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So far I have figured out everything except the angle alpha of f,g. What I tried was I drew this triangle:

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and found side length c by doing dist(f,g) = ||f-g|| = sqrt() and I ended up with c = sqrt(54)

Then I used the law of cosines to find the angle of alpha and got 14.14 degrees but it was the wrong answer.

What is wrong with my approach?

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I'm still getting $\sqrt{54}$, $\int_{0}^{1}((5x^2-3)-(6x+3))*((5x^2-3)-(6x+3)) dx = 54$ then square root is $\sqrt{54}$... – Kyle V. Apr 9 '12 at 2:06
You’re right: I miscopied one of the functions. The problem is in your calculation of $\alpha$. I get $\cos\alpha=\frac{11}{4\sqrt{39}}\approx 0.44$, $\alpha\approx 1.1148$ radians (or about 63.87°). – Brian M. Scott Apr 9 '12 at 2:14
Those answers are incorrect, see my other comment below. – Kyle V. Apr 9 '12 at 2:55
I apologize: I seem to be brain-dead tonight. I dropped a minus sign: $\cos\alpha=\frac{-11/2}{2\sqrt{39}}=-\frac{11}{4\sqrt{39}}$. By the way, I’d expect it to want the angle in radians: that’s a nearly universal convention in mathematics. Out of curiosity, is this WeBWorK? – Brian M. Scott Apr 9 '12 at 3:11
$\frac{-11}{4\sqrt{39}}$ was the correct answer, and yes it is. – Kyle V. Apr 9 '12 at 15:46
up vote 1 down vote accepted

I believe the angle is arccos(| < f, g > | / (||f|| * ||g||)) So assuming your numbers are correct that would be 1.115r or ~64deg

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64 degrees is incorrect too :( – Kyle V. Apr 9 '12 at 2:12
Is this some sort of web page that auto checks it? If so how do you know what units to try and to what precision? 63.87deg? – Jared Kipe Apr 9 '12 at 2:29
Yes, but it does not specify which units. I'm assuming it wants degrees but just in case I tried radians too but it was also incorrect. As for precision it's usually very lenient so I'm assuming the answer 63.87 is just wrong. :/ EDIT: I passed it "arccos(11/(4*sqrt(39)))" but it still says it's incorrect so I assume it isn't an issue of precision. – Kyle V. Apr 9 '12 at 2:53
Ok well how sure are you about the 'other' data you provided? Does it check per field? – Jared Kipe Apr 9 '12 at 3:49
I'm not allowed to comment in the above thread, but you could try it with the negative. That yields 116deg. I thought it was abs on the top value though. – Jared Kipe Apr 9 '12 at 14:42

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