Answer:
Problem 20)

Problem 21)
A)
The velocity function is:

The acceleration function is:

B)

Step-by-step explanation:
Problem 20)
We want to differentiate the equation:

We can take the natural log of both sides. This yields:

Since ln(aᵇ) = bln(a):

Take the derivative of both sides with respect to <em>x: </em>
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Implicitly differentiate the left and use the product rule on the right. Therefore:

Simplify:

Simplify and multiply both sides by <em>y: </em>
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Since <em>y</em> = (cos x)ˣ:
Problem 21)
We are given the position function of a particle:

A)
Recall that the velocity function is the derivative of the position function. Hence:
![\displaystyle v(t)=s'(t)=\frac{d}{dt}[\sin(2\pi t)+2\cos(\pi t)]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20v%28t%29%3Ds%27%28t%29%3D%5Cfrac%7Bd%7D%7Bdt%7D%5B%5Csin%282%5Cpi%20t%29%2B2%5Ccos%28%5Cpi%20t%29%5D)
Differentiate:

The acceleration function is the derivative of the velocity function. Hence:
![\displaystyle a(t)=v'(t)=\frac{d}{dt}[2\pi(\cos(2\pi t)-\sin(\pi t))]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20a%28t%29%3Dv%27%28t%29%3D%5Cfrac%7Bd%7D%7Bdt%7D%5B2%5Cpi%28%5Ccos%282%5Cpi%20t%29-%5Csin%28%5Cpi%20t%29%29%5D)
Differentiate:
![\displaystyle \begin{aligned} a(t)&=2\pi[-2\pi\sin(2\pi t)-\pi\cos(\pi t)]\\&=-2\pi^2(2\sin(2\pi t)+\cos(\pi t))\end{aligned}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cbegin%7Baligned%7D%20a%28t%29%26%3D2%5Cpi%5B-2%5Cpi%5Csin%282%5Cpi%20t%29-%5Cpi%5Ccos%28%5Cpi%20t%29%5D%5C%5C%26%3D-2%5Cpi%5E2%282%5Csin%282%5Cpi%20t%29%2B%5Ccos%28%5Cpi%20t%29%29%5Cend%7Baligned%7D)
B)
The position at <em>t</em> = 0 will be:

The velocity at <em>t</em> = 0 will be:

And the acceleration at <em>t</em> = 0 will be:
