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anonymous
 one year ago
related rate problem:
i got the figure but i am confused on how it goes..
At 4:00Pm boat A left in the direction N 45degrees E. At 4:30 PM boat B left the same pier in the direction S 30 degrees E at 32mph. How fast were they separating at 5:00 Pm in mph?
anonymous
 one year ago
related rate problem: i got the figure but i am confused on how it goes.. At 4:00Pm boat A left in the direction N 45degrees E. At 4:30 PM boat B left the same pier in the direction S 30 degrees E at 32mph. How fast were they separating at 5:00 Pm in mph?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433732518490:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and then? what follows? im confused on the given values

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0familiar with vectors ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i already forgot that sir..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0it would be easy if we use vectors

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How fast was boat A going?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its not given sir. @SithsAndGiggles

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433771683115:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1433771801885:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You can use the law of cosines to set up an equation relating \(z\) to \(x\) and \(y\). \[z^2=x^2+y^22xy\cos75^\circ\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Differentiating with respect to \(t\), you have \[2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}2y\cos75^\circ\frac{dx}{dt}2x\cos75^\circ\frac{dy}{dt}\] We already know that \(\dfrac{dy}{dt}=32\), but we're not given \(\dfrac{dx}{dt}\), which I'm simply setting to be \(k\). We'll need to find \(z\), which isn't too hard because we have an equation for that. Once you have everything you need, you can solve for \(\dfrac{dz}{dt}\).
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