Lecture 2/18/20

 

Lecture 2/16/21

 

Lecture 2/9/21

 

10.9#4 (edited)

I do not understand what the problem wants. I already took the derivative of r(t) and proceeded to take the magnitude of r'(t)=v(t). Then I was given a function in which I solved for t and got 2/7, which is not accepting that answer

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Not to contradict you, but you actually do seem to have a good understanding about what the problem wants.  The reason it's not accepting the answer is because it's the wrong answer.  I can't tell you exactly why, because you didn't send me your calculation. It looks the same way that the right answer does though, so my experienced guess is that you're making a simple algebra or arithmetic mistake.  

By the way, here's a trick: instead of finding the minimum of the speed ||v(t)||, find the minimum of the speed squared ||v(t)||^2=v(t).v(t).  It will give you the same answer but you don't have to deal with the square root (and dealing with the square root is a unnecessary complication that can cause mistakes).

Oh, ok, here we go. You just added an extra ^2 where there was none:

never mind I found the problem. I added a t^(2) to the y component of r(t). Thank you anyway for making me realize that I had a computation error

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You're welcome.  Yes most of the time people make these minor mistakes, and because they're so little they're easy to overlook.

Lecture 2/4/21

 

10.7#3

Hey professor, Question wants me to evaluate the limit of the vector as t approaches 0. <(e^(-2t)-1)/t,t^2/(t^3-t^2),3/(17+t)>. I know part of the equation is right and that’s is <x,y,3/17>. I try evaluating the x and y limits and using L’Hospital’s Rule I get the total <0,0,3/17>. The derivative doesn’t work either. How do I evaluate this equation? Best regards,

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L'Hospital's rule is the right way to go for x(t), but you did it incorrectly--so fixing that is the first step of what you need to do. You could use L'Hospital's rule for y(t) too, but you'd have to use it a couple of times because the numerator and denominator both vanish to second order--BUT--that would be too much work because you could just cancel t^2 from the numerator and denominator and that would make getting the limit easy

Lecture 2/2/21