Partial Derivatives (Chain Rule)

[turophile]

I'm working on a problem involving the chain rule and partial derivatives. Here it is:

Find \(\displaystyle \frac{\partial u}{\partial s}\) where \(\displaystyle u=\frac{1}{\sqrt{x^2+y^2}}\), \(\displaystyle x=s\cos{t}\), \(\displaystyle y=s\sin{t}\), and s > 0.

The chain rule says: \(\displaystyle \frac{\partial u}{\partial s}=\frac{\partial u}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial u}{\partial y}\frac{\partial y}{\partial s}\).

So I calculate:

\(\displaystyle \frac{\partial u}{\partial x}=(-1/2)(x^2+y^2)^{-3/2}(2x)=-\frac{x}{(x^2+y^2)^{3/2}}\)

\(\displaystyle \frac{\partial x}{\partial s}=(s)(0)+(1)(\cos{t})=\cos{t}\)

\(\displaystyle \frac{\partial u}{\partial y}=(-1/2)(x^2+y^2)^{-3/2}(2y)=-\frac{y}{(x^2+y^2)^{3/2}}\)

\(\displaystyle \frac{\partial y}{\partial s}=(s)(0)+(1)(\sin{t})=\sin{t}\)

And I get:

\(\displaystyle \frac{\partial u}{\partial s}=\left(-\frac{x}{(x^2+y^2)^{3/2}}\right)(\cos{t})+\left(-\frac{y}{(x^2+y^2)^{3/2}}\right)(\sin{t})=-\frac{(x\cos{t}+y\sin{t})}{(x^2+y^2)^{3/2}}\)

The answer in my textbook is: \(\displaystyle -\frac{1}{s^2}\). Now if I substitute the respective expressions in s and t for x and y in u, then differentiate u with respect to s, I do get: \(\displaystyle -\frac{1}{s^2}\). I can't see where I'm using the chain rule incorrectly, but I'm obviously missing something. Any hints or other help would be appreciated.

 

[Yogi]

So far so good, just need to keep going a bit. Your expected answer is in terms of s and your current result contains x's and y's. If only there were a way to replace the x's and y's somehow...

 

[turophile]

I see it now:

\(\displaystyle x^2+y^2=(s\cos{t})^2+(s\sin{t})^2=s^2\cos^2t+{s^2}{\sin^2t}=s^2(\cos^2t+\sin^2t)=s^2(1)=s^2\)

\(\displaystyle x\cos{t}+y\sin{t}=(s\cos{t})(\cos{t})+(s\sin{t})({\sin}{t})={s}{\cos^2}{t}+s{\sin^2}{t}=s(\cos^2{t}+\sin^2{t})=s(1)=s\)

Many thanks for the hint, Yogi.