Let $T = x^2y-xy^3+2$ ; where $x = r\cos\theta, y=r\sin\theta$. Find $\frac{\partial T}{\partial r}$ and $\frac{\partial T}{\partial \theta}$

Question

Let $T = x^2y-xy^3+2$ ; where $x = r\cos\theta, y=r\sin\theta$. Find $\dfrac{\partial T}{\partial r}$ and $\dfrac{\partial T}{\partial \theta}$.

My answer:

For $\dfrac{\partial T}{\partial r}$ I get the formula: $$\dfrac{\partial T}{\partial r}= \dfrac{\partial T}{\partial x}\cdot \dfrac{\partial x}{\partial r} + \dfrac{\partial T}{\partial y}\cdot \dfrac{\partial y}{\partial r}$$

Where

$$\dfrac{\partial T}{\partial x} = 2xy-y^3 , \dfrac{\partial T}{\partial y}=x^2-3y^2x , \dfrac{\partial x}{\partial r}= \cos \theta , \dfrac{\partial y}{\partial r} = \sin\theta$$

Substituting $$\dfrac{\partial T}{\partial r} = (2xy-y^3)\cos\theta + (x^2-3y^2x)\sin\theta$$

For $\dfrac{\partial T}{\partial \theta}$ I get the formula: $$\dfrac{\partial T}{\partial \theta}= \dfrac{\partial T}{\partial x}\cdot \dfrac{\partial x}{\partial \theta} + \dfrac{\partial T}{\partial y}\cdot \dfrac{\partial y}{\partial \theta}$$

Where

$$\dfrac{\partial x}{\partial \theta}= -r\sin\theta , \dfrac{\partial y}{\partial \theta} = r\cos\theta$$

Substituting

$$\dfrac{\partial T}{\partial \theta}=(2xy-y^3)(-r\sin\theta) + (x^2-3y^2x)(r\cos\theta)$$

Now here is my problem, when I substitute the original values back in for $x = r\cos\theta$ and $y=r\sin\theta$, it obviously becomes pretty messy and I can't compute it to what the answer in the text says. I imagine it's going to involve some identities but I'm just struggling to see how it can be simplified.

The answer in the textbook is

$\dfrac{\partial T}{\partial r} = 3r^2\sin\theta\cos^2\theta-4r^3\sin^3\theta\cos\theta$

$\dfrac{\partial T}{\partial \theta} = -2r^3\sin^2\theta\cos\theta+r^4\sin^4\theta + r^3cos^3\theta-3r^4sin^2\theta\cos^2\theta$

I need help actually computing this - so showing any steps used in simplifying would be of significant learning benefit to me, thankyou.

Answer 1

We could write, $$T=r^3\sin\theta \cos^2\theta-r^4\cos\theta\sin^3\theta+2$$ Now, $$\frac{\delta T}{\delta r}=3r^2\sin\theta \cos^2\theta-4r^3\cos\theta\sin^3\theta$$ And, $$\frac{\delta T}{\delta\theta}=r^3(-2\cos\theta \sin^2\theta+\cos^3\theta)-r^4(3\sin^2\theta\cos^2\theta-\sin^4\theta)$$ $$\frac{\delta T}{\delta\theta}=-2r^3\cos\theta \sin^2\theta+r^3\cos^3\theta-3r^4\sin^2\theta\cos^2\theta+r^4\sin^4\theta$$

It's just that.