11.6#14

How do I find a generalized "slope" here? Is it related to the idea of gradient?


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There's no generalized "slope". There's just a slope in a direction. It looks like you figured it out though, after pulling the trigger on this question.

Lecture Notes 2/25/19 through 2/27/19

Lecture Notes 2/25/19

Lecture Notes 2/27/19

Lecture Notes 2/18/19 though 2/20/19

Lecture Notes 2/18/19

Lecture Notes 2/20/19

your scores through 2/18/19

Lecture Notes 1/30/19 through 2/13/19

Lecture Notes 1/30/19

Lecture Notes 2/4/19

Lecture Notes 2/6/19

Lecture Notes 2/11/19

Lecture Notes 2/13/19

Exam Review Questions

Exam Review Questions

6.3#11

Mr. Taylor I have tried this problem many times and the answer seems to come up wrong every time. I even checked my answer on Wolfram Alpha and it still came up wrong. I think there is a problem with webwork.

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Ok the only mistake that you're making is that you forgot (and I forgot to mention in class) is that
∫ (1/x) dx is *NOT* exactly ln(x) + C, because ln(x) is only defined for positive values of x, but (1/x) has an antiderivative for all x≠0.
It is more proper to say that ∫ (1/x) dx = ln(|x|) + C, and this works for all x≠0.

Section 6.2 #10

Hi Dr. Taylor,

I have been comparing these indefinite integrals to the known trig substitution rules and I cannot seem to get them to match. What is the trick?

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Well, for example in the second problem, since you have -x^2, your trig substitution would have to be a sin(u), and since you have the 49, you would have to use x=7sin(u). Then dx = 7 cos(u)du and you'd substitute those. The second problem will rearrange tan^2(x) +1 = sec^2(x) to make it sec^2(x)-1=tan^2(x), so that you'd substitute x=7sec(u) and dx = 7sec(u)tan(u)du.  Btw, the textbook section for this part of the homework has a thorough treatment of this. I recommend it.