Given:
The power generated by an electrical circuit (in watts) as a function of its current x (in amperes) is modeled by
To find:
The current which will produce the maximum power.
Solution:
We have,
Differentiate with respect to x.
...(i)
To find the extreme point equate P'(x)=0.
Divide both sides by -30.
Differentiate (i) with respect to x.
(Maximum)
It means, the given function is maximum at x=4.
Therefore, the current of 4 amperes will produce the maximum power.
Answer: 3 and 4
Step-by-step explanation: I took the quiz
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Answer:
Step-by-step explanation:
We have:
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:
![\displaystyle B^\prime(t)=\frac{d}{dt}[24.6\sin(\frac{\pi t}{10})(8-t)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20B%5E%5Cprime%28t%29%3D%5Cfrac%7Bd%7D%7Bdt%7D%5B24.6%5Csin%28%5Cfrac%7B%5Cpi%20t%7D%7B10%7D%29%288-t%29%5D)
We can move the constant outside:
![\displaystyle B^\prime(t)=24.6\frac{d}{dt}[\sin(\frac{\pi t}{10})(8-t)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20B%5E%5Cprime%28t%29%3D24.6%5Cfrac%7Bd%7D%7Bdt%7D%5B%5Csin%28%5Cfrac%7B%5Cpi%20t%7D%7B10%7D%29%288-t%29%5D)
Now, we will utilize the product rule. The product rule is:
We will let:
Then:
(The derivative of u was determined using the chain rule.)
Then it follows that:
![\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}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%20B%5E%5Cprime%28t%29%26%3D24.6%5Cfrac%7Bd%7D%7Bdt%7D%5B%5Csin%28%5Cfrac%7B%5Cpi%20t%7D%7B10%7D%29%288-t%29%5D%20%5C%5C%20%5C%5C%20%26%3D24.6%5B%28%5Cfrac%7B%5Cpi%7D%7B10%7D%5Ccos%28%5Cfrac%7B%5Cpi%20t%7D%7B10%7D%29%29%288-t%29%20-%20%5Csin%28%5Cfrac%7B%5Cpi%20t%7D%7B10%7D%29%5D%20%5Cend%7Baligned%7D)
Therefore:
![\displaystyle B^\prime(6) =24.6[(\frac{\pi}{10}\cos(\frac{\pi (6)}{10}))(8-(6))- \sin(\frac{\pi (6)}{10})]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20B%5E%5Cprime%286%29%20%3D24.6%5B%28%5Cfrac%7B%5Cpi%7D%7B10%7D%5Ccos%28%5Cfrac%7B%5Cpi%20%286%29%7D%7B10%7D%29%29%288-%286%29%29-%20%5Csin%28%5Cfrac%7B%5Cpi%20%286%29%7D%7B10%7D%29%5D)
By simplification:
![\displaystyle B^\prime(6)=24.6 [\frac{\pi}{10}\cos(\frac{3\pi}{5})(2)-\sin(\frac{3\pi}{5})] \approx -28.17](https://tex.z-dn.net/?f=%5Cdisplaystyle%20B%5E%5Cprime%286%29%3D24.6%20%5B%5Cfrac%7B%5Cpi%7D%7B10%7D%5Ccos%28%5Cfrac%7B3%5Cpi%7D%7B5%7D%29%282%29-%5Csin%28%5Cfrac%7B3%5Cpi%7D%7B5%7D%29%5D%20%5Capprox%20-28.17)
So, the slope of the tangent line to the point (6, B(6)) is -28.17.
0.8 is 20% of 4. This is because 20% of something occurs when something is divided by 5, since it is 1/5 of 100. You multiply 0.8*5, and you get 4.
