Q. 4 - Calculus

Question

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

(a) Find $\frac{az}{as}$ by using the Chain Rule, Theorem 12.33.

(b) Find $\frac{\mathrm{az}}{\mathrm{a}s}$ by evaluating $f\left(x\right(s,t),y(s,t\left)\right)=f\left(s\sin t,s^2\cos t\right)$ 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

a. With the chain rule, the statement $\frac{az}{as}$is determined as $-e^{-\left(s^5\sin t{\cos }^{2}t\right)}\left(5s^4\sin t{\cos }^{2}t\right)$

b. The evaluation of $f\left(x\right(s,t),y(s,t\left)\right)=f\left(s\sin t,s^2\cos t\right)$to find $\frac{az}{as}$is determined as $-e^{-\left(s^5\sin t{\cos }^{2}t\right)}\left(5s^4\sin t{\cos }^{2}t\right)$

c. The chain method is considered to be the easiest method

Introduction

The given data is $z=e^{-x{y}^2}$

The objective is to find $\frac{az}{as}$with chain rule and by solving a statement and to find which method was easier

Step 1

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

(a)

The objective is to find $\frac{\partial z}{\partial s}$ using chain rule.

According to 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 $\frac{\partial z}{\partial x}$.

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

$=e^{-x{y}^2}\left(-{\dot{y}}^2\right)\frac{\partial }{\partial x}x$

$=-y^2{e}^{-x{y}^2}$

Step 2

Now, find $\frac{dz}{dy}$

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

$=e^{-x{y}^2}(-x)\frac{\partial }{\partial y}{y}^2$

$=-2xy{e}^{-x{y}^2}$

Step 3

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

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

$=\sin t\frac{\partial }{\partial s}s$

$=\sin t$

To find $\frac{dy}{ds}$

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

$=\cos t\frac{\hat{\partial }}{\partial s}{s}^2$

$=2s\cos t$

Step 4

So,

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

$=\left(-y^2{e}^{-x{y}^2}\right)\sin t+\left(-2xy{e}^{-x{y}^2}\right)\left(2s\cos t\right)$

$=-e^{-x{y}^2}\left(y^2\sin t+4xys\cos t\right)$

$=-e^{-s\sin t{\left(s^2\cos t\right)}^2}\left\{{\left(s^2\cos t\right)}^2\sin t+4\left(s\sin t\right)\left(s^2\cos t\right)s\cos t\right\}$

$=-e^{-s\sin t\left(s^4{\cos }^{2}t\right)}\left(s^4{\cos }^{2}t\sin t+4s^4\sin t{\cos }^{2}t\right)$

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

Step 5

To find $\frac{dz}{ds}$,

The function can be rewritten as

$z=e^{-x{y}^2}$

$=e^{-s\sin r{\left(s^2\cos t\right)}^2}$

$=e^{-s\sin r\left(s^4{\cos }^{2}t\right)}$

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

The equation( 1) can be partially differentiated with respect to $s$.

$\frac{\widehat{\partial z}}{\partial s}=\frac{\hat{\partial }}{\partial s}{e}^{-\left(s^3\sin t{\cos }^{2}t\right)}$

$=e^{-\left(s^5\sin t{\cos }^{2}t\right)}\frac{\partial }{\partial s}\left\{-\left(s^5\sin t{\cos }^{2}t\right)\right\}$

$=e^{-\left(s^5\sin t{\cos }^{2}t\right)}\left(-\sin t{\cos }^{2}t\right)\frac{\partial }{\partial s}{s}^5$

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

Step 6

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)