anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
|dw:1433732518490:dw|
anonymous
  • anonymous
and then? what follows? im confused on the given values
ganeshie8
  • ganeshie8
familiar with vectors ?

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anonymous
  • anonymous
i already forgot that sir..
ganeshie8
  • ganeshie8
it would be easy if we use vectors
anonymous
  • anonymous
how sir?
anonymous
  • anonymous
How fast was boat A going?
anonymous
  • anonymous
its not given sir. @SithsAndGiggles
anonymous
  • anonymous
|dw:1433771683115:dw|
anonymous
  • anonymous
|dw:1433771801885:dw|
anonymous
  • anonymous
You can use the law of cosines to set up an equation relating \(z\) to \(x\) and \(y\). \[z^2=x^2+y^2-2xy\cos75^\circ\]
anonymous
  • anonymous
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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