Chapter 2.10, Problem 22E

To determine

To calculate: The derivative of the function $p\left(t\right)=e^t\left(1+0.2\cos \left(t\right)\right)$ also the value of the function and slope at the points $0,\frac{\pi }{2}\text{ and }\pi$ . Sketch and describe the graph of the function on the domain $0\le t\le 40$ .

Answer to Problem 22E

The derivative of the function is $p\text{'}\left(t\right)=0.2e^t\left(\cos t-\sin t\right)+e^t$ , value and slope of the function at the points $0,\frac{\pi }{2}\text{ and }\pi$ are, $p\left(0\right)=1.2,p\text{'}\left(0\right)=1.2,p\left(\frac{\pi }{2}\right)=e^{\left(\frac{\pi }{2}\right)},p\text{'}\left(\frac{\pi }{2}\right)=0.8e^{\left(\frac{\pi }{2}\right)},p\left(\pi \right)=0.8e^{\pi }\text{ and}p\text{'}\left(\pi \right)=0.8e^{\pi }$.

The graph of the function is provided below,

  

The behavior of the function is interpreted as, the function is an increasing function.

Explanation of Solution

Given information:

The function $p\left(t\right)=e^t\left(1+0.2\cos \left(t\right)\right)$ and the points $0,\frac{\pi }{2}\text{ and }\pi$ .

Formula used:

The slope of a function at a given point is the derivative of the function at that point.

Product rule of differentiation $\frac{d}{dx}f\left(x\right)\cdot g\left(x\right)=f\left(x\right)\frac{d}{dx}g\left(x\right)+g\left(x\right)\frac{d}{dx}f\left(x\right)$ .

Calculation:

Consider the function $p\left(t\right)=e^t\left(1+0.2\cos \left(t\right)\right)$ .

Differentiate the function with respect to t ,

Recall product rule of differentiation $\frac{d}{dx}f\left(x\right)\cdot g\left(x\right)=f\left(x\right)\frac{d}{dx}g\left(x\right)+g\left(x\right)\frac{d}{dx}f\left(x\right)$ .

Apply it,

  $\begin{array}{c}\frac{d}{dt}s\left(t\right)=\frac{d}{dt}\left(e^t\left(1+0.2\cos \left(t\right)\right)\right)\\ =e^t\frac{d}{dt}\left(1+0.2\cos t\right)+\left(1+0.2\cos t\right)\frac{d}{dt}\left(e^t\right)\\ =-e^{t}0.2\sin t+\left(1+0.2\cos t\right)e^t\\ =0.2e^t\left(\cos t-\sin t\right)+e^t\end{array}$

Therefore, derivative of the function of the function is $p\text{'}\left(t\right)=0.2e^t\left(\cos t-\sin t\right)+e^t$ .

Recall that the slope of a function at a given point is the derivative of the function at that point.

Evaluate the value of the function and slope at the points $0,\frac{\pi }{2}\text{ and }\pi$ .

At the point $t=0$ ,

Substitute $t=0$ to evaluate the value of the function $p\left(t\right)=e^t\left(1+0.2\cos \left(t\right)\right)$ ,

  $\begin{array}{c}p\left(0\right)=e^0\left(1+0.2\cos \left(0\right)\right)\\ =1\left(1+0.2\right)\\ =1.2\end{array}$

Substitute $t=0$ in $p\text{'}\left(t\right)=0.2e^t\left(\cos t-\sin t\right)+e^t$ to evaluate the slope of the function $p\left(t\right)=e^t\left(1+0.2\cos \left(t\right)\right)$ ,

  $\begin{array}{c}p\text{'}\left(0\right)=0.2e^{\left(0\right)}\left(\cos \left(0\right)-\sin \left(0\right)\right)+e^{\left(0\right)}\\ =0.2\left(1-0\right)+1\\ =1.2\end{array}$

At the point $t=\frac{\pi }{2}$ ,

Substitute $t=\frac{\pi }{2}$ to evaluate the value of the function $p\left(t\right)=e^t\left(1+0.2\cos \left(t\right)\right)$ ,

  $\begin{array}{c}p\left(\frac{\pi }{2}\right)=e^{\frac{\pi }{2}}\left(1+0.2\cos \left(\frac{\pi }{2}\right)\right)\\ =e^{\frac{\pi }{2}}\end{array}$

Substitute $t=\frac{\pi }{2}$ in $p\text{'}\left(t\right)=0.2e^t\left(\cos t-\sin t\right)+e^t$ to evaluate the slope of the function $p\left(t\right)=e^t\left(1+0.2\cos \left(t\right)\right)$ ,

  $\begin{array}{c}p\text{'}\left(\frac{\pi }{2}\right)=0.2e^{\left(\frac{\pi }{2}\right)}\left(\cos \left(\frac{\pi }{2}\right)-\sin \left(\frac{\pi }{2}\right)\right)+e^{\left(\frac{\pi }{2}\right)}\\ =0.2e^{\left(\frac{\pi }{2}\right)}\left(0-1\right)+e^{\left(\frac{\pi }{2}\right)}\\ =0.8e^{\left(\frac{\pi }{2}\right)}\end{array}$

At the point $t=\pi$ ,

Substitute $t=\pi$ to evaluate the value of the function $p\left(t\right)=e^t\left(1+0.2\cos \left(t\right)\right)$ ,

  $\begin{array}{c}p\left(\pi \right)=e^{\left(\pi \right)}\left(1+0.2\cos \left(\pi \right)\right)\\ =e^{\left(\pi \right)}\left(1-0.2\right)\\ =0.8e^{\left(\pi \right)}\end{array}$

Substitute $t=\pi$ in $p\text{'}\left(t\right)=0.2e^t\left(\cos t-\sin t\right)+e^t$ to evaluate the slope of the function $p\left(t\right)=e^t\left(1+0.2\cos \left(t\right)\right)$ ,

  $\begin{array}{c}p\text{'}\left(\pi \right)=0.2e^{\left(\pi \right)}\left(\cos \left(\pi \right)-\sin \left(\pi \right)\right)+e^{\left(\pi \right)}\\ =0.2e^{\left(\pi \right)}\left(-1-0\right)+e^{\left(\pi \right)}\\ =0.8e^{\pi }\end{array}$

The graph of the function is provided below,

  

The behavior of the function is interpreted as, the functionis an increasing function.

Thus, derivative of the function is $p\text{'}\left(t\right)=0.2e^t\left(\cos t-\sin t\right)+e^t$ , value and slope of the function at the points $0,\frac{\pi }{2}\text{ and }\pi$ are, $p\left(0\right)=1.2,p\text{'}\left(0\right)=1.2,p\left(\frac{\pi }{2}\right)=e^{\left(\frac{\pi }{2}\right)},p\text{'}\left(\frac{\pi }{2}\right)=0.8e^{\left(\frac{\pi }{2}\right)},p\left(\pi \right)=0.8e^{\pi }\text{ and}p\text{'}\left(\pi \right)=0.8e^{\pi }$ .