The velocity function of the car is equal to the first derivative of the position function of the car, and is equal to

\(\displaystyle -\sin(x)e^{11x}+11\cos(x)e^{11x}-\csc(x)\cot(x)\)

The derivative was found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u \frac{\mathrm{du} }{\mathrm{d} x}\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)\),  \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \csc(x)=-\csc(x)\cot(x)\)

In order to solve for the first and second derivative, we must use the chain rule.

The chain rule states that if

\(\displaystyle y=f(u)\) 

and 

\(\displaystyle u=g(x)\)

then the derivative is

\(\displaystyle \frac{dy}{dx}=\frac{dy}{du}*\frac{du}{dx}\)

In order to find the first derviative of the function

\(\displaystyle f(x)=e^{\pi x}\)

we set

\(\displaystyle y=e^u\)

and

\(\displaystyle u=\pi x\)

Because the derivative of the exponential function is the exponential function itself, we get

\(\displaystyle \frac{dy}{du}=\frac{d}{du}[e^u]\)

\(\displaystyle \frac{dy}{du}=e^u\)

And differentiating \(\displaystyle u\) we use the power rule which states

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}\)

As such

\(\displaystyle \frac{du}{dx}=\frac{\mathrm{d} }{\mathrm{d} x}[\pi x]\)

\(\displaystyle \frac{du}{dx}=\pi x^{1-1}=\pi\)

And so

\(\displaystyle \frac{dy}{dx}=e^{\pi x}*\pi=\pi e^{\pi x}\)

\(\displaystyle f'(x)=\pi e^{\pi x}\)

To solve for the second derivative we set 

\(\displaystyle y=\pi e^{u}\)

and 

\(\displaystyle u=\pi x\)

Because the derivative of the exponential function is the exponential function itself, we get

\(\displaystyle \frac{dy}{du}=\frac{d}{du}[\pi e^u]\)

\(\displaystyle \frac{dy}{du}=\pi e^u\)

And differentiating \(\displaystyle u\) we use the power rule which states

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}\)

As such

\(\displaystyle \frac{du}{dx}=\frac{\mathrm{d} }{\mathrm{d} x}[\pi x]\)

\(\displaystyle \frac{du}{dx}=\pi x^{1-1}=\pi\)

And so the second derivative becomes

\(\displaystyle \frac{dy}{dx}=\pi e^{\pi x}*\pi=\pi^2 e^{\pi x}\)

\(\displaystyle f''(x)=\pi^2e^{\pi x}\)

The velocity function is given by the first derivative of the position function:

\(\displaystyle v(t)=s'(t)=e^t\tan(t)+e^t\sec^2(t)+5t^4\)

and was found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \tan(t)=\sec^2(t)\), \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x\)

We must find the first and second derivatives.

We use the properties that

• The derivative of  \(\displaystyle sec(x)\)  is  \(\displaystyle sec(x)tan(x)\)
• The derivative of \(\displaystyle tan(x)\)  is  \(\displaystyle sec^2(x)\)

As such

\(\displaystyle f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[sec(x)]=sec(x)tan(x)\)

To find the second derivative we differentiate again and use the product rule which states

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}[uv]=uv'+vu'\)

Setting

\(\displaystyle u=sec(x)\)  and  \(\displaystyle v=tan(x)\)

we find that

\(\displaystyle f''(x)=\frac{\mathrm{d} }{\mathrm{d} x}[sec(x)tan(x)]=sec(x)*sec^2(x)+tan(x)*sec(x)tan(x)\)

\(\displaystyle =sec^3(x)+sec(x)tan^2(x)\)

As such

\(\displaystyle f'(x)=sec(x)tan(x)\)

\(\displaystyle f''(x)=sec^3(x)+sec(x)tan^2(x)\)

We can find the acceleration function \(\displaystyle a(x)\) from the velocity function by taking the derivative:

\(\displaystyle a(x)=v'(x)\)

We can view the function

\(\displaystyle v(x)=\int_{0}^{x^2+2x+1} z^d\sin(z^5) dz\)

as the composition of the following functions

 \(\displaystyle g(x)=\int_{0}^{x} z^d\sin(z^5) dz\)

\(\displaystyle h(x)=x^2+2x+1=(x+1)^2\)

so that \(\displaystyle v(x)=g(h(x))\). This means we use the chain rule

\(\displaystyle [g(h(x))]'=g'(h(x))h'(x)\)

to find the derivative. We have \(\displaystyle g'(x)=x^d\sin x^5\) and \(\displaystyle h'(x)=2x+2\), so we have

\(\displaystyle a(x)=v'(x)=[(x+1)^2]^d\sin [(x+1)^2]^5\cdot (2x+2)\)

\(\displaystyle =(2x+2)(x+1)^{2d}\sin (x+1)^{10}\)

If this function gives the position, the first derivative will give its speed and the second derivative will give its acceleration. 

\(\displaystyle y ' = t^2 - t -2\)

\(\displaystyle y'' = 2t - 1\)

Now plug in 2 for t: 

\(\displaystyle y'' = 2(2) - 1 = 4- 1 = 3\)

If this function gives the position, the first derivative will give its speed and the second derivative will give its acceleration. 

\(\displaystyle y' = 2 t^3 +10\)

\(\displaystyle y'' = 6t^2\)

Plug in 3 for t: 

\(\displaystyle 6(3)^2 = 6(9) = 54\)

If this function gives the position, the first derivative will give its speed.

\(\displaystyle y' = \cos (2t ) \cdot 2\)

Plug in \(\displaystyle \pi\) for t: 

\(\displaystyle 2 \cos (2 \pi ) = 2\cdot 1 = 2\)

If this function gives the position, the first derivative will give its speed. To differentiate, use the chain rule: \(\displaystyle f(g(x)) ' = f'(g(x))\cdot g'(x)\). In this case, \(\displaystyle f(x) = \frac{1}{2} x^2\) and \(\displaystyle g(x) = \cos x\). Since \(\displaystyle f'(x) = x\) and \(\displaystyle g'(x ) = - \sin (x)\), the first derivative is \(\displaystyle \cos t \cdot - \sin t\). 

Plug in \(\displaystyle \frac{\pi}{4}\) for t: 

\(\displaystyle \cos (\frac{\pi}{4} ) \cdot - \sin (\frac{\pi}{4} ) = \frac{1}{\sqrt{2}} \cdot - \frac{1}{\sqrt2} = -\frac{1}{2}\)

To find when the particle changes direction, we need to find the critical values of \(\displaystyle f(t)\). This is done by finding the velocity function, setting it equal to \(\displaystyle 0\), and solving for \(\displaystyle t.\)

\(\displaystyle 0=v(t) = f'(t) = \cos(t)-1\).

Hence \(\displaystyle \cos(t)=1\).

The solutions to this on the unit circle are \(\displaystyle t =0, \pi\), so these are the values of \(\displaystyle t\) where the particle would normally change direction. However, our given interval is \(\displaystyle 0 < t < \pi\), which does not contain \(\displaystyle 0, \pi\). Hence the particle does not change direction on the given interval.