Triangles, Rates, Angles, oh my!

[animalplanet]

An isosceles triangle has equal sides 6 inches long.
If the angle, theta, between the equal sides is changing at a rate of 2 degrees per minute,
how fast is the area of the triangle changing when theta = 30 degrees?

A problem for a chapter on Related Rates. Please help! I would very much appreciate it

 

[mmm4444bot]

Start by writing a function that inputs theta and outputs the area.

 

[soroban]

Hello, animalplanet!

We must be careful.
Calculus requires radians . . . not degrees.

An isosceles triangle has equal sides 6 inches long.
If the angle \(\displaystyle \theta\), between the equal sides, is changing at a rate of \(\displaystyle 2^o\) per minute,
how fast is the area of the triangle changing when \(\displaystyle \theta = 30^o\)?

[SIZE=3]
            C
            *
           * *
          * θ *
       6 *     * 6
        *       *
       *         *
    A *  *  *  *  * B[/SIZE]

Formula: .\(\displaystyle A \;=\;\frac{1}{2}ab\sin C\)

The area of a triangle equals one-half the product of two sides and the sine of the included angle.


We have: .\(\displaystyle A \;=\;\frac{1}{2}(6^2)\sin\theta \;=\;18\sin\theta\)

Differentiate with respect to time: .\(\displaystyle \frac{dA}{dt} \;=\;18\cos\theta\frac{d\theta}{dt}\) .[1]


Rate of change of the angle: .\(\displaystyle \frac{d\theta}{dt} \:=\:2^o\text{/min} \:=\:\frac{\pi}{90}\text{ radians/min} \)

The angle is: .\(\displaystyle \theta \:=\:30^o \:=\:\frac{\pi}{6}\text{ radians}\)


Substitute into [1]: .\(\displaystyle \frac{dA}{dt} \;=\;18\cos\left(\frac{\pi}{6}\right)\cdot \frac{\pi}{90} \;=\; 18\cdot\frac{\sqrt{3}}{2}\cdot\frac{\pi}{90} \)

Therefore: .\(\displaystyle \dfrac{dA}{dt}\;=\;\frac{\sqrt{3}}{10}\!\pi\; \text{in}^2\!\text{/min} \)

 

[animalplanet]

Thank you so much! I really appreciate the guidance you've given me. I have more of these problems in a packet but now that you've provided a sort of guideline of steps to follow I hopefully won't have much trouble now. Thanks!