MATH200 December 2011
• Q1 (a) • Q1 (b) • Q1 (c) • Q2 (a) • Q2 (b) • Q3 (a) • Q3 (b) • Q4 • Q5 (a) • Q5 (b) • Q6 • Q7 • Q8 (a) • Q8 (b) i • Q8 (b) ii • Q8 (b) iii •
Question 02 (b)

Suppose z = f(x,y) has continuous second order partial derivatives and x = r cos(t), y = r sin(t). Express the following partial derivatives in terms of r,t and the partial derivatives of f.
b)
 ${\frac {\partial ^{2}z}{\partial t^{2}}}$

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If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

You should use your answer from part (a) to solve this problem:
 ${\frac {\partial ^{2}z}{\partial t^{2}}}={\frac {\partial }{\partial t}}\left({\frac {\partial z}{\partial t}}\right).$

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Solution

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To evaluate this derivative we merely need to recognize that we apply the first derivative operator to our answer from part (a):
 ${\frac {\partial ^{2}z}{\partial t^{2}}}={\frac {\partial }{\partial t}}\left({\frac {\partial z}{\partial t}}\right).$
We need to apply the chain rule again as in part (a):
 ${\begin{aligned}{\frac {\partial ^{2}z}{\partial t^{2}}}={\frac {\partial }{\partial t}}\left({\frac {\partial z}{\partial t}}\right)&={\frac {\partial }{\partial x}}\left({\frac {\partial z}{\partial t}}\right){\frac {\partial x}{\partial t}}+{\frac {\partial }{\partial y}}\left({\frac {\partial z}{\partial t}}\right){\frac {\partial y}{\partial t}}\\&={\frac {\partial }{\partial x}}\left({\frac {\partial z}{\partial t}}\right)\cdot (r\sin(t))+{\frac {\partial }{\partial y}}\left({\frac {\partial z}{\partial t}}\right)\cdot r\cos(t)\\&=y{\frac {\partial }{\partial x}}\left({\frac {\partial z}{\partial t}}\right)+x{\frac {\partial }{\partial y}}\left({\frac {\partial z}{\partial t}}\right).\end{aligned}}$
where we have use the fact that x = r cos(t), y = r sin(t) to write the last equation above. From part (a), we found that
 ${\begin{aligned}{\frac {\partial z}{\partial t}}&={\frac {\partial f}{\partial x}}\cdot (r\sin(t))+{\frac {\partial f}{\partial y}}\cdot r\cos(t)\\&=y{\frac {\partial f}{\partial x}}+x{\frac {\partial f}{\partial y}}.\end{aligned}}$
Using this, we continue to evaluate the partial derivatives above
 ${\begin{aligned}{\frac {\partial }{\partial x}}\left({\frac {\partial z}{\partial t}}\right)&=y{\frac {\partial ^{2}f}{\partial x^{2}}}+{\frac {\partial f}{\partial y}}+x{\frac {\partial ^{2}f}{\partial y\partial x}}\\{\frac {\partial }{\partial y}}\left({\frac {\partial z}{\partial t}}\right)&={\frac {\partial f}{\partial x}}y{\frac {\partial ^{2}f}{\partial y\partial x}}+x{\frac {\partial ^{2}f}{\partial y^{2}}}\end{aligned}}$
finally giving us
 ${\begin{aligned}{\frac {\partial ^{2}z}{\partial t^{2}}}&=y{\frac {\partial }{\partial x}}\left({\frac {\partial z}{\partial t}}\right)+x{\frac {\partial }{\partial y}}\left({\frac {\partial z}{\partial t}}\right)\\&=y\left(y{\frac {\partial ^{2}f}{\partial x^{2}}}+{\frac {\partial f}{\partial y}}+x{\frac {\partial ^{2}f}{\partial y\partial x}}\right)+x\left({\frac {\partial f}{\partial x}}y{\frac {\partial ^{2}f}{\partial y\partial x}}+x{\frac {\partial ^{2}f}{\partial y^{2}}}\right)\\&=y^{2}{\frac {\partial ^{2}f}{\partial x^{2}}}+x^{2}{\frac {\partial ^{2}f}{\partial y^{2}}}2xy{\frac {\partial ^{2}f}{\partial y\partial x}}y{\frac {\partial f}{\partial y}}x{\frac {\partial f}{\partial x}}\end{aligned}}$
Writing the result above in terms of r, t gives the final answer:
 ${\color {blue}{\frac {\partial ^{2}z}{\partial t^{2}}}=r^{2}\sin ^{2}(t){\frac {\partial ^{2}f}{\partial x^{2}}}+r^{2}\cos ^{2}(t){\frac {\partial ^{2}f}{\partial y^{2}}}2r^{2}\cos(t)\sin(t){\frac {\partial ^{2}f}{\partial y\partial x}}r\sin(t){\frac {\partial f}{\partial y}}r\cos(t){\frac {\partial f}{\partial x}}}$

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