• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q19E

Expert-verified Found in: Page 656 ### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280 # Find the value of $$\frac{{\partial w}}{{\partial r}}$$ and$$\frac{{\partial w}}{{\partial \theta }}$$, using the chain rule if $$w = xy + yz + zx,x = r\cos \theta ,y = r\sin \theta$$ and $$z = r\theta$$ when $$r = 2$$ and $$\theta = \frac{\pi }{2}.$$

The value of $$\frac{{\partial w}}{{\partial r}}$$ at $$r = 2$$ and $$\theta = \frac{\pi }{2}$$ is $$2\pi .$$

The value of $$\frac{{\partial w}}{{\partial \theta }}$$at $$r = 2$$and $$\theta = \frac{\pi }{2}$$is $$- 2\pi$$

See the step by step solution

## Step 1: The chain rule.

With the chain rule, the partial derivative of function $${\rm{f(x, y)}}$$ with respect to $${\rm{t}}$$ becomes $$\frac{{\partial f}}{{\partial t}} = \frac{{\partial f}}{{\partial x}} \cdot \frac{{\partial x}}{{\partial t}} + \frac{{\partial f}}{{\partial y}} \cdot \frac{{\partial y}}{{\partial t}}.$$

## Step 2: Use the chain rule for calculation.

The function is, $${\rm{w = xy + yz + zx}}{\rm{.}}$$

Given that the values of $$r = 2$$ and$$\theta = \frac{\pi }{2}.$$.

Substitute $$r = 2$$ and $$\theta = \frac{\pi }{2}.$$in$$x,$$

Thus, the value of$$x = 0.$$

Substitute $$r = 2$$ and $$\theta = \frac{\pi }{2}.$$in $$y,$$

Thus, the value of $$y = 2.$$

The partial derivative $$\frac{{\partial w}}{{\partial r}}$$ using chain rule is computed as follows,

\begin{aligned}{l}\frac{{\partial w}}{{\partial r}} = \frac{{\partial w}}{{\partial x}} \cdot \frac{{\partial x}}{{\partial r}} + \frac{{\partial w}}{{\partial y}} \cdot \frac{{\partial y}}{{\partial r}} + \frac{{\partial w}}{{\partial z}} \cdot \frac{{\partial z}}{{\partial r}}\\ = \frac{\partial }{{\partial x}}(xy + yz + zx) \cdot \frac{\partial }{{\partial r}}(r\cos \theta ) + \frac{\partial }{{\partial y}}(xy + yz + zx) \cdot \frac{\partial }{{\partial r}}(r\sin \theta ) + \frac{\partial }{{\partial z}}(xy + yz + zx) \cdot \frac{\partial }{{\partial r}}(r\theta )\\ = (y + z)(\cos \theta ) + (x + z)(\sin \theta ) + (x + y)(\theta )\end{aligned}

Thus, the partial derivative, $$\frac{{\partial w}}{{\partial r}} = (y + z)(\cos \theta ) + (x + z)(\sin \theta ) + (x + y)(\theta ).$$

Substitute the respective values in $$\frac{{\partial w}}{{\partial r}}$$ and obtain the required value.

\begin{aligned}{l}\frac{{\partial w}}{{\partial r}} = (y + z)(\cos \theta ) + (x + z)(\sin \theta ) + (y + x)(\theta )\\ = (2 + \pi ) \cdot \cos \left( {\frac{\pi }{2}} \right) + (0 + \pi ) \cdot \sin \left( {\frac{\pi }{2}} \right) + (2 + 0)\left( {\frac{\pi }{2}} \right)\\ = (2\pi ) \cdot (0) + (\pi ) \cdot (1) + \pi \\ = 2\pi \end{aligned}

Thus, the required value is, $$\frac{{\partial w}}{{\partial r}} = 2\pi .$$

The partial derivatives $$\frac{{\partial w}}{{\partial \theta }}$$ using chain rule is computed as follows.

\begin{aligned}{l}\frac{{\partial w}}{{\partial \theta }} = \frac{{\partial w}}{{\partial x}} \cdot \frac{{\partial x}}{{\partial \theta }} + \frac{{\partial w}}{{\partial y}} \cdot \frac{{\partial y}}{{\partial \theta }} + \frac{{\partial w}}{{\partial z}} \cdot \frac{{\partial z}}{{\partial \theta }}\\ = \frac{\partial }{{\partial x}}(xy + yz + zx) \cdot \frac{\partial }{{\partial \theta }}(r\cos \theta ) + \frac{\partial }{{\partial y}}(xy + yz + zx) \cdot \frac{\partial }{{\partial \theta }}(r\sin \theta ) + \frac{\partial }{{\partial z}}(xy + yz + zx) \cdot \frac{\partial }{{\partial \theta }}(r\theta )\\ = (y + z)( - r\sin \theta ) + (x + z)(r\cos \theta ) + (x + y)(r)\end{aligned}

Thus, the partial derivative, $$\frac{{\partial w}}{{\partial \theta }} = (y + z)( - r\sin \theta ) + (x + z)(r\cos \theta ) + (x + y)(r).$$

Substitute the respective values in $$\frac{{\partial w}}{{\partial \theta }}$$ and obtain the required value.

\begin{aligned}{l}\frac{{\partial w}}{{\partial \theta }} = (y + z)( - r\sin \theta ) + (x + z)(r\cos \theta ) + (y + x)(r)\\ = (2 + \pi )\left( { - (2)\sin \left( {\frac{\pi }{2}} \right)} \right) + (0 + \pi )\left( {(2)\cos \left( {\frac{\pi }{2}} \right)} \right) + (2 + 0)(2)\\ = (2 + \pi )( - 2) + \pi (2 \cdot 0) + 4 = - 4 - 2\pi + (\pi )(0) + 4\end{aligned}

Thus, the required value is, $$\frac{{\partial w}}{{\partial \theta }} = - 2\pi .$$ ### Want to see more solutions like these? 