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Q19E

Expert-verifiedFound in: Page 656

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 \( - 2\pi \)

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

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, \(\frac{{\partial w}}{{\partial \theta }} = - 2\pi .\)

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