Chapter 3.5, Problem 50E

a.

To determine

To graph: The function $f(h)=\frac{\sin (h)}{h}$ and estimate ${\lim }_{h\to 0}f(h)$ and then to compare the estimate with $\frac{\pi }{180}$ and then to state if there is a reason to believe that limit should be $\frac{\pi }{180}$ .

Answer to Problem 50E

Estimate is equal to $\frac{\pi }{180}$ .

Explanation of Solution

Given Information: $\underset{h\to 0}{\lim }\frac{\sin (h)}{h}$ is to be estimated.

Graph:

  

Observing the graph, it can inferred that the limit is near about $0.0175$ , comparing with $\frac{\pi }{180}$ , it is found that the estimate is equal to $\frac{\pi }{180}$ . It is so because $\frac{\pi }{180}$ is the conversion factor for converting degrees to radians.

b.

To determine

To estimate: $\underset{h\to 0}{\lim }\frac{\cos (h)-1}{h}$ using graphing utility in degree mode.

Answer to Problem 50E

  $\underset{h\to 0}{\lim }\frac{\cos (h)-1}{h}=0$

Explanation of Solution

Given Information: $\underset{h\to 0}{\lim }\frac{\cos (h)-1}{h}$ is to be estimated.

Graph:

  

Observing the graph it can be easily estimated that $\underset{h\to 0}{\lim }\frac{\cos (h)-1}{h}=0$ .

c.

To determine

The formula obtained for the derivative of $\sin x$ .

Answer to Problem 50E

The formula is $\frac{d}{dx}(\sin x)=\frac{\pi }{180}(\cos x)$ .

Explanation of Solution

Given Information: Degree mode limits are to be used.

Calculation:

  $\begin{array}{l}\frac{d}{dx}(\sin x)=\underset{h\to 0}{\lim }\frac{\sin (x+h)-\sin x}{h}\\ \\ =\underset{h\to 0}{\lim }\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}\\ \\ =\underset{h\to 0}{\lim }\frac{\sin x(\cos h-1)+\cos x\sin h}{h}\\ \\ =\left(\underset{h\to 0}{\lim }\sin x\right)\left(\underset{h\to 0}{\lim }\frac{\cos h-1}{h}\right)+\left(\underset{h\to 0}{\lim }\cos x\right)\left(\underset{h\to 0}{\lim }\frac{\sin h}{h}\right)\\ \\ =(\sin x)(0)+(\cos x)\left(\frac{\pi }{180}\right)\\ \\ =\frac{\pi }{180}(\cos x)\end{array}$

Thus, the formula is $\frac{d}{dx}(\sin x)=\frac{\pi }{180}(\cos x)$ .

d.

To determine

The formula obtained for the derivative of $\cos x$ .

Answer to Problem 50E

The formula is $\frac{d}{dx}(\cos x)=-\frac{\pi }{180}(\sin x)$ .

Explanation of Solution

Given Information: Degree mode limits are to be used.

Calculation:

  $\begin{array}{l}\frac{d}{dx}(\cos x)=\underset{h\to 0}{\lim }\frac{\cos (x+h)-\cos x}{h}\\ \\ =\underset{h\to 0}{\lim }\frac{\cos x\cos h-\sin x\sin h-\cos x}{h}\\ \\ =\underset{h\to 0}{\lim }\frac{\cos x(\cos h-1)-\sin x\sin h}{h}\\ \\ =\left(\underset{h\to 0}{\lim }\cos x\right)\left(\underset{h\to 0}{\lim }\frac{\cos h-1}{h}\right)-\left(\underset{h\to 0}{\lim }\sin x\right)\left(\underset{h\to 0}{\lim }\frac{\sin h}{h}\right)\\ \\ =(\cos x)(0)-(\sin x)\left(\frac{\pi }{180}\right)\\ \\ =-\frac{\pi }{180}(\sin x)\end{array}$

Thus, the formula is $\frac{d}{dx}(\cos x)=-\frac{\pi }{180}(\sin x)$ .

e.

To determine

To calculate: The second and third degree-mode derivatives of $\sin x$ and $\cos x$ .

Answer to Problem 50E

The derivatives are $\frac{d^2}{d{x}^2}\sin x=-{\left(\frac{\pi }{180}\right)}^2\sin x,\frac{d^3}{d{x}^3}\sin x=-\frac{{\pi }^3}{{180}^3}\cos x$ and $\frac{d^2}{d{x}^2}\cos x=-{\left(\frac{\pi }{180}\right)}^2\cos x,\frac{d^3}{d{x}^3}\cos x=\frac{{\pi }^3}{{180}^3}\sin x$ .

Explanation of Solution

Given Information: Degree mode derivatives are to be calculated.

Calculation:

  $\begin{array}{l}\frac{d^2}{d{x}^2}\sin x=\frac{d}{dx}(\frac{\pi }{180}\cos x)\\ \\ =\frac{\pi }{180}\left(-\frac{\pi }{180}\sin x\right)\\ \\ =-{\left(\frac{\pi }{180}\right)}^2\sin x\\ \\ \frac{d^3}{d{x}^3}\sin x=\frac{d}{dx}\left[-{\left(\frac{\pi }{180}\right)}^2\sin x\right]\\ \\ =-\frac{{\pi }^2}{{180}^2}\left(\frac{\pi }{180}\cos x\right)\\ \\ =-\frac{{\pi }^3}{{180}^3}\cos x\end{array}$

  $\begin{array}{l}\frac{d^2}{d{x}^2}\cos x=\frac{d}{dx}(-\frac{\pi }{180}\sin x)\\ \\ =-\frac{\pi }{180}\left(\frac{\pi }{180}\cos x\right)\\ \\ =-{\left(\frac{\pi }{180}\right)}^2\cos x\end{array}$

  $\begin{array}{l}\frac{d^3}{d{x}^3}\cos x=\frac{d}{dx}\left[{\left(-\frac{\pi }{180}\right)}^2\cos x\right]\\ \\ =-\frac{{\pi }^2}{{180}^2}\left(-\frac{\pi }{180}\sin x\right)\\ \\ =\frac{{\pi }^3}{{180}^3}\sin x\end{array}$

Thus, the derivatives are $\frac{d^2}{d{x}^2}\sin x=-{\left(\frac{\pi }{180}\right)}^2\sin x,\frac{d^3}{d{x}^3}\sin x=-\frac{{\pi }^3}{{180}^3}\cos x$ and $\frac{d^2}{d{x}^2}\cos x=-{\left(\frac{\pi }{180}\right)}^2\cos x,\frac{d^3}{d{x}^3}\cos x=\frac{{\pi }^3}{{180}^3}\sin x$ .