Answer:
[tex]B^\prime(6) \approx -28.17[/tex]
Step-by-step explanation:
We have:
[tex]\displaystyle B(t)=24.6\sin(\frac{\pi t}{10})(8-t)[/tex]
And we want to find B’(6).
So, we will need to find B(t) first. To do so, we will take the derivative of both sides with respect to x. Hence:
[tex]\displaystyle B^\prime(t)=\frac{d}{dt}[24.6\sin(\frac{\pi t}{10})(8-t)][/tex]
We can move the constant outside:
[tex]\displaystyle B^\prime(t)=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)][/tex]
Now, we will utilize the product rule. The product rule is:
[tex](uv)^\prime=u^\prime v+u v^\prime[/tex]
We will let:
[tex]\displaystyle u=\sin(\frac{\pi t}{10})\text{ and } \\ \\ v=8-t[/tex]
Then:
[tex]\displaystyle u^\prime=\frac{\pi}{10}\cos(\frac{\pi t}{10})\text{ and } \\ \\ v^\prime= -1[/tex]
(The derivative of u was determined using the chain rule.)
Then it follows that:
[tex]\displaystyle \begin{aligned} B^\prime(t)&=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)] \\ \\ &=24.6[(\frac{\pi}{10}\cos(\frac{\pi t}{10}))(8-t) - \sin(\frac{\pi t}{10})] \end{aligned}[/tex]
Therefore:
[tex]\displaystyle B^\prime(6) =24.6[(\frac{\pi}{10}\cos(\frac{\pi (6)}{10}))(8-(6))- \sin(\frac{\pi (6)}{10})][/tex]
By simplification:
[tex]\displaystyle B^\prime(6)=24.6 [\frac{\pi}{10}\cos(\frac{3\pi}{5})(2)-\sin(\frac{3\pi}{5})] \approx -28.17[/tex]
So, the slope of the tangent line to the point (6, B(6)) is -28.17.
