Free math worksheets

- Write both requested derivatives in terms of the parameter. - Solve their equality before evaluating coordinates. - Include every parameter value that satisfies the equation.

1. Because \(x=t\), \(\frac{dy}{dx}=3t^2\) and \(\frac{d^2y}{dx^2}=6t\). 2. Setting the derivatives equal gives \(6t=3t^2\), or \(3t(t-2)=0\). 3. Thus, \(t=0\) or \(t=2\). 4. The corresponding points are \((0,0)\) and \((2,8)\).

\((0,0)\) and \((2,8)\)

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

\(1\)

- First form \(\frac{dy}{dx}\) as a function of \(t\). - Differentiate that expression with respect to \(t\). - Divide by \(\frac{dx}{dt}\) once more before evaluating at \(t=1\).

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

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

- Use the product rule when differentiating the y-coordinate. - Keep \(\frac{dy}{dx}\) in terms of \(t\), then differentiate it with respect to \(t\). - Convert the result to an x-derivative by dividing by \(\frac{dx}{dt}\).

1. The first derivative is \(\frac{dy}{dx}=\frac{e^{t}+te^{t}}{e^{t}}=t+1\). 2. Differentiate with respect to \(t\): \(\frac{d}{dt}\left(\frac{dy}{dx}\right)=1\). 3. Divide by \(\frac{dx}{dt}=e^{t}\): \(\frac{d^{2}y}{dx^{2}}=e^{-t}\). 4. At \(t=0\), the value is \(1\).

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

- Simplify the first derivative quotient before differentiating it again. - Differentiate \(\frac{dy}{dx}\) with respect to \(t\), not directly with respect to \(x\). - Divide by the horizontal component derivative and then evaluate the exact trigonometric values.

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 \(-1\).

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

- Form and simplify the first derivative as a rational function of \(t\). - Differentiate that rational function using an appropriate quotient rule or algebraic form. - Divide by \(\frac{dx}{dt}\) and check that it is nonzero at \(t=2\).

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 \(-\frac{10}{243}\).

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

- Differentiate the logarithmic x-coordinate and polynomial y-coordinate first. - Simplify \(\frac{dy}{dx}\) before taking its derivative with respect to \(t\). - Divide by \(\frac{dx}{dt}\) to obtain the second derivative with respect to \(x\).

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

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

- Differentiate the reciprocal terms with their correct signs. - Treat \(\frac{dy}{dx}\) as a quotient in \(t\) and simplify before differentiating. - Divide the t-derivative of the slope by \(\frac{dx}{dt}\) and then substitute \(t=2\).

1. The first derivative is \(\frac{dy}{dx}=\frac{t^{2}+1}{t^{2}-1}\). 2. Differentiate with respect to \(t\): \(\frac{d}{dt}\left(\frac{dy}{dx}\right)=-\frac{4t}{(t^{2}-1)^{2}}\). 3. Divide by \(\frac{dx}{dt}=1-\frac{1}{t^{2}}\): \(\frac{d^{2}y}{dx^{2}}=-\frac{4t^{3}}{(t^{2}-1)^{3}}\). 4. At \(t=2\), the value is \(-\frac{32}{27}\).

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

- 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.

- Use exponential rules to simplify \(\frac{dy}{dx}\) before differentiating again. - Divide the t-derivative of the slope by \(\frac{dx}{dt}\). - Check whether the exponential horizontal derivative can ever equal zero.

1. The first derivative is \(\frac{dy}{dx}=\frac{e^{-t}}{2}\). 2. Differentiating with respect to \(t\) and dividing by \(\frac{dx}{dt}=2e^{2t}\) gives \(\frac{d^{2}y}{dx^{2}}=-\frac{e^{-3t}}{4}\). 3. Because exponential functions are nonzero for every real input, the expression is defined for every real parameter value.

\(\frac{d^{2}y}{dx^{2}}=-\frac{e^{-3t}}{4}\); defined for all real \(t\).

- Simplify the first derivative using a basic trigonometric quotient. - Differentiate the slope with respect to \(t\), then divide by \(\frac{dx}{dt}\). - Solve where the cosine factor in the denominator is zero.

1. The first derivative is \(\frac{dy}{dx}=-\tan(t)\). 2. Differentiating with respect to \(t\) and dividing by \(\frac{dx}{dt}=\cos(t)\) gives \(\frac{d^{2}y}{dx^{2}}=-\frac{1}{\cos^{3}(t)}\). 3. The expression is undefined when \(\cos(t)=0\), which occurs at \(t=\frac{\pi}{2}+k\pi\), where \(k\in\mathbb{Z}\).

\(\frac{d^{2}y}{dx^{2}}=-\frac{1}{\cos^{3}(t)}\), undefined at \(t=\frac{\pi}{2}+k\pi\), 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; \(\frac{d^{2}y}{dx^{2}}=\frac{9}{32}\).

- Simplify the exponential derivative quotient before differentiating it again. - Evaluate the second derivative at \(t=0\). - Interpret its sign using the standard concavity convention.

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

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

- 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 domain restriction from the logarithmic x-coordinate. - Simplify the first derivative in terms of \(t\), then differentiate and divide by \(\frac{dx}{dt}\). - Use the sign of the exact value at \(t=2\) to state concavity.

1. The first derivative is \(\frac{dy}{dx}=t-\frac{1}{t}\). 2. Differentiating with respect to \(t\) and dividing by \(\frac{dx}{dt}=\frac{1}{t}\) gives \(\frac{d^{2}y}{dx^{2}}=t+\frac{1}{t}\). 3. At \(t=2\), the value is \(\frac{5}{2}>0\), so the curve is concave up.

The curve is concave up; \(\frac{d^{2}y}{dx^{2}}=\frac{5}{2}\).

- Find where the sign of the second-order expression could change. - Exclude values where the curve does not have the required derivative. - Check signs on both sides of each candidate.

1. Because \(x=t\), \(\frac{d^{2}y}{dx^{2}}=\frac{d^{2}y}{dt^{2}}=6t-6\). 2. Setting \(6t-6=0\) gives the candidate \(t=1\). 3. The second derivative changes from negative to positive at \(t=1\), so the curve changes concavity there.

\(t=1\)

- 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.

- Obtain a single second-derivative expression and evaluate it at both parameters. - A zero second derivative alone does not determine local concavity. - Check the sign immediately before and after the zero to decide whether concavity changes.

1. The second derivative is \(\frac{d^{2}y}{dx^{2}}=2(1-t)e^{-2t}\). 2. At \(t=0\), \(\frac{d^{2}y}{dx^{2}}=2>0\), so the curve is concave up. 3. At \(t=1\), \(\frac{d^{2}y}{dx^{2}}=0\). The expression changes from positive to negative as \(t\) passes through \(1\), so the point is an inflection point where the curve changes from concave up to concave down.

At \(t=0\), \(\frac{d^{2}y}{dx^{2}}=2\), and the curve is concave up. At \(t=1\), \(\frac{d^{2}y}{dx^{2}}=0\), and the curve changes from concave up to concave down.

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

- Evaluate the exponential coordinates at the given parameter to get the point. - Use the product rule in the y-coordinate when finding the tangent slope. - Derive the second derivative as a separate step and use its sign for concavity.

1. The point is \((e, e)\), and the tangent slope is \(2\). 2. The tangent line is \(y-e=2(x-e)\). 3. The second derivative at the point is \(e^{-1}>0\), so the curve is concave up.

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

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

- Use the inflection parameter to impose a zero second-derivative condition. - Substitute the resulting constant back into the second derivative. - Check signs on both sides rather than relying only on a zero value.

1. Because \(x=t\), the second derivative is \(\frac{d^2y}{dx^2}=6t+2a\). 2. Requiring this expression to equal \(0\) at \(t=1\) gives \(6+2a=0\), so \(a=-3\). 3. With \(a=-3\), \(\frac{d^2y}{dx^2}=6(t-1)\), which is negative for \(t<1\) and positive for \(t>1\). 4. Therefore, the concavity changes at \(t=1\).

\(a=-3\)

- Form the first derivative from the two component rates. - Convert the rate of change of the slope from the parameter to x. - Use the logarithmic coordinate to eliminate the parameter at the end.

1. The component derivatives are \(\frac{dx}{dt}=\frac{1}{t}\) and \(\frac{dy}{dt}=2t\), so \(\frac{dy}{dx}=2t^2\). 2. Differentiate the slope with respect to \(t\), then divide by \(\frac{dx}{dt}\): \(\frac{d^2y}{dx^2}=\frac{4t}{1/t}=4t^2\). 3. Since \(x=\ln t\), the parameter is \(t=e^x\). 4. Therefore, \(\frac{d^2y}{dx^2}=4e^{2x}\).

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

- Find the first derivative away from the parameter value where the horizontal rate vanishes. - Convert its parameter-rate change into a rate with respect to x. - Relate the failure at \(t=0\) to the tangent’s orientation.

1. For \(t\ne0\), \(\frac{dy}{dx}=\frac{1}{2t}\). 2. Differentiating with respect to \(t\) and dividing by \(\frac{dx}{dt}=2t\) gives \(\frac{d^2y}{dx^2}=-\frac{1}{4t^3}\). 3. At \(t=0\), \(\frac{dx}{dt}=0\) and \(\frac{dy}{dt}=1\), so the curve has a vertical tangent. 4. The derivative of \(y\) with respect to \(x\), and therefore its second derivative with respect to \(x\), is not finite there.

\(\frac{d^2y}{dx^2}=-\frac{1}{4t^3}\) for \(t\ne0\); it is undefined at \(t=0\).

- 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, \(k=1\).

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

- A zero second derivative is only a candidate condition. - Check the sign of the second derivative immediately to the left and right. - An inflection point requires a change in concavity.

1. Because \(x=t\), \(\frac{d^2y}{dx^2}=12t^2\). 2. The second derivative equals \(0\) at \(t=0\). 3. However, \(12t^2\) is positive on both sides of \(0\), so the curve is concave up on both sides. 4. The concavity does not change, so the origin is not an inflection point.

The claim is incorrect. The second derivative is zero at \(t=0\), but the curve is concave up on both sides, so there is no inflection point.

- First locate the parameter values where the tangent is horizontal. - Evaluate the coordinates for each parameter value. - Use the second derivative’s sign to distinguish the two local shapes.

1. Horizontal tangents occur when \(\frac{dy}{dx}=3t^2-3=0\), so \(t=-1\) or \(t=1\). 2. The corresponding points are \((-1,2)\) and \((1,-2)\). 3. The second derivative is \(6t\). 4. At \(t=-1\), the curve is concave down. At \(t=1\), the curve is concave up.

At \((-1,2)\), the curve is concave down. At \((1,-2)\), the curve is 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 second derivative for each parameterization rather than comparing only first derivatives. - Track how each coordinate scaling changes its component rate. - Compare the two exact values after the calculation.

1. For the original curve, \(\frac{d^2y}{dx^2}=2\). 2. For the stretched curve, \(\frac{dy}{dx}=\frac{4t}{3}\). 3. Differentiating with respect to \(t\) and dividing by \(\frac{dx}{dt}=3\) gives \(\frac{d^2y}{dx^2}=\frac{4}{9}\). 4. The horizontal stretch reduces the second derivative more strongly than the vertical stretch increases it.

The stretched curve has \(\frac{d^2y}{dx^2}=\frac{4}{9}\), compared with \(2\) for the original curve.

- 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.

- Compute the second derivative only where the component-rate quotient is defined. - Examine its sign separately on the two branches. - An inflection point requires a concavity change, not merely an undefined derivative.

1. For \(t\ne0\), \(\frac{dy}{dx}=\frac{2}{3t}\). 2. The second derivative is \(\frac{d^2y}{dx^2}=-\frac{2}{9t^4}\). 3. This value is negative for both \(t<0\) and \(t>0\), so both branches are concave down. 4. The second derivative is undefined at \(t=0\), and the concavity does not change across the origin, so the origin is not an inflection point.

Both branches are concave down. The origin is not an inflection point.

- Locate horizontal tangents before classifying them. - Evaluate the coordinates at each critical parameter value. - Use the sign of the second derivative to distinguish maximum from minimum behavior.

1. The first derivative is \(\frac{dy}{dx}=3t^2-12t+9=3(t-1)(t-3)\). 2. Horizontal tangents occur at \(t=1\) and \(t=3\), giving points \((1,4)\) and \((3,0)\). 3. The second derivative is \(6t-12\). 4. At \(t=1\), the second derivative is negative, so \((1,4)\) is a local maximum. At \(t=3\), it is positive, so \((3,0)\) is a local minimum.

\((1,4)\) is a local maximum, and \((3,0)\) is a 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 \((2,2)\).

\((2,2)\)

- Simplify the first derivative before differentiating again. - Use the constant-second-derivative condition to determine one coefficient first. - Apply the tangent-slope condition afterward.

1. The first derivative is \(\frac{dy}{dx}=2at^2+b\). 2. The second derivative is the constant \(2a\), so \(2a=6\) and \(a=3\). 3. At \(t=1\), the slope condition gives \(2a+b=5\). 4. Substituting \(a=3\) gives \(b=-1\).

\(a=3\) and \(b=-1\)

- 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.

- Compute the second derivative in parameter form first. - Look for a power of the given y-coordinate within the exponential expression. - Replace the parameter-dependent factor only after simplifying.

1. The first derivative is \(\frac{dy}{dx}=-2e^{-3t}\). 2. Differentiating with respect to \(t\) and dividing by \(\frac{dx}{dt}=e^t\) gives \(\frac{d^2y}{dx^2}=6e^{-4t}\). 3. Since \(y=e^{-2t}\), \(e^{-4t}=y^2\). 4. Therefore, \(\frac{d^2y}{dx^2}=6y^2\).

\(\frac{d^2y}{dx^2}=6y^2\)

- Find the second derivative over the stated interval. - Separate interior zeros from interval endpoints. - Verify a sign change before reporting the corresponding point.

1. Because \(x=t\), \(\frac{d^2y}{dx^2}=-4\sin(2t)\). 2. In the interior, this expression is zero at \(t=\frac{\pi}{2}\). 3. It is negative on \(0<t<\frac{\pi}{2}\) and positive on \(\frac{\pi}{2}<t<\pi\), so the concavity changes. 4. The inflection point is \(\left(\frac{\pi}{2},0\right)\).

\(\left(\frac{\pi}{2},0\right)\)

- 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 \(t=0\).

\(a=3\)

- Identify the student’s derivative as \(\frac{d}{dt}\left(\frac{dy}{dx}\right)\), not yet \(\frac{d^{2}y}{dx^{2}}\). - Convert from a t-derivative to an x-derivative by dividing by \(\frac{dx}{dt}\). - Evaluate the corrected expression and use its sign for 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, \(\frac{d^{2}y}{dx^{2}}=\frac{3}{4t}\) 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, \(\frac{d^{2}y}{dx^{2}}=\frac{3}{4t}\) for \(t\ne0\); at \(t=1\), the value is \(\frac{3}{4}\), so the curve is concave up.

- Recompute \(\frac{dy}{dx}\) from the two exponential component derivatives. - After differentiating the slope with respect to \(t\), divide by the horizontal component derivative. - Compare the corrected value at \(t=0\) with zero to classify concavity.

1. The first derivative is \(\frac{dy}{dx}=2e^{t}\). 2. Differentiating with respect to \(t\) gives \(2e^{t}\), which must then be divided by \(\frac{dx}{dt}=e^{t}\). 3. Thus, \(\frac{d^{2}y}{dx^{2}}=2\). At \(t=0\), it equals \(2\). 4. Because the value is positive, the curve is concave up there.

The student omitted division by \(\frac{dx}{dt}\). The correct second derivative is \(\frac{d^{2}y}{dx^{2}}=2\); at \(t=0\), the curve is concave up.

- A negative t-derivative of the slope does not by itself determine concavity in \(x\). - Divide by \(\frac{dx}{dt}\), whose sign may affect the result. - Evaluate the complete second derivative at the specified angle before interpreting its sign.

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, \(\frac{d^{2}y}{dx^{2}}=-\frac{1}{\cos^{3}(t)}\). Its sign depends on the sign of \(\cos(t)\), so the student’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.

The student omitted division by \(\frac{dx}{dt}\). The correct second derivative is \(\frac{d^{2}y}{dx^{2}}=-\frac{1}{\cos^{3}(t)}\); at \(t=\frac{\pi}{4}\), it is \(-2\sqrt{2}\), so the curve is concave down there.

- Derive the first parametric slope and differentiate it with respect to \(t\). - Divide by \(\frac{dx}{dt}\) to determine which proposed second derivative has the correct power of \(t\). - Use the sign at \(t=1\) to state concavity.

1. For \(t\ne0\), the first derivative is \(\frac{dy}{dx}=\frac{2}{3t}\). 2. Differentiating with respect to \(t\) gives \(-\frac{2}{3t^{2}}\), which must then be divided by \(\frac{dx}{dt}=3t^{2}\). 3. Thus, \(\frac{d^{2}y}{dx^{2}}=-\frac{2}{9t^{4}}\) for \(t\ne0\). At \(t=1\), it equals \(-\frac{2}{9}\). 4. Because the value is negative, the curve is concave down there.

\(\frac{d^{2}y}{dx^{2}}=-\frac{2}{9t^{4}}\) for \(t\ne0\); at \(t=1\), the value is \(-\frac{2}{9}\), so the curve is concave down.

- 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 three matching requirements produce three separate equations. - Compare the two pieces at the joining parameter before solving. - Work from the highest derivative condition downward.

1. Matching position gives \(a+b+c=1\). 2. Matching first derivatives gives \(3a+b=2\). 3. Matching second derivatives gives \(6a=2\), so \(a=\frac{1}{3}\). 4. Then \(b=1\), and \(c=-\frac{1}{3}\).

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

- Compute the second derivative in parameter form on the stated branch. - Identify where its sign can change within the interval. - Confirm the sign change before converting the parameter to a point.

1. The first derivative is \(\frac{dy}{dx}=\frac{5}{2}t(t^2-3)\). 2. The second derivative is \(\frac{15(t^2-1)}{4t}\). 3. On the stated interval, the second derivative changes from negative to positive at \(t=1\). 4. The corresponding point is \((1,-4)\).

\((1,-4)\)

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

- Differentiate each coordinate relation twice with respect to the parameter. - Use the positive-value condition to identify the coordinates at \(t=0\). - Combine the four component derivatives only after evaluating them.

1. At \(t=0\), \(x=1\) and \(y=2\). 2. Differentiating the coordinate relations gives \(x'=\frac{1}{2}\), \(x''=-\frac{1}{4}\), \(y'=-\frac{1}{12}\), and \(y''=-\frac{1}{144}\) at \(t=0\). 3. Therefore, \(\frac{d^2y}{dx^2}=\frac{y''x'-y'x''}{(x')^3}\). 4. Substitution gives \(\frac{d^2y}{dx^2}=-\frac{7}{36}\).

\(-\frac{7}{36}\)

- 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.