SOLVED Help with stationary points for a function with in integral included

Aug 2010
7
0
I need help solving the following problem. I have already worked it out, but I am getting stuck on the partial derivative relative to y because I am getting a negative square root somewhere along the way...

Here is the problem: classify the stationary points of:
f(x,y) = -2x + 3y + [ (integral) upperbound = y , lowerbound = x ] e^t^2 dt

I hope it makes sense as written...

Thank you.
 

Ackbeet

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Jun 2010
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Just to clarify: you're asked to classify the stationary points of

\(\displaystyle \displaystyle{f(x,y)=-2x+3y+\int_{x}^{y}e^{t^{2}}\,dt.}\)

Is that correct? If so, what's your partial w.r.t. \(\displaystyle x\), and what's your partial w.r.t. \(\displaystyle y\)?
 
Aug 2010
7
0
That is correct. My partial relative to x is: -2 + e^x^2; I was able to get the 0 of fx as : x= sqrt(ln 2)

Relative to y: 3+ e^y^2. I get the 0 of fy as: y = sqrt(ln (-3))

What I attempted to do is get the 0s for the first derivatives relative to x and to y. Then get the second derivatives to form the Hessian and test for the points obtained to classify them as maxima or minima, etc... When getting the derivatives, our prof instructed/suggested that we do not integrate first. I guess that would be the long way, the desperate way...

By the way, I'm new here. How did you get the function input in that form?

Thank you.
 

Ackbeet

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Jun 2010
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Your x partial derivative is incorrect. Because x is the lower limit, there's a minus sign associated with it (just think Fundamental Theorem of the Calculus Part II:

\(\displaystyle \displaystyle{\int_{a}^{b}f'(x)\,dx=f(b)-f(a).}\)

To get equations to look like this, you have to use LaTeX. You can find out more about that here in the stickies.

So, your x partial derivative is

\(\displaystyle \displaystyle{f_{x}(x,y)=-2-e^{x^{2}},}\) and the \(\displaystyle y\) partial is, as you've said,

\(\displaystyle \displaystyle{f_{y}(x,y)=3+e^{y^{2}}.}\)

Query: if you set either of these equal to zero, are there any real solutions? Your "solution" unfortunately has you taking the logarithm of a negative number. Negative numbers are not in the domain of the logarithm function, unless you're thinking a different branch of the complex logarithm function. You're going to run into this problem for both derivatives. What does this tell you?
 

Ackbeet

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Jun 2010
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Incidentally, for these functions, your instructor is correct. Don't integrate first! There are no known elementary antiderivatives for those functions.
 
Aug 2010
7
0
Thank you! That was an oversight for the lower limit. I forgot about that. Indeed I was running into the log of a negative number. My conclusion: "no solutions", therefore no stationary points??? I think that is my final answer!
 

Ackbeet

MHF Hall of Honor
Jun 2010
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CT, USA
I agree: no stationary points.