Related rates finding the dA/dt.

  • Thread starter athamz
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  • #1
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Homework Statement


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.


Homework Equations





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?
 

Answers and Replies

  • #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
 
  • #3
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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] ?
 
  • #4
lanedance
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well you're given theta
 
  • #5
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oh.. thank you so much..
Is it alright if I will get a negative answer?
 
  • #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].
 
  • #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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