[Math] Related Rates Triangle Problem

calculusderivatives

Part of a homework problem I have for my Calc class.
In a right triangle with base = 10, height = h, and the angle across from the hight = theta, theta is increasing at a rate of 15/26 rad/min

At what rate is the area of the triangle changing when h = 24.

  1. 338/4
  2. 39
  3. 195/4
  4. 182
  5. 195

What I tried to do:

Area=(1/2)(base)(height)*sin(theta)

A=.5*10*24*sin(theta)

A=120*sin(theta)

(dA/dt)=120cos(theta)*(d(theta)/dt)

Then I did the inverse tangent of 24/10 to get the radians of theta when h=24. This means theta = 1.176 when h=24

Then I substituted in 1.176 for theta, 15/26 for (d(theta)/dt) and solved for (dA/dt)

I got (dA/dt)=26.63

What did I do wrong??

Best Answer

The area is just $\frac 12$base*height. There should not be a factor $\sin \theta$ You need to write $h=$ some function of $\theta$, then take the derivative of the area with respect to $\theta$ You can calculate $\theta$ from the fact that $h=24$. Then you are given $\frac {d\theta}{dt}$

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