MATH105 April 2012
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Question 05 (b)

Now define
 $H(x,y)=\int _{y}^{x}\ln(2+\sin t)\,dt.$
Find the first partial derivatives of H and use this to compute $\displaystyle \nabla H\left({\frac {\pi }{2}},{\frac {\pi }{2}}\right)$.

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

Recall the definition of the gradient:
 $\nabla H(x,y)=\left\langle {\frac {\partial H}{\partial x}}(x,y),{\frac {\partial H}{\partial y}}(x,y)\right\rangle$
To use part (a), write H in terms of F and G.

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Solution

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We write H as a sum of F and G:
 $H(x,y)=\int _{y}^{0}\ln(2+\sin t)dt+\int _{0}^{x}\ln(2+\sin t)dt=G(y)+F(x)$.
Then
 $\displaystyle H_{x}={\frac {\partial }{\partial x}}(F(x)+G(y))=F'(x)+0=\ln(2+\sin(x))$
and similarly
 $\displaystyle H_{y}={\frac {\partial }{\partial y}}(F(x)+G(y))=0+G'(y)=\ln(2+\sin(y))$
where we used the knowledge about the derivatives F'(x) and G'(y) from part (a). Finally, all that is left to do is to plug in the given values to find that
 $\nabla H\left({\frac {\pi }{2}},{\frac {\pi }{2}}\right)=\left\langle F'\left({\frac {\pi }{2}}\right),G'\left({\frac {\pi }{2}}\right)\right\rangle =\langle \ln 3,\ln 3\rangle$

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