Question

Use the chain rule to find dw/dt. w = ln x2 + y2 + z2 , x = 7 sin(t), y = 9 cos(t), z = 6 tan(t)

Answer 1

By the chain rule,

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

We have

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

$\dfrac{\mathrm dx}{\mathrm dt}=7\cos t$
$\dfrac{\mathrm dy}{\mathrm dt}=-9\sin t$
$\dfrac{\mathrm dz}{\mathrm dt}=6\sec^2t$

So we have

$\dfrac{\mathrm dw}{\mathrm dt}=\dfrac2{x^2+y^2+z^2}\left(7x\cos t-9y\sin t+6z\sec^2t\right)$
$\dfrac{\mathrm dw}{\mathrm dt}=\dfrac{2\left(49\sin t\cos t-81\cos t\sin t+36\tan t\sec^2t\right)}{49\sin^2t+81\cos^2t+36\tan^2t}$
$\dfrac{\mathrm dw}{\mathrm dt}=\dfrac{8\sin t(9-16\cos^4t)}{\cos t(36+13\cos^2t+32\cos^4t)}$

Answer 2

Using the chain rule, we find that:

$\frac{dw}{dt} = \frac{1}{49\sin^{2}{t} + 81\cos^2t + 36\tan^2t}(-64\sin{t}\cos{t} + 72\tan{t}\sec^2{t})$

The function w is:

$w(x,y,z) = \ln{x^2 + y^2 + z^2}$

x, y and z are functions of t, given by:

$x(t) = 7\sin{t}$

$y(t) = 9\cos{t}$

$z(t) = 6\tan{t}$

Thus, by the chain rule:

$\frac{dw}{dt} = \frac{dw}{dx}\frac{dx}{dt} + \frac{dw}{dy}\frac{dy}{dt} + \frac{dw}{dz}\frac{dz}{dt}$

The derivatives are:

$\frac{dw}{dx} = \frac{2x}{x^2 + y^2 + z^2} = \frac{14\sin{t}}{49\sin^{2}{t} + 81\cos^2t + 36\tan^2t}$

$\frac{dx}{dt} = 7\cos{t}$

$\frac{dw}{dy} = \frac{2y}{x^2 + y^2 + z^2} = \frac{18\cos{t}}{49\sin^{2}{t} + 81\cos^2t + 36\tan^2t}$

$\frac{dy}{dt} = -9\sin{t}$

$\frac{dw}{dz} = \frac{2z}{x^2 + y^2 + z^2} = \frac{12\tan{t}}{49\sin^{2}{t} + 81\cos^2t + 36\tan^2t}$

$\frac{dy}{dt} = 6\sec^2{t}$

Then:

$\frac{dw}{dt} = \frac{dw}{dx}\frac{dx}{dt} + \frac{dw}{dy}\frac{dy}{dt} + \frac{dw}{dz}\frac{dz}{dt}$

$\frac{dw}{dt} = \frac{1}{49\sin^{2}{t} + 81\cos^2t + 36\tan^2t}(98\sin{t}\cos{t} - 162\sin{t}\cos{t} + 72\tan{t}\sec^2{t})$

$\frac{dw}{dt} = \frac{1}{49\sin^{2}{t} + 81\cos^2t + 36\tan^2t}(-64\sin{t}\cos{t} + 72\tan{t}\sec^2{t})$

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