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expeople1 [14]
3 years ago
12

Use chain rule to find dw/dt w=ln(x^2 + y^2 + z^2)^(1/2) , x=sint, y=cost, z=tant

Mathematics
1 answer:
son4ous [18]3 years ago
4 0
The chain rule states that

\dfrac{\mathrm dw}{\mathrm dt}=\dfrac{\partial w}{\partial x}\dfrac{\mathrm dx}{\mathrm dt}+\dfrac{\partial w}{\partial y}\dfrac{\mathrm dy}{\mathrm dt}+\dfrac{\partial w}{\partial z}\dfrac{\mathrm dz}{\mathrm dt}

Since \ln(x^2+y^2+z^2)^{1/2}=\dfrac12\ln(x^2+y^2+z^2), you have

\dfrac{\partial w}{\partial x}=\dfrac x{x^2+y^2+z^2}
\dfrac{\partial w}{\partial x}=\dfrac y{x^2+y^2+z^2}
\dfrac{\partial w}{\partial z}=\dfrac z{x^2+y^2+z^2}

and

\dfrac{\mathrm dx}{\mathrm dt}=\cos t
\dfrac{\mathrm dy}{\mathrm dt}=-\sin t
\dfrac{\mathrm dz}{\mathrm dt}=\sec^2t

Also, since x^2+y^2+z^2=\sin^2t+\cos^2t+\tan^2t=1+\tan^2t=\sec^2t, the derivative is

\dfrac{\mathrm dw}{\mathrm dt}=\dfrac{\sin t\cos t}{\sec^2t}-\dfrac{\sin t\cos t}{\sec^2t}+\dfrac{\tan t\sec^2t}{\sec^2t}
\dfrac{\mathrm dw}{\mathrm dt}=\tan t
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