Science:Math Exam Resources/Courses/MATH200/December 2011/Question 02 (b)/Solution 1

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:

${\color {blue}{\frac {\partial ^{2}z}{\partial t^{2}}}=r^{2}\sin ^{2}(t){\frac {\partial ^{2}f}{\partial x^{2}}}+r^{2}\cos ^{2}(t){\frac {\partial ^{2}f}{\partial y^{2}}}-2r^{2}\cos(t)\sin(t){\frac {\partial ^{2}f}{\partial y\partial x}}-r\sin(t){\frac {\partial f}{\partial y}}-r\cos(t){\frac {\partial f}{\partial x}}}$