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Related rates finding the dA/dt.

  1. Jul 5, 2011 #1
    1. The problem statement, all variables and given/known data
    The measure of one of the acute angles of a right triangle is decreasing at the rate 1/36 pi rad/sec. If the length of the hypotenuse is constant at 40cm, find how fast the area is changing when the measure of the acute angle is 1/6 pi.


    2. Relevant equations



    3. The attempt at a solution
    I used the formula for the area of a triangle which is 1/2 base times the height, using trigonometry function, I substitute the value of base with 40costeta and the height is 40sinteta. How will I differentiated A = 40costeta*40sinteta? Will I use the product rule to differentiate? where will I include or put the rate given?
     
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  3. Jul 5, 2011 #2

    lanedance

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    so you have
    [tex] A(\theta) = 40 cos(\theta) sin(\theta) [/tex]

    differentiate both sides with respect to t, you will need to use both chain & product rules
     
  4. Jul 5, 2011 #3
    oh.. meaning

    [tex] dA/dt = (cos^2 [/theta] - sin^2 [/theta]) d[/theta]/dt [/tex]

    Where will I get the value for cos[/theta] and sin[/theta] ?
     
  5. Jul 5, 2011 #4

    lanedance

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    well you're given theta
     
  6. Jul 5, 2011 #5
    oh.. thank you so much..
    Is it alright if I will get a negative answer?
     
  7. Jul 5, 2011 #6

    Char. Limit

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    A note: This equation actually becomes very easy to differentiate if you recognize that [itex]40 sin(\theta) cos(\theta) = 20 sin(2 \theta)[/itex].
     
  8. Jul 6, 2011 #7

    lanedance

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    i haven't checked if it is the correct answer, but assuming you've done the maths correctly a negative answer is fine, it just means the area is decreasing
     
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