Science:Math Exam Resources/Courses/MATH105/April 2011/Question 09 (g)

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MATH105 April 2011

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Question 09 (g)
Each of the short-answer questions below is worth 5 points. Put your answer in the box provided and show your work. No credit will be given for the answer without the correct accompanying work. You are given a function $f$ satisfying $f(1,0)=0,$ $f_{x}(1,0)=3,$ and $f_{y}(1,0)=-1.$ Find $z'(0)$ if $\displaystyle {z(t)=f(e^{3t},\sin(2t))}.$

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 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
How does the chain rule works when dealing with functions of several variables?

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Solution

The function f depends on the variables x and y which themselves are functions of the variable t. We have $\displaystyle z(t)=f(x(t),y(t))\quad {\text{and}}\quad x(t)=e^{3t}\quad {\text{and}}\quad y(t)=\sin(2t)$ Now, using the chain rule for multivariable functions we obtain ${\frac {dz}{dt}}(t)={\frac {\partial f}{\partial x}}(x(t),y(t)){\frac {dx}{dt}}(t)+{\frac {\partial f}{\partial y}}(x(t),y(t)){\frac {dy}{dt}}(t)$ which using the notation as in the question statement gives $\displaystyle z'(t)=f_{X}(x(t),y(t))\,x'(t)+f_{Y}\left(x(t),y(t)\right)\,y'(t)$ Now we just want to know what's going on at t = 0. We have $\displaystyle x(0)=e^{0}=1\quad {\text{and}}\quad y(0)=\sin(0)=0$ and since $\displaystyle x'(t)=3e^{3t}\quad {\text{and}}\quad y'(t)=2\cos(2t)$ we have $\displaystyle x'(0)=3\quad {\text{and}}\quad y'(0)=2$ So we can conclude that $\displaystyle {\begin{aligned}z'(0)&=f_{X}(1,0)\cdot 3+f_{Y}(1,0)\cdot 2\\&=3\cdot 3+(-1)\cdot 2\\&=7\end{aligned}}$. Notice that the provided information for $f(1,0)=0$ was unnecessary.

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