Science:Math Exam Resources/Courses/MATH200/April 2012/Question 03

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MATH200 April 2012

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  •  April 2012  •  December 2011  •  

Question 03
Let $\displaystyle z=f(x,y)$, where $\displaystyle f(x,y)$ has continuous second order partial derivatives, and ${\begin{aligned}x&=2t^{2},&y&=t^{3},&&\\f_{x}(2,1)&=5,&f_{y}(2,1)&=-2,&&\\f_{xx}(2,1)&=2,&f_{xy}(2,1)&=1,&f_{yy}(2,1)&=-4\end{aligned}}$ Find ${\frac {{\textrm {d}}^{2}z}{{\textrm {d}}t^{2}}}$ when t=1.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
To evaluate the derivative of a multivariable function g = g(x,y) with respect to t where x,y are functions of t, we need to use the chain rule for multivariable functions: ${\frac {dg}{dt}}={\frac {\partial g}{\partial x}}{\frac {dx}{dt}}+{\frac {\partial g}{\partial y}}{\frac {dy}{dt}}=g_{x}{\frac {dx}{dt}}+g_{y}{\frac {dy}{dt}}.$

Hint 2
After applying the multivariable chain rule to ƒ once, you will need to apply it again to dƒ/dt since the partial derivatives of ƒ are also functions of x, y. You will also need the product rule.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

First we evaluate the first derivative of z with respect to t using the multivariable chain rule. ${\begin{aligned}{\frac {dz}{dt}}&=f_{x}{\frac {dx}{dt}}+f_{y}{\frac {dy}{dt}}&=f_{x}\cdot 4t+f_{y}\cdot 3t^{2}\end{aligned}}$ We then take the next derivative of z by taking the derivative with respect to t of the above result, using the multivariable chain rule again as well as the product rule. ${\begin{aligned}{\frac {d^{2}z}{dt^{2}}}&={\frac {d}{dt}}\left(f_{x}\cdot 4t+f_{y}\cdot 3t^{2}\right)\\&=4t(f_{xx}\cdot 4t+f_{xy}\cdot 3t^{2})+4f_{x}+3t^{2}(f_{yx}\cdot 4t+f_{yy}3t^{2})+6t\cdot f_{y}\end{aligned}}$ Next, we recognize that if t = 1, then (x,y) = (2,1). Thus, we can use the information given in the question about the partial derivatives of f to evaluate $d^{2}z/dt^{2}$ at t = 1: ${\begin{aligned}{\frac {d^{2}z}{dt^{2}}}&=4t(f_{xx}\cdot 4t+f_{xy}\cdot 3t^{2})+4f_{x}+3t^{2}(f_{yx}\cdot 4t+f_{yy}3t^{2})+6t\cdot f_{y}\\{\frac {d^{2}z}{dt^{2}}}{\Big \vert }_{t=1}&=4(2(4)+3)+4(5)+3(4-4(3))+6(-2)\\&=4(11)+20+3(-8)-12=28\end{aligned}}$ Therefore, ${\color {blue}{\frac {d^{2}z}{dt^{2}}}{\Big \vert }_{t=1}=28}.$

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