QUESTION IMAGE
Question
- compute all of the first and second order partial derivatives for the following functions which are related to polar coordinates. verify clairaut’s theorem. simplify when reasonable.
(a) (x(r,\theta)=rcos\theta).
(b) (y(r,\theta)=rsin\theta).
(c) (r(x,y)=sqrt{x^{2}+y^{2}}).
(d) (\theta(x,y)=arctan(y / x)).
Step1: First - order partial derivatives of \(x(r,\theta)=r\cos\theta\)
- \(\frac{\partial x}{\partial r}=\cos\theta\) (Treat \(\theta\) as a constant)
- \(\frac{\partial x}{\partial\theta}=-r\sin\theta\) (Treat \(r\) as a constant)
Step2: Second - order partial derivatives of \(x(r,\theta)\)
- \(\frac{\partial^{2}x}{\partial r^{2}} = 0\) (Differentiate \(\frac{\partial x}{\partial r}=\cos\theta\) with respect to \(r\))
- \(\frac{\partial^{2}x}{\partial\theta\partial r}=-\sin\theta\) (Differentiate \(\frac{\partial x}{\partial r}=\cos\theta\) with respect to \(\theta\))
- \(\frac{\partial^{2}x}{\partial r\partial\theta}=-\sin\theta\) (Differentiate \(\frac{\partial x}{\partial\theta}=-r\sin\theta\) with respect to \(r\))
- \(\frac{\partial^{2}x}{\partial\theta^{2}}=-r\cos\theta\) (Differentiate \(\frac{\partial x}{\partial\theta}=-r\sin\theta\) with respect to \(\theta\))
Clairaut's Theorem is verified since \(\frac{\partial^{2}x}{\partial\theta\partial r}=\frac{\partial^{2}x}{\partial r\partial\theta}=-\sin\theta\)
Step3: First - order partial derivatives of \(y(r,\theta)=r\sin\theta\)
- \(\frac{\partial y}{\partial r}=\sin\theta\) (Treat \(\theta\) as a constant)
- \(\frac{\partial y}{\partial\theta}=r\cos\theta\) (Treat \(r\) as a constant)
Step4: Second - order partial derivatives of \(y(r,\theta)\)
- \(\frac{\partial^{2}y}{\partial r^{2}} = 0\) (Differentiate \(\frac{\partial y}{\partial r}=\sin\theta\) with respect to \(r\))
- \(\frac{\partial^{2}y}{\partial\theta\partial r}=\cos\theta\) (Differentiate \(\frac{\partial y}{\partial r}=\sin\theta\) with respect to \(\theta\))
- \(\frac{\partial^{2}y}{\partial r\partial\theta}=\cos\theta\) (Differentiate \(\frac{\partial y}{\partial\theta}=r\cos\theta\) with respect to \(r\))
- \(\frac{\partial^{2}y}{\partial\theta^{2}}=-r\sin\theta\) (Differentiate \(\frac{\partial y}{\partial\theta}=r\cos\theta\) with respect to \(\theta\))
Clairaut's Theorem is verified since \(\frac{\partial^{2}y}{\partial\theta\partial r}=\frac{\partial^{2}y}{\partial r\partial\theta}=\cos\theta\)
Step5: First - order partial derivatives of \(r(x,y)=\sqrt{x^{2}+y^{2}}=(x^{2}+y^{2})^{\frac{1}{2}}\)
- \(\frac{\partial r}{\partial x}=\frac{x}{\sqrt{x^{2}+y^{2}}}\) (Using the chain - rule: \(\frac{\partial r}{\partial x}=\frac{1}{2}(x^{2}+y^{2})^{-\frac{1}{2}}\cdot2x\))
- \(\frac{\partial r}{\partial y}=\frac{y}{\sqrt{x^{2}+y^{2}}}\) (Using the chain - rule: \(\frac{\partial r}{\partial y}=\frac{1}{2}(x^{2}+y^{2})^{-\frac{1}{2}}\cdot2y\))
Step6: Second - order partial derivatives of \(r(x,y)\)
- \(\frac{\partial^{2}r}{\partial x^{2}}=\frac{\sqrt{x^{2}+y^{2}}-\frac{x^{2}}{\sqrt{x^{2}+y^{2}}}}{x^{2}+y^{2}}=\frac{y^{2}}{(x^{2}+y^{2})^{\frac{3}{2}}}\)
- \(\frac{\partial^{2}r}{\partial y\partial x}=\frac{-xy}{(x^{2}+y^{2})^{\frac{3}{2}}}\)
- \(\frac{\partial^{2}r}{\partial x\partial y}=\frac{-xy}{(x^{2}+y^{2})^{\frac{3}{2}}}\)
- \(\frac{\partial^{2}r}{\partial y^{2}}=\frac{x^{2}}{(x^{2}+y^{2})^{\frac{3}{2}}}\)
Clairaut's Theorem is verified since \(\frac{\partial^{2}r}{\partial y\partial x}=\frac{\partial^{2}r}{\partial x\partial y}=\frac{-xy}{(x^{2}+y^{2})^{\frac{3}{2}}}\)
Step7: First - order partial derivatives of \(\theta(x,y)=\arctan(\frac{y}{x})\)
- \(\frac{\partial\theta}{\partial x}=\frac{-y}{x^{2}+y^{2}}\) (Using the formula for the derivative of \(\arctan(u)\) where \(u = \frac{y}{x}\) and the quotient - rule)
- \(\frac{\partial\theta}{\partial y}=\frac{x}{x^{2}+y^{2}}\)
Step8: Second - order partial derivatives of \(\theta(x,y)\)
- \(\frac{\partial^{2}\theta}{\partial x^{2}}=\frac…
Snap & solve any problem in the app
Get step-by-step solutions on Sovi AI
Photo-based solutions with guided steps
Explore more problems and detailed explanations
Step1: First - order partial derivatives of \(x(r,\theta)=r\cos\theta\)
- \(\frac{\partial x}{\partial r}=\cos\theta\) (Treat \(\theta\) as a constant)
- \(\frac{\partial x}{\partial\theta}=-r\sin\theta\) (Treat \(r\) as a constant)
Step2: Second - order partial derivatives of \(x(r,\theta)\)
- \(\frac{\partial^{2}x}{\partial r^{2}} = 0\) (Differentiate \(\frac{\partial x}{\partial r}=\cos\theta\) with respect to \(r\))
- \(\frac{\partial^{2}x}{\partial\theta\partial r}=-\sin\theta\) (Differentiate \(\frac{\partial x}{\partial r}=\cos\theta\) with respect to \(\theta\))
- \(\frac{\partial^{2}x}{\partial r\partial\theta}=-\sin\theta\) (Differentiate \(\frac{\partial x}{\partial\theta}=-r\sin\theta\) with respect to \(r\))
- \(\frac{\partial^{2}x}{\partial\theta^{2}}=-r\cos\theta\) (Differentiate \(\frac{\partial x}{\partial\theta}=-r\sin\theta\) with respect to \(\theta\))
Clairaut's Theorem is verified since \(\frac{\partial^{2}x}{\partial\theta\partial r}=\frac{\partial^{2}x}{\partial r\partial\theta}=-\sin\theta\)
Step3: First - order partial derivatives of \(y(r,\theta)=r\sin\theta\)
- \(\frac{\partial y}{\partial r}=\sin\theta\) (Treat \(\theta\) as a constant)
- \(\frac{\partial y}{\partial\theta}=r\cos\theta\) (Treat \(r\) as a constant)
Step4: Second - order partial derivatives of \(y(r,\theta)\)
- \(\frac{\partial^{2}y}{\partial r^{2}} = 0\) (Differentiate \(\frac{\partial y}{\partial r}=\sin\theta\) with respect to \(r\))
- \(\frac{\partial^{2}y}{\partial\theta\partial r}=\cos\theta\) (Differentiate \(\frac{\partial y}{\partial r}=\sin\theta\) with respect to \(\theta\))
- \(\frac{\partial^{2}y}{\partial r\partial\theta}=\cos\theta\) (Differentiate \(\frac{\partial y}{\partial\theta}=r\cos\theta\) with respect to \(r\))
- \(\frac{\partial^{2}y}{\partial\theta^{2}}=-r\sin\theta\) (Differentiate \(\frac{\partial y}{\partial\theta}=r\cos\theta\) with respect to \(\theta\))
Clairaut's Theorem is verified since \(\frac{\partial^{2}y}{\partial\theta\partial r}=\frac{\partial^{2}y}{\partial r\partial\theta}=\cos\theta\)
Step5: First - order partial derivatives of \(r(x,y)=\sqrt{x^{2}+y^{2}}=(x^{2}+y^{2})^{\frac{1}{2}}\)
- \(\frac{\partial r}{\partial x}=\frac{x}{\sqrt{x^{2}+y^{2}}}\) (Using the chain - rule: \(\frac{\partial r}{\partial x}=\frac{1}{2}(x^{2}+y^{2})^{-\frac{1}{2}}\cdot2x\))
- \(\frac{\partial r}{\partial y}=\frac{y}{\sqrt{x^{2}+y^{2}}}\) (Using the chain - rule: \(\frac{\partial r}{\partial y}=\frac{1}{2}(x^{2}+y^{2})^{-\frac{1}{2}}\cdot2y\))
Step6: Second - order partial derivatives of \(r(x,y)\)
- \(\frac{\partial^{2}r}{\partial x^{2}}=\frac{\sqrt{x^{2}+y^{2}}-\frac{x^{2}}{\sqrt{x^{2}+y^{2}}}}{x^{2}+y^{2}}=\frac{y^{2}}{(x^{2}+y^{2})^{\frac{3}{2}}}\)
- \(\frac{\partial^{2}r}{\partial y\partial x}=\frac{-xy}{(x^{2}+y^{2})^{\frac{3}{2}}}\)
- \(\frac{\partial^{2}r}{\partial x\partial y}=\frac{-xy}{(x^{2}+y^{2})^{\frac{3}{2}}}\)
- \(\frac{\partial^{2}r}{\partial y^{2}}=\frac{x^{2}}{(x^{2}+y^{2})^{\frac{3}{2}}}\)
Clairaut's Theorem is verified since \(\frac{\partial^{2}r}{\partial y\partial x}=\frac{\partial^{2}r}{\partial x\partial y}=\frac{-xy}{(x^{2}+y^{2})^{\frac{3}{2}}}\)
Step7: First - order partial derivatives of \(\theta(x,y)=\arctan(\frac{y}{x})\)
- \(\frac{\partial\theta}{\partial x}=\frac{-y}{x^{2}+y^{2}}\) (Using the formula for the derivative of \(\arctan(u)\) where \(u = \frac{y}{x}\) and the quotient - rule)
- \(\frac{\partial\theta}{\partial y}=\frac{x}{x^{2}+y^{2}}\)
Step8: Second - order partial derivatives of \(\theta(x,y)\)
- \(\frac{\partial^{2}\theta}{\partial x^{2}}=\frac{2xy}{(x^{2}+y^{2})^{2}}\)
- \(\frac{\partial^{2}\theta}{\partial y\partial x}=\frac{y^{2}-x^{2}}{(x^{2}+y^{2})^{2}}\)
- \(\frac{\partial^{2}\theta}{\partial x\partial y}=\frac{y^{2}-x^{2}}{(x^{2}+y^{2})^{2}}\)
- \(\frac{\partial^{2}\theta}{\partial y^{2}}=\frac{-2xy}{(x^{2}+y^{2})^{2}}\)
Clairaut's Theorem is verified since \(\frac{\partial^{2}\theta}{\partial y\partial x}=\frac{\partial^{2}\theta}{\partial x\partial y}=\frac{y^{2}-x^{2}}{(x^{2}+y^{2})^{2}}\)
For \(x(r,\theta)\):
- First - order: \(\frac{\partial x}{\partial r}=\cos\theta,\frac{\partial x}{\partial\theta}=-r\sin\theta\)
- Second - order: \(\frac{\partial^{2}x}{\partial r^{2}} = 0,\frac{\partial^{2}x}{\partial\theta\partial r}=-\sin\theta,\frac{\partial^{2}x}{\partial r\partial\theta}=-\sin\theta,\frac{\partial^{2}x}{\partial\theta^{2}}=-r\cos\theta\)
For \(y(r,\theta)\):
- First - order: \(\frac{\partial y}{\partial r}=\sin\theta,\frac{\partial y}{\partial\theta}=r\cos\theta\)
- Second - order: \(\frac{\partial^{2}y}{\partial r^{2}} = 0,\frac{\partial^{2}y}{\partial\theta\partial r}=\cos\theta,\frac{\partial^{2}y}{\partial r\partial\theta}=\cos\theta,\frac{\partial^{2}y}{\partial\theta^{2}}=-r\sin\theta\)
For \(r(x,y)\):
- First - order: \(\frac{\partial r}{\partial x}=\frac{x}{\sqrt{x^{2}+y^{2}}},\frac{\partial r}{\partial y}=\frac{y}{\sqrt{x^{2}+y^{2}}}\)
- Second - order: \(\frac{\partial^{2}r}{\partial x^{2}}=\frac{y^{2}}{(x^{2}+y^{2})^{\frac{3}{2}}},\frac{\partial^{2}r}{\partial y\partial x}=\frac{-xy}{(x^{2}+y^{2})^{\frac{3}{2}}},\frac{\partial^{2}r}{\partial x\partial y}=\frac{-xy}{(x^{2}+y^{2})^{\frac{3}{2}}},\frac{\partial^{2}r}{\partial y^{2}}=\frac{x^{2}}{(x^{2}+y^{2})^{\frac{3}{2}}}\)
For \(\theta(x,y)\):
- First - order: \(\frac{\partial\theta}{\partial x}=\frac{-y}{x^{2}+y^{2}},\frac{\partial\theta}{\partial y}=\frac{x}{x^{2}+y^{2}}\)
- Second - order: \(\frac{\partial^{2}\theta}{\partial x^{2}}=\frac{2xy}{(x^{2}+y^{2})^{2}},\frac{\partial^{2}\theta}{\partial y\partial x}=\frac{y^{2}-x^{2}}{(x^{2}+y^{2})^{2}},\frac{\partial^{2}\theta}{\partial x\partial y}=\frac{y^{2}-x^{2}}{(x^{2}+y^{2})^{2}},\frac{\partial^{2}\theta}{\partial y^{2}}=\frac{-2xy}{(x^{2}+y^{2})^{2}}\)