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

 

Lecture 1/28/21

 

Lecture 1/26/21

 

10.4#1

 Vector a <-1,-3,5> Vector b <0, 0, -1> Did the cross product and got <3-0, 1-0, 0-0> = <3, 1, 0> My y-value is apparently wrong. Did I multiply incorrectly? I thought it would have been j(az-cx)

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Looks like you figured it out.  You were forgetting about the - sign in the template 

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10.4#5

The question wants me to find two vectors orthogonal to <5,3,3> and <1,-1,-3>. I first tried the cross-product only to get the vectors <6,-18,8> and it's opposite <-6,18,-8> but after some research I had to include 1/sqrt(6^2+(-18)^2+8^2) = 1/sqrt(424) and multiply it by the vectors so they would come out as 1/sqrt(424)<6,-18,8> and to it's negative counterpart. Why did I have to include the magnitude of the new vector to get the exact answer?

 

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See that part I circled in red on the problem?  You need to divide by the magnitude of the new vector to get a unit vector.

The right hand rule ( section 10.4)

Dear Dr. Taylor

In the picture I provided, I am confused as to how to solve as this does not give any number to calculate. I also searched online and it states to use the right hand rule in which I am having trouble understanding how to know which finger points which direction? 

Regards,