## University Calculus: Early Transcendentals (3rd Edition)

$\frac{ds}{dt} = \frac{x{\frac{dx}{dt}+y\frac{dy}{dt}+z\frac{dz}{dt}}}{\sqrt{(x^2+y^2+z^2)}}$ $\frac{ds}{dt} = \frac{(y\frac{dy}{dt}+z\frac{dz}{dt})}{\sqrt{(x^2+y^2+z^2)}}$ $x{\frac{dx}{dt}=-(y\frac{dy}{dt}+z\frac{dz}{dt}})$
Given $s={\sqrt{(x^2+y^2+z^2)}}$ on differentiating both sides: $\frac{ds^2}{dt}=\frac{d{{(x^2+y^2+z^2)}}}{dt}$ $2s\frac{ds}{dt}=2x{\frac{dx}{dt}+2y\frac{dy}{dt}+2z\frac{dz}{dt}}$ $\frac{ds}{dt} = \frac{x{\frac{dx}{dt}+y\frac{dy}{dt}+z\frac{dz}{dt}}}{\sqrt{(x^2+y^2+z^2)}}$ taking x as a constant, the above equation becomes: $s\frac{ds}{dt}=y\frac{dy}{dt}+z\frac{dz}{dt}$ $\frac{ds}{dt} = \frac{(y\frac{dy}{dt}+z\frac{dz}{dt})}{\sqrt{(x^2+y^2+z^2)}}$ on taking s as a constant: $0=x{\frac{dx}{dt}+y\frac{dy}{dt}+z\frac{dz}{dt}}$ $x{\frac{dx}{dt}=-(y\frac{dy}{dt}+z\frac{dz}{dt}})$ Thus the final answer is: $\frac{ds}{dt} = \frac{x{\frac{dx}{dt}+y\frac{dy}{dt}+z\frac{dz}{dt}}}{\sqrt{(x^2+y^2+z^2)}}$ $\frac{ds}{dt} = \frac{(y\frac{dy}{dt}+z\frac{dz}{dt})}{\sqrt{(x^2+y^2+z^2)}}$ $x{\frac{dx}{dt}=-(y\frac{dy}{dt}+z\frac{dz}{dt}})$