Question

Please help me with solving these. Thank you very much. Have a great day!

Answer 1

Problem 20)

`\displaystyle (dy)/(dx)=(\cos x)^x\left(\ln \cos x-x\tan x\right)`

Problem 21)

A)

The velocity function is:

`\displaystyle v(t) =2\pi(\cos(2\pi t)-\sin(\pi t))`

The acceleration function is:

`\displaystyle a(t)=-2\pi^2(2\sin(2\pi t)+\cos(\pi t))`

B)

`s(0)=2\text{, }v(0) = 2\pi \text{ m/s}\text{, and } a(0) = -2\pi^2\text{ m/s$^2$}`

Explanation:

Problem 20)

We want to differentiate the equation:

`\displaystyle y=\left(\cos x\right)^x`

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

`\displaystyle \ln y = \ln((\cos x)^x)`

Since ln(aᵇ) = bln(a):

`\displaystyle \ln y =x\ln \cos x`

Take the derivative of both sides with respect to x:

`\displaystyle (d)/(dx)\left[\ln y \right]=(d)/(dx)\left[x \ln \cos x\right]`

Implicitly differentiate the left and use the product rule on the right. Therefore:

`\displaystyle (1)/(y)(dy)/(dx)=\ln \cos x+x\left((1)/(\cos x)\cdot -\sin(x)\right)`

Simplify:

`\displaystyle (1)/(y)(dy)/(dx)=\ln \cos x-(x\sin x)/(\cos x)`

Simplify and multiply both sides by y:

`\displaystyle (dy)/(dx)=y\left(\ln \cos x-x \tan x\right)`

Since y = (cos x)ˣ:

`\displaystyle (dy)/(dx)=(\cos x)^x\left(\ln \cos x-x\tan x\right)`

Problem 21)

We are given the position function of a particle:

`\displaystyle s(t)= \sin (2\pi t)+2\cos(\pi t)`

A)

Recall that the velocity function is the derivative of the position function. Hence:

`\displaystyle v(t)=s'(t)=(d)/(dt)[\sin(2\pi t)+2\cos(\pi t)]`

Differentiate:

`\displaystyle \begin{aligned} v(t) &= 2\pi \cos(2\pi t)-2\pi \sin(\pi t)\\&=2\pi(\cos(2\pi t)-\sin(\pi t))\end{aligned}`

The acceleration function is the derivative of the velocity function. Hence:

`\displaystyle a(t)=v'(t)=(d)/(dt)[2\pi(\cos(2\pi t)-\sin(\pi t))]`

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}`

B)

The position at t = 0 will be:

`\displaystyle \begin{aligned} s(0)&=\sin(2\pi(0))+2\cos(\pi(0))\\&=\sin(0)+2\cos(0)\\&=(1)+2(1)\\&=2\end{aligned}`

The velocity at t = 0 will be:

`\displaystyle \begin{aligned} v(0)&=2\pi(\cos(2\pi (0)-\sin(\pi(0))\\&=2\pi(\cos(0)-\sin(0))\\&=2\pi((1)-(0))\\&=2\pi \text{ m/s}\end{aligned}`

And the acceleration at t = 0 will be:

`\displaystyle \begin{aligned} a(0) &= -2\pi ^2(2\sin(2\pi(0))+\cos(\pi(0)) \\ & = -2\pi ^2(2\sin(0)+\cos(0)) \\ &= -2\pi ^2(2(0)+(1)) \\ &= -2\pi^2(1) \\ &= -2\pi^2\text{ m/s$^2$} \end{aligned}`