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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?

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  1. anonymous
    • one year ago
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    |dw:1433732518490:dw|

  2. anonymous
    • one year ago
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    and then? what follows? im confused on the given values

  3. ganeshie8
    • one year ago
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    familiar with vectors ?

  4. anonymous
    • one year ago
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    i already forgot that sir..

  5. ganeshie8
    • one year ago
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    it would be easy if we use vectors

  6. anonymous
    • one year ago
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    how sir?

  7. anonymous
    • one year ago
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    How fast was boat A going?

  8. anonymous
    • one year ago
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    its not given sir. @SithsAndGiggles

  9. anonymous
    • one year ago
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    |dw:1433771683115:dw|

  10. anonymous
    • one year ago
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    |dw:1433771801885:dw|

  11. anonymous
    • one year ago
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    You can use the law of cosines to set up an equation relating \(z\) to \(x\) and \(y\). \[z^2=x^2+y^2-2xy\cos75^\circ\]

  12. anonymous
    • one year ago
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    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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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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