Question

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

Answer 1

Answer:

Problem 20)

$\displaystyle \frac{dy}{dx}=(\cos x)^x\left(\ln \cos x-x\tan x\right)$

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)

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

Step-by-step 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:

$\displaystyle \frac{d}{dx}\left[\ln y \right]=\frac{d}{dx}\left[x \ln \cos x\right]$

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

$\displaystyle \frac{1}{y}\frac{dy}{dx}=\ln \cos x+x\left(\frac{1}{\cos x}\cdot -\sin(x)\right)$

Simplify:

$\displaystyle \frac{1}{y}\frac{dy}{dx}=\ln \cos x-\frac{x\sin x}{\cos x}$

Simplify and multiply both sides by y:

$\displaystyle \frac{dy}{dx}=y\left(\ln \cos x-x \tan x\right)$

Since y = (cos x)ˣ:

$\displaystyle \frac{dy}{dx}=(\cos x)^x\left(\ln \cos x-x\tan x\right)$

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:

$\displaystyle v(t)=s'(t)=\frac{d}{dt}[\sin(2\pi t)+2\cos(\pi t)]$

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:

$\displaystyle a(t)=v'(t)=\frac{d}{dt}[2\pi(\cos(2\pi t)-\sin(\pi t))]$

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:

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