Q 4. - Calculus

Question

Let $z = e-xy2 , x = s \sin t,$ and $y = s2 \cos t$

(a) Find $\frac{\partial z}{\partial s}$

by using the Chain Rule, Theorem $12.33$

(b) Find $\frac{\partial z}{\partial s}$ by evaluating $f \left(x\right(s, t), y(s, t\left)\right)=f (s \sin t, s2\cos t)$ and taking the partial derivative with respect to $s$ of the resulting function.

(c) Show that your answers from parts (a) and (b) are the same. Which method was easier?

Step-by-Step Solution

Answer

Part (a)  $-e^{-\left(s^3{\sin }^{2}t{\cos }^{2}t\right)}\left(5s^4\sin t{\cos }^{2}t\right)$

Part (b) $-e^{-\left(s^5\sin t{\cos }^{2}t\right)}\left(5s^4\sin t{\cos }^{2}t\right)$

Part (c) The method based on the chain rule is less difficult than the method used in part (b).

Part (a) Step 1: Given information

$z=e^{-x{y}^2},x=s\sin t$ and $y=s^2\cos t$

Part (a) Step 1: Explanation

Let $z=e^{-x{y}^2},x=s\sin t$ and $y=s^2\cos t$

The goal is to locate $\frac{\partial z}{\partial s}$ using chain rule.

According to the chain rule

$\frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial s}$

where $z=f(x, y), x=u(s, t)$ and $y=v(s, t)$

First, find $\frac{\partial z}{\partial x}$

$$\begin{gathered} \frac{\partial z}{\partial x}=\frac{\partial }{\partial x}\left(e^{-x{y}^2}\right) \\ =e^{-x{y}^2}\left(-y^2\right)\frac{\partial }{\partial x}x \\ =-y^2{e}^{-x{y}^2} \end{gathered}$$

Next find $\frac{\partial z}{\partial y}$

$$\begin{gathered} \frac{\partial z}{\partial y}=\frac{\partial }{\partial y}\left(e^{-x{y}^2}\right) \\ =e^{-x{y}^2}(-x)\frac{\partial }{\partial y}{y}^2 \\ =-2xy{e}^{-x{y}^2} \end{gathered}$$

Again, find $\frac{\partial x}{\partial s}$

$$\begin{gathered} \frac{\partial x}{\partial s}=\frac{\partial }{\partial s}\left(s\sin t\right) \\ =\sin t\frac{\partial }{\partial s}s \\ =\sin t \end{gathered}$$

Also, find $\frac{\partial y}{\partial s}$

$$\begin{gathered} \frac{\partial y}{\partial s}=\frac{\partial }{\partial s}\left(s^2\cos t\right) \\ =\cos t\frac{\partial }{\partial s}{s}^2 \\ =2s\cos t \end{gathered}$$

Thus,

$$\begin{gathered} \frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial s} \\ =\left(-y^2{e}^{-x{y}^2}\right)\sin t+\left(-2xy{e}^{-x{y}^2}\right)\left(2s\cos t\right) \\ =-e^{-x{y}^2}\left(y^2\sin t+4xys\cos t\right) \\ =-e^{-x\sin t{\left(x^2\cos t\right)}^2}\left\{{\left(s^2\cos t\right)}^2\sin t+4\left(s\sin t\right)\left(s^2\cos t\right)s\cos t\right\} \\ =-e^{-s\sin \left(\left(s^4{con}^{2}t\right)\right.}\left(s^4{\cos }^{2}t\sin t+4s^4\sin t{\cos }^{2}t\right) \\ =-e^{-\left(s^3{\sin }^{2}t{\cos }^{2}t\right)}\left(5s^4\sin t{\cos }^{2}t\right) \end{gathered}$$

Part (b) Step 1: Explanation

The goal is to find $\frac{\partial z}{\partial s}$

Rewrite the function as

$$\begin{gathered} z=e^{-x{y}^2} \\ =e^{-s\sin {\left(s^2{\cos }^2\right)}^2} \\ =e^{-s\sin \left(s^2{\cos }^{2}t\right)} \\ =e^{-\left(x^3\sin \left({\cos }^{2}t\right)\right.}\dots \dots .\left(1\right) \end{gathered}$$

Differentiate ( 1 ) partially with respect to $s$

$$\begin{gathered} \frac{\partial z}{\partial s}=\frac{\partial }{\partial s}{e}^{-\left(s^3{\sin }^2{\cos }^{2}t\right)} \\ =e^{-\left(s^3{\sin }^{2}f{\cos }^{2}t\right)}\frac{\partial }{\partial s}\left\{-\left(s^5\sin t{\cos }^{2}t\right)\right\} \\ =e^{-\left(s^5\sin t{\cos }^{2}t\right)}\left(-\sin t{\cos }^{2}t\right)\frac{\partial }{\partial s}{s}^5 \\ =-e^{-\left(s^5{\sin }^{2}t{\cos }^{2}t\right)}\left(5s^4\sin t{\cos }^{2}t\right) \end{gathered}$$

Part (c) Step 1: Explanation

Parts (a) and (b) show that the outcome is the same. The method based on the chain rule is less difficult than the method used in part (b).