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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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?

Mathematics
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and then? what follows? im confused on the given values
familiar with vectors ?

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i already forgot that sir..
it would be easy if we use vectors
how sir?
How fast was boat A going?
its not given sir. @SithsAndGiggles
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You can use the law of cosines to set up an equation relating \(z\) to \(x\) and \(y\). \[z^2=x^2+y^2-2xy\cos75^\circ\]
Differentiating 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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