Q. 3 - Calculus

Question

3. Let $z=e^{-x}(3xy-4x+y^2)$, and$x=\sin t, y=\cos t$

(a) Find$\frac{dz}{dt}$  by using the Chain Rule, Theorem 12.32 .

(b) Find $\frac{dz}{dt}$by evaluating  $f\left(x\right(t),y(t\left)\right)=f(\sin t,\cos t)$and taking the derivative 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. The example constructed for the statement of $f$is not differentiable at $(0,0)$is determined as $e^{-\sin t}\left(2\sin t\cos t-3\sin t{\cos }^{2}t-{\cos }^{3}t+3{\cos }^{2}t-3{\sin }^{2}t-4\cos t\right)$

b. The example for the statement of every point in ${\mathrm{ℝ}}^2$is determined as $e^{-\sin t}\left(2\sin t\cos t-3\sin t{\cos }^{2}t-{\cos }^{3}t+3{\cos }^{2}t-3{\sin }^{2}t-4\cos t\right)$

c. The statement of a unit vector that$D_{u}f(0,0)\ne \nabla f(0,0)\times u$, it is determined that chains rule is simpler than the other method applied

Introduction

The given data is a function '$z=e^{-x}(3xy-4x+y^2)$

The objective is to determine the statement $\frac{dz}{dt}$

Step 1

Let $z=e^{-x}\left(3xy-4x+y^2\right),x=\sin t$ and $y=\cos t$.

(a)

Find the function$\frac{dz}{dt}$ using chain rule

By the chain rule

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

Here,

$z=f(x, y), x=u\left(t\right)$ and $y=v\left(t\right)$.

Find $\frac{\partial z}{\partial x}$.

$\frac{\partial z}{\partial x}=\frac{\partial }{\partial x}\left\{e^{-x}\left(3xy-4x+y^2\right)\right\}$

$=e^{-x}\left(3y\frac{\partial }{\partial x}x-4\frac{\partial }{\partial x}x+\frac{\partial }{\partial x}{y}^2\right)+\left(3xy-4x+y^2\right)e^{-x}\frac{\partial }{\partial x}(-x)$

$=e^{-x}(3y-4·1+0)+\left(3xy-4x+y^2\right)e^{-x}(-1)$

$=e^{-x}(3y-4)-\left(3xy-4x+y^2\right)e^{-x}$

Step 2

Now, 

Find $\frac{\partial z}{\partial y}$,

$=e^{-x}\frac{\partial }{\partial y}\left(3xy-4x+y^2\right)$

$=e^{-x}\left(3x\frac{\partial }{\partial y}y-4\frac{\partial }{\partial y}x+\frac{\partial }{\partial y}{y}^2\right)$

$=e^{-x}(3x·1-4·0+2y)$

$=e^{-x}(3x+2y)$

Step 3

Again,

$$\begin{gathered} \frac{dx}{dt}=\frac{d}{dt}\sin t \\ =\cos t \end{gathered}$$

and, 

$$\begin{gathered} \frac{dy}{dt}=\frac{d}{dt}\cos t \\ =-\sin t \end{gathered}$$

Now,

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$=\left\{e^{-x}(3y-4)-\left(3xy-4x+y^2\right)e^{-x}\right\}\cos t+e^{-x}(3x+2y)(-\sin t)$

$=e^{-x}\left[\right\{(3y-4)-(3xy-4x+y^2)e^{-x}\}\cos t+e^{-x}(3x+2y\left)\right(-\sin t)$

$=e^{-\sin t}\left(3{\cos }^{2}t-4\cos t-3\sin t{\cos }^{2}t+4\sin t\cos t-{\cos }^{3}t\right.$

$\left.-3{\sin }^{2}t-2\sin t\cos t\right)$

$=e^{-\sin t}\left(3{\cos }^{2}t-4\cos t-3\sin t{\cos }^{2}t+2\sin t\cos t-{\cos }^{3}t-3{\sin }^{2}t\right)$

$=e^{-\sin t}\left(2\sin t\cos t-3\sin t{\cos }^{2}t-{\cos }^{3}t+3{\cos }^{2}t-3{\sin }^{2}t-4\cos t\right)$

Step 4

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

The function can be rewritten as

$z=e^{-\sin t}\left(3\sin t\cos t-4\sin t+{\cos }^{2}t\right)$

The equation ( 1 ) is differentiated with respect to $t$.

$\frac{dz}{dt}=\frac{d}{dt}\left\{e^{-\sin t}\left(3\sin t\cos t-4\sin t+{\cos }^{2}t\right)\right\}$

$=e^{-\sin t}\frac{d}{dt}\left(3\sin t\cos t-4\sin t+{\cos }^{2}t\right)$

$+\left(3\sin t\cos t-4\sin t+{\cos }^{2}t\right)\frac{d}{dt}{e}^{-\sin t}$

$=e^{-\sin t}(3\frac{d}{dt}\sin t\cos t-4\frac{d}{dt}\sin t+\frac{d}{dt}{\cos }^{2}t)+(3\sin t\cos t-4\sin t+{\cos }^{2}t)e^{-\sin t}\frac{d}{dt}(-\sin t)$

Step 5

$$\begin{gathered} =e^{-\sin t}\left\{3\right(\sin t\frac{d}{dt}\cos t+\cos t\frac{d}{dt}\sin t)-4\cos t+2\cos t\frac{d}{dt}\cos t\}+\left(3\right(\sin t\cos t-4\sin t+{\cos }^{2}t\left)e^{-\sin t}\right(-\cos t) \\ =e^{-\sin t}\{3\left(\sin t\right(-\sin t)+\cos t-\cos t)-4\cos t-2\cos t\sin t\}-\cos t·e^{-\sin t}(3\sin t\cos t-4\sin t+{\cos }^{2}t) \end{gathered}$$

$$\begin{gathered} =e^{-\sin t}\left[\right\{3(-{\sin }^{2}t+{\cos }^{2}t)-4\cos t-2\cos t\sin t\}-\cos t(3\sin t\cos t-4\sin t+{\cos }^{2}t\left)\right\} \\ =e^{-\sin t}\left[\right\{3(-{\sin }^{2}t+{\cos }^{2}t)-4\cos t-2\cos t\sin t\}-3\sin t{\cos }^{2}t+4\sin t\cos t+{\cos }^{3}t] \\ =e^{-\sin t}\left[\right\{-3{\sin }^{2}t+3{\cos }^{2}t-4\cos t-2\cos t\sin t\}-3\sin t{\cos }^{2}t-4\sin t\cos t-{\cos }^{3}t] \\ =e^{-\sin t}\left[\right\{-3{\sin }^{2}t+3{\cos }^{2}t-4\cos t+2\cos t\sin t\}-3\sin t{\cos }^{2}t-{\cos }^{3}t)\} \\ =e^{-\sin t}(2\sin t\cos t-3\sin t{\cos }^{2}t-{\cos }^{3}t+3{\cos }^{3}t-3{\sin }^{2}t-4\cos t) \end{gathered}$$

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)