Science:Math Exam Resources/Courses/MATH200/December 2011/Question 02 (b)

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  •  

   Other MATH200 Exams

  •  April 2012  •  December 2011  •  

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

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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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