Jenna P. answered • 03/04/15
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If you draw the right triangle, you can find the relationship between the altitude a, and the angle theta is tantheta=a/4. Take the derivative of both sides noting theta is dependent on t. You can then solve for dtheta/dt in terms of theta and t. But you are given that t=1 and from this you can also find theta at t=1. Plug in and solve...does this give you enough to setup the problem?
Jenna P.
tutor
I wish I could show you a photo, would be much easier to explain. But so if the relationship between the altitude and theta is clear then the math follows like this:
tan(theta)=20t2/4=5t2
sec2(theta) dtheta/dt=10t
then dtheta/dt=10t/sec2(theta)
t=1
theta =arctan(20t^2/4)
can you take it from there?
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03/04/15
Raj K.
ok then what do I do with the derivative we found? can you just finish it because I'm really struggling on this one and want to see all the steps!
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03/04/15
Jenna P.
tutor
The derivative is what the problem is asking you to find. You just plug in the values of t and theta. :)
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03/04/15
Raj K.
So if I plug in 1 for t=1 you get arctan(5) correct and that's the answer?
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03/04/15
Jenna P.
tutor
Yes, then plug arctan in for theta and solve for dtheta/dt. This link will help you. It has a bunch of problems with solutions. One of them is verry similar to the problem you are looking at. Maybe it can help you understand how to set them up.
http://www.barrington220.org/cms/lib2/IL01001296/Centricity/ModuleInstance/10306/Related%20Rates%20Packet.pdf
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03/04/15
Jenna P.
tutor
yes, then plug that value is for theta to get a value for dtheta/dt. This link has a bunch of worked out problems, one of which is nearly the same problem you are looking at. Hopefully it can help you to see how to set them up.
http://www.barrington220.org/cms/lib2/IL01001296/Centricity/ModuleInstance/10306/Related%20Rates%20Packet.pdf
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03/04/15
Raj K.
03/04/15