Q. 48 - Calculus

Question

In Exercises 45-48 use Example 6 to find $\frac{d^{2}y}{d{x}^2}$ for the parametric curve at the given value of t. Note that these are the same parametric equations as in Exercises 41-44.

$x={\cos }^{3}t,y={\sin }^{3}t,t=\frac{\pi }{4}$.

Step-by-Step Solution

Answer

The second derivative of the given parametric curve$t=\frac{\pi }{4}$ is $-2$.

Step 1: Given information

The  parametric equations $x={\cos }^{3}t,y={\sin }^{3}t,t=\frac{\pi }{4}$.

Step 2: Calculation

Consider the parametric curves $x={\cos }^{3}t,y={\sin }^{3}t$ at $t=\frac{\pi }{2}$.

The objective is to find the second derivative that is $\frac{d^{2}y}{d{x}^2}$.

The formula to find the second derivative is $\frac{d^{2}y}{d{x}^2}=\frac{\frac{dx}{dt}·\frac{d^{2}y}{d{t}^2}-\frac{d^{2}x}{d^{2}t}·\frac{dy}{dt}}{{\left(\frac{dx}{dt}\right)}^2}$.

First find the derivatives $\frac{dx}{dt},\frac{dy}{dt},\frac{d^{2}y}{d{t}^2},\frac{d^{2}x}{d^{2}t}$ and substitute in the formula.

Now take the parametric equation $x={\cos }^{3}t$.

Differentiate the curve with respect to t.

Then

$$\begin{gathered} \frac{dx}{dt}=\frac{d}{dt}{\cos }^{3}t\left[\sin cex={\cos }^{3}t\right] \\ \frac{dx}{dt}=3{\cos }^{2}t·\frac{d}{dt}\cos t \\ \frac{dx}{dt}=3{\cos }^{2}t·-\sin t\left[\text{ since }\frac{d{\cos }^{3}t}{dt}=-3{\cos }^{2}t\sin t\right] \end{gathered}$$

The equation $3{\cos }^{2}t\sin t$can be written in terms of $\sin t$in the following way.

$$\begin{gathered} \frac{dx}{dt}=-3\left(1-{\sin }^{2}t\right)\sin t\left[\sin ce {\cos }^{2}t=1-{\sin }^{2}t\right] \\ \frac{dx}{dt}=-3\sin t+3{\sin }^{3}t \end{gathered}$$

Again differentiate with respect to t,

$\frac{d^{2}x}{d{t}^2}=\frac{d}{dt}\left(-3\sin t+3{\sin }^{3}t\right)$

On further simplification,

$$\begin{gathered} \frac{d^{2}x}{d{t}^2}=\frac{d}{dt}(-3\sin t)+\frac{d}{dt}\left(3{\sin }^{3}t\right) \\ \frac{d^{2}x}{d{t}^2}=-3\cos t+3·3{\sin }^{2}t·\cos t \\ \frac{d^{2}x}{d{t}^2}=-3\cos t+9{\sin }^{2}t\cos t \end{gathered}$$

Now take the parametric equation $y={\sin }^{3}t$.

Differentiate the curve with respect to t.

$$\begin{gathered} \frac{dy}{dt}=\frac{d}{dt}\left({\sin }^{3}t\right) \\ \frac{dy}{dt}=3{\sin }^{2}t\frac{d}{dt}\left(\sin t\right) \\ \frac{dy}{dt}=3{\sin }^{2}t·\cos t \\ \frac{dy}{dt}=3{\sin }^{2}t\cos t \end{gathered}$$

The equation $3{\sin }^{2}t\cos t$can be written in terms of$\cos t$ in the following way.

$$\begin{gathered} \frac{dy}{dt}=3\left(1-{\cos }^{2}t\right)\cos t\left[Since {\sin }^{2}t=1-{\cos }^{2}t\right] \\ \frac{dy}{dt}=3\cos t-3{\cos }^{3}t \end{gathered}$$

Again differentiate with respect to t,

$\frac{d^{2}y}{d{t}^2}=\frac{d}{dt}\left(3\cos t-3{\cos }^{3}t\right)$

Then,

$$\begin{gathered} \frac{d^{2}y}{d{t}^2}=\frac{d}{dt}3\cos t-\frac{d}{dt}3{\cos }^{3}t \\ \frac{d^{2}y}{d{t}^2}=3\frac{d}{dt}\cos t-3\frac{d}{dt}{\cos }^{3}t \\ \frac{d^{2}y}{d{t}^2}=3(-\sin t)-3·3{\cos }^{2}t·-\sin t \\ \frac{d^{2}y}{d{t}^2}=-3\sin t+9{\cos }^{2}t\sin t \end{gathered}$$

Now substitute the values in the formula $\frac{d^{2}y}{d{x}^2}=\frac{\frac{dx}{dt}\times \frac{d^{2}y}{d{t}^2}-\frac{d^{2}x}{d^{2}t}\times \frac{dy}{dt}}{{\left(\frac{dx}{dt}\right)}^2}$.

$\frac{d^{2}y}{d{x}^2}=\frac{\left(-3{\cos }^{2}t\sin t\right)·\left(-3\sin t+9{\cos }^{2}t\sin t\right)-\left(-3\cos t+9{\sin }^{2}t\cos t\right)·\left(3{\sin }^{2}t\cos t\right)}{{\left(-3{\cos }^{2}t\sin t\right)}^2}$

Then,

$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=\frac{9{\sin }^{2}t{\cos }^{2}t-27{\sin }^{2}t{\cos }^{4}t-\left(-9{\sin }^{2}t{\cos }^{2}t+27{\sin }^{4}t{\cos }^{2}t\right.}{{\left(-3{\cos }^{2}t\sin t\right)}^2} \\ \frac{d^{2}y}{d{x}^2}=\frac{9{\sin }^{2}t{\cos }^{2}t-27{\sin }^{2}t{\cos }^{4}t+9{\sin }^{2}t{\cos }^{2}t-27{\sin }^{4}t{\cos }^{2}t}{{\left(-3{\cos }^{2}t\sin t\right)}^2} \\ \frac{d^{2}y}{d{x}^2}=\frac{18{\sin }^{2}t{\cos }^{2}t-27{\sin }^{2}t{\cos }^{4}t-27{\sin }^{4}t{\cos }^{2}t}{{\left(-3{\cos }^{2}t\sin t\right)}^2} \\ \frac{d^{2}y}{d{x}^2}=\frac{18{\sin }^{2}t{\cos }^{2}t-27{\sin }^{2}t{\cos }^{2}t\left({\cos }^{2}t+{\sin }^{2}t\right)}{{\left(-3{\cos }^{2}t\sin t\right)}^2} \end{gathered}$$

Step 3: Calculation

On further simplification,

$\frac{d^{2}y}{d{x}^2}=\frac{18{\sin }^{2}t{\cos }^{2}t-27{\sin }^{2}t{\cos }^{2}t·1}{{\left(-3{\cos }^{2}t\sin t\right)}^2}\left[since{\sin }^{2}t+{\cos }^{2}t=1\right]$

$\frac{d^{2}y}{d{x}^2}=\frac{-9{\sin }^{2}t{\cos }^{2}t}{9{\cos }^{4}t{\sin }^{2}t}$

$\frac{d^{2}y}{d{x}^2}=\frac{9{\sin }^{2}t{\cos }^{2}t}{y{\cos }^{2}t{\sin }^{2}t}$

$\frac{d^{2}y}{d{x}^2}=\frac{-1}{{\cos }^{2}t}$

Now at an angle $t=\frac{\pi }{4}$the second derivative is,

$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=\frac{-1}{{\left(\cos \frac{\pi }{4}\right)}^2} \\ \frac{d^{2}y}{d{x}^2}=\frac{-1}{{\left(\frac{1}{\sqrt{2}}\right)}^2}\left[\text{ since }\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}\right] \\ \frac{d^{2}y}{d{x}^2}=-2 \end{gathered}$$

At $t=\frac{\pi }{4}$ the second derivative is $\frac{d^{2}y}{d{x}^2}=-2$,

Therefore, the second derivative of the given parametric curve $t=\frac{\pi }{4}$ is $-2$.