1. Rate of Change with radians

This confused me a little bit:

Suppose that side a is increasing at a rate of 1.5 meters/second and c is fixed at 8 meters.
i) Find the rate of change of the measure of angle A when b=4 meters.
ii) Find the rate of change of the measure of angle B when b=4 meters.

2. We first write everything in mathematical notation:

$\frac{da}{dt}=1.5\;\;\;\;\;\;\;\;\;\;c=8 \;\;\;\;\;\;\;\;\;\;\sin A=\cos B=\frac{a}{c}\;\;\;\;\;\;\;\;\;\;\cos A=\sin B=\frac{b}{c}$

Now we may use the Chain Rule, remembering that $a$, $b$, and $A$ are all functions of $t$.

3. Would the answer be sin(sqrt(48)/8) * 1.5? For the first one

4. Originally Posted by Velvet Love
Would the answer be sin(sqrt(48)/8) * 1.5? For the first one
For the answer to the first, we may differentiate both sides of $\sin A=\frac{a}{c}$:

\begin{aligned}
\frac{d}{dt}(\sin A)&=\frac{d}{dt}\left(\frac{a}{c}\right)\\
\cos A\,\frac{dA}{dt}&=\frac{1}{c}\frac{da}{dt}=\frac{1 .5}{8}.
\end{aligned}

Now we may solve for $\frac{dA}{dt}$ knowing the value of $\cos A$.

5. so cosA is cos(sqrt(48)/8)

6. Not quite. From the definition of cosine, we know that

$\cos A=\frac{b}{c}$.

7. Oh ok. I think i got it now.
It should be (1.5/8)/(cos(4/8)) right?

8. Originally Posted by Velvet Love
Oh ok. I think i got it now.
It should be (1.5/8)/(cos(4/8)) right?
Not quite. We know that $c=8$ from the statement of the problem. When $b=4$, what is $\frac{b}{c}$?

Here is the next step:


\begin{aligned}
\cos A\,\frac{dA}{dt}&=\frac{1.5}{8}\\
\frac{b}{c} \frac{dA}{dt}&=\frac{1.5}{8}\\
\frac{c}{b}\cdot\frac{b}{c}\frac{dA}{dt}&=\frac{c} {b}\cdot\frac{1.5}{8}\\
\frac{dA}{dt}&=\frac{c}{b}\cdot\frac{1.5}{8}.
\end{aligned}

9. 4/8 or 1/2

10. If the first answer is (3/8), then the second one is -(3/8) right?

11. Originally Posted by Velvet Love
4/8 or 1/2
Correct. The value of $\frac{dA}{dt}$ is therefore

$\frac{dA}{dt}=\frac{2}{1}\cdot\frac{1.5}{8}.$