Can I get some up please?

Mathematics

1 answer:

qwelly [4]3 years ago

5 0

The answer is a it cant be anything else

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

son4ous [18]

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

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