Free math worksheets

- 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 \(x\): \(\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}(dy/dx)}{dx/dt}\). 2. Thus, \(\frac{d^2y}{dx^2}=\frac{-6}{2}=-3\).

\(-3\)

- Read the sign of the slope of \(x(t)\) at \(t=0\). - Read whether the tangent-slope function \(m(t)\) is increasing or decreasing there. - Concavity depends on how \(m\) changes as \(x\), rather than merely as \(t\), increases.

1. At \(t=0\), \(x(t)\) is increasing, so \(\frac{dx}{dt}>0\). 2. The slope function \(m(t)\) is decreasing, so \(\frac{dm}{dt}<0\). 3. Therefore, \(\frac{d^2y}{dx^2}=\frac{dm/dt}{dx/dt}<0\), so the curve is concave down.

The curve is concave down at \(t=0\).

- 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 \(\left(\frac{4}{3\sqrt3},\frac13\right)\)

- 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 \(\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}(dy/dx)}{dx/dt}\). 2. Substitution gives \(3=\frac{-6}{dx/dt}\). 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 \(1\).

\(1\)

- 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 \(\frac{5}{4}\).

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

- Track both the first and second parameter-rates of the two coordinates at \(t=0\). - A second derivative with respect to \(x\) is not the same quantity as \(\frac{d^2y}{dt^2}\). - Check that the horizontal rate is nonzero at the evaluation point.

1. At \(t=0\), \(x'=2\), \(x''=1\), \(y'=1\), and \(y''=3\), where primes here denote 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 \(\frac{3\cdot2-1\cdot1}{2^3}=\frac58\).

\(\frac{d^2y}{dx^2}=\frac58\)

- 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 restricts the allowed parameter values, but \(t=1\) is valid. - Record the first and second parameter-rates of both coordinates before combining them. - Distinguish curvature in the plane from the ordinary second derivative of either component alone.

1. At \(t=1\), \(x'=1+\frac1t=2\), \(x''=-\frac1{t^2}=-1\), \(y'=2t=2\), and \(y''=2\). 2. Therefore, \(\frac{d^2y}{dx^2}=\frac{y''x'-y'x''}{(x')^3}\). 3. Substitution gives \(\frac{2\cdot2-2(-1)}{2^3}=\frac34\).

\(\frac{d^2y}{dx^2}=\frac34\)

- Differentiate the reciprocal terms with particular attention to their signs. - At the requested parameter, gather the four component-rate values before forming the planar second derivative. - Check that \(\frac{dx}{dt}\) is nonzero.

1. At \(t=2\), \(x'=1-\frac1{t^2}=\frac34\) and \(x''=\frac{2}{t^3}=\frac14\). 2. Also, \(y'=2t+\frac1{t^2}=\frac{17}{4}\) and \(y''=2-\frac{2}{t^3}=\frac74\). 3. Thus \(\frac{d^2y}{dx^2}=\frac{y''x'-y'x''}{(x')^3}=\frac{\frac74\cdot\frac34-\frac{17}{4}\cdot\frac14}{(\frac34)^3}=\frac{16}{27}\).

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

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

1. At \(t=2\), \(x'=1+\frac12=\frac32\) and \(x''=-\frac14\). 2. Also, \(y'=1-\frac14=\frac34\) and \(y''=\frac14\). 3. Therefore, \(\frac{d^2y}{dx^2}=\frac{\frac14\cdot\frac32-\frac34(-\frac14)}{(\frac32)^3}=\frac16\). 4. Since \(\frac16>0\), the curve is concave up.

The curve is concave up; \(\frac{d^2y}{dx^2}=\frac16\).

- 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 \((e,2e)\).

- Derive one expression for \(\frac{d^{2}y}{dx^{2}}\) before substituting either parameter value. - Evaluate that expression separately at \(t=1\) and \(t=2\). - Compare signs to classify concavity; magnitudes only describe relative curvature strength.

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

At \(t=1\), \(\frac{d^{2}y}{dx^{2}}=\frac{3}{2}\), and the curve is concave up. At \(t=2\), \(\frac{d^{2}y}{dx^{2}}=\frac{15}{32}\), and the curve is also 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 \(\frac{d^2y}{dx^2}=-\frac12\), 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 \(t=0\), \(\frac{d^2y}{dx^2}=-\frac12\), 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 \(\frac12\). 2. The tangent line is \(y=\frac12(x-1)\). 3. At \(t=0\), \(x''=1\) and \(y''=2\). Thus \(\frac{d^2y}{dx^2}=\frac{2\cdot2-1\cdot1}{2^3}=\frac38\). 4. Since \(\frac38>0\), the curve is concave up.

a) \(y=\frac12(x-1)\) b) Concave up; \(\frac{d^2y}{dx^2}=\frac38\)

- 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; \(\frac{d^{2}y}{dx^{2}}=-\frac{32}{27}\)

- 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, \(\frac{d^2y}{dx^2}=\frac{11}{8}\).

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

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

- 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 \(t=1\).

- 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 \(\frac{d^2y}{dx^2}=\frac12\) 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 \(t=\frac1{\sqrt3}\), \(x'=0\) while \(y'=1+\frac{2}{\sqrt3}\ne0\), 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 \(t\ne\pm\frac1{\sqrt3}\); it is undefined at \(t=\frac1{\sqrt3}\) because \(\frac{dx}{dt}=0\) there while \(\frac{dy}{dt}\ne0\).

- 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 \(-\frac32\) and the second derivative is \(-\frac38\), so the curve is concave down. 5. At \(t=1\), the corresponding values are \(\frac32\) and \(\frac38\), so the curve is concave up.

At \(t=-1\): point \((-2,2)\), \(\frac{d}{dt}(\frac{dy}{dx})=-\frac32\), \(\frac{d^2y}{dx^2}=-\frac38\), concave down. At \(t=1\): point \((2,-2)\), \(\frac{d}{dt}(\frac{dy}{dx})=\frac32\), \(\frac{d^2y}{dx^2}=\frac38\), 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 \(\frac{d}{dt}(\frac{dy}{dx})=\frac{2(1-3t^2)}{(3t^2+1)^2}\). 2. At \(t=1\), the t-derivative of slope is \(-\frac14\). 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 \(\frac23\). Thus, at \(t=1\), the stretched t-derivative of slope is \(-\frac16\). 4. The stretched horizontal rate is \(12\), so the stretched second derivative is \(-\frac{1}{72}\). 5. The stretched value is \(\frac29\) of the original second derivative.

Original at \(t=1\): \(\frac{d}{dt}(\frac{dy}{dx})=-\frac14\), \(\frac{d^2y}{dx^2}=-\frac{1}{16}\) Stretched at \(t=1\): \(\frac{d}{dt}(\frac{dy}{dx})=-\frac16\), \(\frac{d^2y}{dx^2}=-\frac{1}{72}\) The stretched second derivative is \(\frac29\) 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.

- 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.

- 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 \((2,2)\).

\((2,2)\)

- 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 \(t=1\).

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

- First estimate how the tangent slope changes with the parameter. - The requested derivative measures slope change with respect to x instead. - Use the sign of the estimate to classify concavity.

1. The centered estimate of the parameter-rate of slope change is \(\frac{3.7-4.1}{2.1-1.9}=-2\). 2. Dividing by \(\frac{dx}{dt}=5\) gives \(\frac{d^2y}{dx^2}\approx-\frac{2}{5}\). 3. The negative estimate indicates that the curve is concave down at \(t=2\).

\(\frac{d^2y}{dx^2}\approx-\frac{2}{5}\); concave down.

- 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, \(\frac{d}{dt}(\frac{dy}{dx})=-2\sin(2t)\). 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 \(t=\frac\pi2\). 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 \(t=\frac\pi2\)

- 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 \(\frac32\), 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 \(\frac34\). 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 \(\frac34\), 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 \(\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac32\) 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 \(\frac{d^2y}{dx^2}=\frac{\frac32}{2}=\frac34\). 4. Because \(\frac34>0\), the curve is concave up.

The report is incorrect. \(\frac{d^2y}{dx^2}=\frac34\) 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 \(-2\sqrt2<0\), 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 \(-2\sqrt2\), so the curve is concave down there.

- Ask what the value \(-\frac14\) 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 \(t=1\), \(\frac{d}{dt}(\frac{dy}{dx})=-\frac14\). 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.

\(-\frac1{16}\) 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.

- The two pieces cannot be reduced to convenient explicit functions of x, so compare their 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 \(\frac12\). On the right, \(x'=2\) and \(y'=b\), so matching slopes gives \(b=1\). 3. On the left, \(x''=1\) and \(y''=2\), giving \(\frac{d^2y}{dx^2}=\frac{2\cdot2-1\cdot1}{2^3}=\frac38\). 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 \(a=\frac34\).

\(a=\frac34\), \(b=1\), 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 \([\frac12,2]\), 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 \((2,\frac52)\).

\((2,\frac52)\)

- 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 \(b=\frac{2}{3}\)

- 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 \(x'=-\frac14\) and \(x''=-\frac1{32}\) at \(t=0\). 3. Differentiating \(y^3+e^t=9\) twice gives \(y'=-\frac1{12}\) and \(y''=-\frac{13}{144}\) at \(t=0\). 4. Therefore, \(\frac{d^2y}{dx^2}=\frac{y''x'-y'x''}{(x')^3}=\frac{23}{1152}\div\left(-\frac{1}{64}\right)=-\frac{23}{18}\).

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

- 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 \(\frac{d}{dt}(dy/dx)\) 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.