fbpx
calculus online - Free exercises and solutions to help you succeed!

Multivariable Chain Rule – Exercise 3350

Exercise

Given

z(x,y)=\arctan (\frac{x}{y})

f(u,v)=u\sin v

g(u,v)=u\cos v

Z(u,v)=z(f(u,v),g(u,v))

Calculate the derivative

Z'_u,Z'_v

Final Answer

Z'_u=0

Z'_v=1

Solution

We will use the chain rule to calculate the derivative

Z'_u=z'_f\cdot f'_u+z'_g\cdot g'_u

We have

Z(u,v)=z(f(u,v),g(u,v))

And function u is

z(x,y)=\arctan(\frac{x}{y})

We put it in the function above and get

z(f,g)=\arctan(\frac{f}{g})

Hence, the partial derivatives of z are

z'_f=\frac{1}{1+{(\frac{f}{g})}^2}\cdot \frac{1}{g}

z'_g=\frac{1}{1+{(\frac{f}{g})}^2}\cdot \frac{-f}{g^2}

Also, we calculate the derivatives of f and g

f'_u=\sin v

g'_u=\cos v

We put all the derivatives and get

Z'_u=z'_f\cdot f'_u+z'_g\cdot g'_u=

=\frac{1}{1+{(\frac{f}{g})}^2}\cdot \frac{1}{g}\cdot \sin v+\frac{1}{1+{(\frac{f}{g})}^2}\cdot \frac{-f}{g^2}\cdot \cos v=

=\frac{1}{1+{(\frac{u\sin v}{u\cos v})}^2}\cdot \frac{1}{u\cos v}\cdot \sin v+\frac{1}{1+{(\frac{u\sin v}{u\cos v})}^2}\cdot \frac{-u\sin v}{{(u\cos v)}^2}\cdot \cos v=

=\frac{1}{1+{(\frac{u\sin v}{u\cos v})}^2}(\frac{\sin v}{u\cos v}+\frac{-u\sin v}{u^2\cos^2 v}\cdot \cos v)=

=\frac{1}{1+{(\frac{u\sin v}{u\cos v})}^2}(\frac{\sin v}{u\cos v}-\frac{\sin v}{u\cos v})=

=\frac{1}{1+{(\frac{u\sin v}{u\cos v})}^2}\cdot 0=0

We will use the chain rule to calculate the partial derivatives of Z.

Z'_v=z'_f\cdot f'_v+z'_g\cdot g'_v

We got above

z'_f=\frac{1}{1+{(\frac{f}{g})}^2}\cdot \frac{1}{g}

z'_g=\frac{1}{1+{(\frac{f}{g})}^2}\cdot \frac{-f}{g^2}

Alao, we calculate the derivatives of f and g.

f'_v=u\cos v

g'_v=-u\sin v

We put the derivatives and get

Z'_v=z'_f\cdot f'_v+z'_g\cdot g'_v

=\frac{1}{1+{(\frac{f}{g})}^2}\cdot \frac{1}{g}\cdot u\cos v+\frac{1}{1+{(\frac{f}{g})}^2}\cdot \frac{-f}{g^2}\cdot(-u\sin v)=

=\frac{1}{1+{(\frac{u\sin v}{u\cos v})}^2}\cdot \frac{1}{u\cos v}\cdot u\cos v+\frac{1}{1+{(\frac{u\sin v}{u\cos v})}^2}\cdot \frac{-u\sin v}{{(u\cos v)}^2}\cdot(-u\sin v)=

=\frac{1}{1+{(\frac{u\sin v}{u\cos v})}^2}(\frac{u\cos v}{u\cos v} +\frac{-u\sin v}{{(u\cos v)}^2}\cdot(-u\sin v))=

=\frac{1}{1+\frac{\sin^2 v}{\cos^2 v}}(1+\frac{u^2\sin^2 v}{u^2\cos^2 v})=

=\frac{1}{1+\tan^2 v}(1+\tan^2 v)=

=\frac{1+\tan^2 v}{1+\tan^2 v}=1

Have a question? Found a mistake? – Write a comment below!
Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions! 

Share with Friends

Leave a Reply