine which remaining factors can affect the sign of the second derivative. - Test the sign on each side of the interior candidate and confirm that \(\frac{dx}{dt}\ne0\) there.

1. The second derivative simplifies to \(\frac{d^{2}y}{dx^{2}}=\frac{15(t^{2}-1)}{4t}\). 2. In the given interval, the only candidate is \(t=1\); also, \(\frac{dx}{dt}=2t\ne0\) there. 3. The second derivative changes from negative to positive at \(t=1\), so the curve changes concavity there.

\(t=1\)

- Use the exponential horizontal coordinate to simplify the denominator and note whether it can vanish. - Factor the remaining expression so its possible sign changes are easy to identify. - Check the sign immediately before and after each candidate in the stated interval.

1. The second derivative simplifies to \(\frac{d^{2}y}{dx^{2}}=3t(2-t)e^{-2t}\). 2. In the given interval, the only candidate is \(t=2\); also, \(\frac{dx}{dt}=e^{t}\ne0\) for every real \(t\). 3. The second derivative changes from positive to negative at \(t=2\), so the curve changes concavity there.

\(t=2\)

- Derive and factor the parametric second derivative completely. - Separate factors that are always positive from the factors whose signs can change. - Test every interval determined by the candidates, and verify that the horizontal component rate is nonzero at each one.

1. The second derivative simplifies to \(\frac{d^{2}y}{dx^{2}}=\frac{10t(t-1)(t+1)(t^{2}+3)}{9(t^{2}+1)^{3}}\). 2. The candidates are \(t=-1\), \(t=0\), and \(t=1\). The horizontal component rate \(\frac{dx}{dt}=3(t^{2}+1)\) is nonzero at each candidate. 3. The sign changes at all three candidates, so the curve changes concavity at \(t=-1\), \(t=0\), and \(t=1\).

\(t=-1\), \(t=0\), \(t=1\)

- Derive \(\frac{d^{2}y}{dx^{2}}\) while keeping \(a\) symbolic. - Substitute \(t=1\) and set the result equal to the stated value. - Apply \(a\ge0\) and verify that \(\frac{dx}{dt}\ne0\) for the candidate.

1. Form the first derivative \(\frac{dy}{dx}=\frac{3t^{2}}{a+2t}\). 2. The second derivative is \(\frac{d^{2}y}{dx^{2}}=\frac{6t(a+t)}{(a+2t)^{3}}\). 3. At \(t=1\), solve \(\frac{6(a+1)}{(a+2)^{3}}=\frac{3}{4}\). This gives \(a=0\), \(a=-3-\sqrt{5}\), or \(a=-3+\sqrt{5}\). 4. The condition \(a\ge0\) leaves \(a=0\), and \(\frac{dx}{dt}=2\ne0\) for this value at \(t=1\).

\(a=0\)

- Keep the exponential terms exact while deriving the second derivative. - Evaluate the symbolic expression at \(t=0\) before solving for \(a\). - Check both the nonnegative constraint and the horizontal-rate denominator.

1. Form the first derivative \(\frac{dy}{dx}=\frac{2e^{2t}}{a+e^{t}}\). 2. The second derivative is \(\frac{d^{2}y}{dx^{2}}=\frac{2(2a+e^{t})e^{2t}}{(a+e^{t})^{3}}\). 3. At \(t=0\), solve \(\frac{2(2a+1)}{(a+1)^{3}}=2\). This gives \(a=0\) or \(a=\frac{-3\pm\sqrt{5}}{2}\). 4. The condition \(a\ge0\) leaves \(a=0\), and \(\frac{dx}{dt}=1\ne0\) for this value at \(t=0\).

\(a=0\)

- Differentiate the product in the x-coordinate with \(a\) treated as constant. - Convert the stated second-derivative value into an algebraic equation at \(t=1\). - Test all real candidates against \(a\ge0\) and \(\frac{dx}{dt}\ne0\).

1. Form the first derivative \(\frac{dy}{dx}=\frac{2t}{a+3t^{2}}\). 2. The second derivative is \(\frac{d^{2}y}{dx^{2}}=\frac{2(a-3t^{2})}{(a+3t^{2})^{3}}\). 3. At \(t=1\), solve \(\frac{2(a-3)}{(a+3)^{3}}=-\frac{1}{16}\). The resulting cubic factors as \((a-1)(a^{2}+10a+69)=0\). 4. The only real solution is \(a=1\), and \(\frac{dx}{dt}=4\ne0\) at \(t=1\).

\(a=1\)

- Build the first and second parametric derivatives with \(a\) left symbolic. - Evaluate at \(t=0\), where many terms simplify. - Solve the resulting equation and verify the derivative formula remains defined.

1. Form the first derivative \(\frac{dy}{dx}=\frac{3t^{2}-1}{2at+1}\). 2. The second derivative is \(\frac{d^{2}y}{dx^{2}}=\frac{2(3at^{2}+3t+a)}{(2at+1)^{3}}\). 3. At \(t=0\), this becomes \(2a\). Solving \(2a=6\) gives \(a=3\). 4. For \(a=3\), \(\frac{dx}{dt}=1\ne0\) at \(t=0\).

\(a=3\)

- Ask what variable the final derivative is taken with respect to. - The requested second derivative measures how tangent slope changes as \(x\) changes, not merely as \(t\) changes. - After correcting the derivative, use its sign at the specified parameter to interpret concavity.

1. For \(t\ne0\), the first derivative is \(\frac{dy}{dx}=\frac{3t}{2}\). 2. Differentiating with respect to \(t\) gives \(\frac{3}{2}\), which must then be divided by \(\frac{dx}{dt}=2t\) to convert from differentiation with respect to \(t\) to differentiation with respect to \(x\). 3. Thus, \(\frac{d^2y}{dx^2}=\frac{3}{4t}\) for \(t\ne0\). At \(t=1\), it equals \(\frac{3}{4}\). 4. Because the value is positive, the curve is concave up there.

The missing step is division by \(\frac{dx}{dt}\). Thus, \(\frac{d^2y}{dx^2}=\frac{3}{4t}\) for \(t\ne0\); at \(t=1\), the value is \(\frac{3}{4}\), so the curve is concave up.

- Identify the variable with respect to which each reported rate is taken. - Decide whether the second number is already an x-rate. - Use the sign only after the derivative has been converted to the correct variable.

1. Marta's \(\frac{dy}{dx}=1\) and \(\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{3}{2}\) at \(t=0\) are correct. 2. However, the second quantity is a rate with respect to \(t\). At \(t=0\), \(\frac{dx}{dt}=2\). 3. Hence \(\frac{d^2y}{dx^2}=\frac{\frac{3}{2}}{2}=\frac{3}{4}\). 4. Because \(\frac{3}{4}>0\), the curve is concave up.

The report is incorrect. \(\frac{d^2y}{dx^2}=\frac{3}{4}\) at \(t=0\), so the curve is concave up.

- Identify the variable with respect to which the negative derivative is taken. - Ask whether the direction in which \(x\) changes can affect the sign of change with respect to \(x\). - Evaluate the complete second derivative only after resolving that distinction.

1. The first derivative is \(\frac{dy}{dx}=-\tan t\). 2. Differentiating with respect to \(t\) gives \(-\frac{1}{\cos^2 t}\), but this must then be divided by \(\frac{dx}{dt}=\cos t\). 3. Thus, \(\frac{d^2y}{dx^2}=-\frac{1}{\cos^3 t}\). Its sign depends on the sign of \(\cos t\), so Yuki's global conclusion is not justified. 4. At \(t=\frac{\pi}{4}\), the value is \(-2\sqrt{2}<0\), so the curve is concave down there.

Yuki's reasoning is incomplete because \(\frac{d}{dt}\left(\frac{dy}{dx}\right)\) is not yet \(\frac{d^2y}{dx^2}\). The correct second derivative is \(\frac{d^2y}{dx^2}=-\frac{1}{\cos^3 t}\); at \(t=\frac{\pi}{4}\), it is \(-2\sqrt{2}\), so the curve is concave down there.

- Ask what the value \(-\frac{1}{4}\) represents before choosing between the proposals. - The two proposed values differ by a factor related to horizontal motion at \(t=1\). - Concavity follows from the sign of the x-based second derivative.

1. The tangent slope is \(\frac{dy}{dx}=\frac{2t}{1+3t^2}\). 2. At \(t=1\), \(\frac{d}{dt}\left(\frac{dy}{dx}\right)=-\frac{1}{4}\). This is not yet the second derivative with respect to \(x\). 3. Since \(\frac{dx}{dt}=4\) at \(t=1\), \(\frac{d^2y}{dx^2}=\frac{-\frac{1}{4}}{4}=-\frac{1}{16}\). 4. The negative value means the curve is concave down.

\(-\frac{1}{16}\) is correct; the curve is concave down at \(t=1\).

- Treat the two parameter values as separate visits to the same point. - Find the change in tangent slope with respect to the parameter before converting to change with respect to x. - Use the sign of each result to classify the branch’s concavity.

1. The first derivative is \(\frac{dy}{dx}=\frac{3t^2-3}{2t}\). 2. Differentiating with respect to \(t\) and dividing by \(\frac{dx}{dt}=2t\) gives \(\frac{d^2y}{dx^2}=\frac{3(t^2+1)}{4t^3}\). 3. At \(t=-\sqrt{3}\), the second derivative is \(-\frac{1}{\sqrt{3}}\), so that branch is concave down. 4. At \(t=\sqrt{3}\), the second derivative is \(\frac{1}{\sqrt{3}}\), so that branch is concave up.

At \(t=-\sqrt{3}\): \(\frac{d^2y}{dx^2}=-\frac{1}{\sqrt{3}}\), concave down. At \(t=\sqrt{3}\): \(\frac{d^2y}{dx^2}=\frac{1}{\sqrt{3}}\), concave up.

- Keep the unknown coefficient through the product-rule differentiation of y. - After the common exponential factor cancels in the first derivative, remember that \(dx/dt\) is still exponential for the second conversion. - An inflection requires the final second derivative to change sign across the stated parameter.

1. The component rates are \(\frac{dx}{dt}=e^t\) and \(\frac{dy}{dt}=e^t(t^2+(a+2)t+a)\), so \(\frac{dy}{dx}=t^2+(a+2)t+a\). 2. Thus, \(\frac{d}{dt}(\frac{dy}{dx})=2t+a+2\), and \(\frac{d^2y}{dx^2}=(2t+a+2)e^{-t}\). 3. An inflection at \(t=1\) requires \(a+4=0\), so \(a=-4\). 4. Then \(\frac{d^2y}{dx^2}=2(t-1)e^{-t}\), which changes from negative to positive at \(t=1\).

\(a=-4\) \(\frac{dy}{dx}=t^2-2t-4\) \(\frac{d}{dt}\left(\frac{dy}{dx}\right)=2t-2\) \(\frac{d^2y}{dx^2}=2(t-1)e^{-t}\), which changes sign at \(t=1\).

- Compare the parameter-based geometric data at the join. - Position, tangent slope, and second derivative impose three separate conditions. - The second parameter derivatives of the two x-coordinates differ at \(t=0\), so matching y-derivatives alone is not enough.

1. The left piece reaches \((0,0)\) at \(t=0\), so continuity of the right piece gives \(c=0\). 2. On the left at \(t=0\), \(x'=2\) and \(y'=1\), so the tangent slope is \(\frac{1}{2}\). On the right, \(x'=2\) and \(y'=b\), so matching slopes gives \(b=1\). 3. On the left, \(x''=1\) and \(y''=2\), giving \(\frac{d^2y}{dx^2}=\frac{2\cdot2-1\cdot1}{2^3}=\frac{3}{8}\). 4. On the right at \(t=0\), \(x''=0\) and \(y''=2a\), so \(\frac{d^2y}{dx^2}=\frac{(2a)(2)}{2^3}=\frac{a}{2}\). Matching gives \(a=\frac{3}{4}\).

\(a=\frac{3}{4}\), \(b=1\), and \(c=0\)

- An inflection point requires a sign change in the second derivative, not merely a zero. - Work with the parameter-form second derivative on the stated branch. - Convert the qualifying parameter to a point only after confirming the concavity change.

1. The parametric second derivative simplifies to \(\frac{d^2y}{dx^2}=-\frac{3(t-1)(3t+1)}{(3t^2+1)^3}\). 2. On \([\frac{1}{2},2]\), the denominator and \(3t+1\) are positive, so the sign changes at \(t=1\). 3. The curve is concave up before \(t=1\) and concave down after it, so this is an inflection point. 4. At \(t=1\), the point is \((2,\frac{5}{2})\).

\((2,\frac{5}{2})\)

- The two derivative conditions provide two equations for the constants. - Use the first-derivative condition to express one constant in terms of the other. - Substitute into the second-derivative condition and check the resulting pair.

1. The tangent-slope condition is \(\frac{3b}{a+2}=1\). 2. The second derivative at \(t=1\) is \(\frac{6b(a+1)}{(a+2)^3}\), so \(\frac{6b(a+1)}{(a+2)^3}=\frac{1}{2}\). 3. From the slope condition, \(b=\frac{a+2}{3}\). 4. Substitution into the second condition gives \(a=0\), and then \(b=\frac{2}{3}\).

\(a=0\) and \(b=\frac{2}{3}\)

- The horizontal rate never vanishes, so locate horizontal tangents from \(dy/dt=0\). - At such a point, the quotient-rule expression for the t-rate of slope simplifies substantially. - Divide by the horizontal rate one more time before using the sign to classify the extremum.

1. Because \(\frac{dx}{dt}=3t^2+1>0\), horizontal tangents occur when \(\frac{dy}{dt}=3(t-1)(t-3)=0\), so \(t=1\) and \(t=3\). 2. The points are \((2,4)\) and \((30,0)\). 3. At a horizontal tangent, differentiating \(\frac{dy/dt}{dx/dt}\) simplifies to \(\frac{d}{dt}(\frac{dy}{dx})=\frac{y''(t)}{x'(t)}\). 4. At \(t=1\), this is \(-\frac64=-\frac32\), and dividing again by \(x'(1)=4\) gives \(\frac{d^2y}{dx^2}=-\frac38\), so \((2,4)\) is a local maximum. 5. At \(t=3\), the t-derivative of slope is \(\frac6{28}=\frac3{14}\), and \(\frac{d^2y}{dx^2}=\frac{3}{392}\), so \((30,0)\) is a local minimum.

\((2,4)\) at \(t=1\): \(\frac{d}{dt}(\frac{dy}{dx})=-\frac32\), \(\frac{d^2y}{dx^2}=-\frac38\), local maximum. \((30,0)\) at \(t=3\): \(\frac{d}{dt}(\frac{dy}{dx})=\frac3{14}\), \(\frac{d^2y}{dx^2}=\frac{3}{392}\), local minimum.

- Use the shared parameter to differentiate each coordinate relation; neither relation should be converted into a Cartesian equation first. - The positive-value condition identifies the coordinate values needed after differentiation. - Keep first and second parameter-rates separate until the final x-based second derivative.

1. At \(t=0\), positivity gives \(x=2\) and \(y=2\). 2. Differentiating \(x^2+\sin(t)=4\) twice gives \(x'=-\frac{1}{4}\) and \(x''=-\frac{1}{32}\) at \(t=0\). 3. Differentiating \(y^3+e^t=9\) twice gives \(y'=-\frac{1}{12}\) and \(y''=-\frac{13}{144}\) at \(t=0\). 4. Therefore, \(\frac{d^2y}{dx^2}=\frac{y''x'-y'x''}{(x')^3}=\frac{23}{1152}\div\left(-\frac{1}{64}\right)=-\frac{23}{18}\).

\(-\frac{23}{18}\)

- Keep both unknown coefficients through the first-derivative calculation. - The second derivative requires dividing the t-derivative of the slope by \(dx/dt\). - The two conditions at \(t=1\) should produce two different linear equations in \(a\) and \(b\).

1. Since \(\frac{dx}{dt}=2t\) and \(\frac{dy}{dt}=4at^3+3bt^2\), \(\frac{dy}{dx}=2at^2+\frac32bt\). 2. Therefore, \(\frac{d}{dt}(\frac{dy}{dx})=4at+\frac32b\), and \(\frac{d^2y}{dx^2}=2a+\frac{3b}{4t}\). 3. At \(t=1\), the slope condition gives \(2a+\frac32b=5\), and the second-derivative condition gives \(2a+\frac34b=6\). 4. Subtracting gives \(\frac34b=-1\), so \(b=-\frac43\). Then \(2a-2=5\), so \(a=\frac72\). 5. With these values, \(\frac{dy}{dx}=7t^2-2t\), \(\frac{d}{dt}(\frac{dy}{dx})=14t-2\), and \(\frac{d^2y}{dx^2}=7-\frac1t\), which give \(5\) and \(6\) at \(t=1\).

\(\frac{dy}{dx}=2at^2+\frac32bt\) \(\frac{d}{dt}(\frac{dy}{dx})=4at+\frac32b\) \(\frac{d^2y}{dx^2}=2a+\frac{3b}{4t}\) \(a=\frac72\), \(b=-\frac43\)

- Work with nonzero parameter values before taking the limit. - Simplify each derivative enough to reveal whether a finite limit exists. - Compare the limiting results with the geometric curve near the origin.

1. For \(t\ne0\), \(\frac{dy}{dx}=\frac{6t^5}{3t^2}=2t^3\), whose limit at \(0\) is \(0\). 2. For \(t\ne0\), \(\frac{d^2y}{dx^2}=\frac{6t^2}{3t^2}=2\). 3. The limiting tangent slope is \(0\), so the tangent line is horizontal. 4. The limiting second derivative is \(2\), consistent with the Cartesian relation \(y=x^2\).

The tangent slope is \(0\), and \(\frac{d^2y}{dx^2}=2\).

- Compare change with respect to the parameter against change with respect to \(x\). - Check the sign of the horizontal component rate on the interval. - A reversed \(x\)-direction reverses how a slope trend should be interpreted.

1. The tangent slope is \(\frac{dy}{dx}=\frac{2t}{3(t^2-1)}\). 2. Its parameter derivative is \(-\frac{2(t^2+1)}{3(t^2-1)^2}\), which is negative on \((-1,1)\). 3. The horizontal component rate is \(\frac{dx}{dt}=3(t^2-1)\), which is also negative on this interval. 4. Dividing the negative slope-rate by the negative horizontal rate gives \(\frac{d^2y}{dx^2}>0\), so the curve is concave up. As \(t\) increases, \(x\) decreases, reversing the comparison direction.

Both \(\frac{d}{dt}\left(\frac{dy}{dx}\right)\) and \(\frac{dx}{dt}\) are negative on \((-1,1)\), so their quotient \(\frac{d^2y}{dx^2}\) is positive. The statements are consistent because \(x\) decreases as \(t\) increases.