Q. 2 - Calculus

Question

Examples: Construct examples of the thing(s) described in the following. Try to find examples that are different than any in the reading.

(a) A graph of a function f(x) on [a, b] whose arc length is poorly approximated by four line segments.

(b) An equation of a function f(x) on [a, b] that gives rise to an arc length integral that we do not know how to solve.

(c) An equation of a surface of revolution defined by revolving a function f(x) on [a, b] around the x-axis that gives rise to a surface area integral that we do not know how to solve.

Step-by-Step Solution

Answer

Part (a) The function is $f\left(x\right)=\sin x$ on the interval $\left[0,4\mathrm{\pi }\right]$ whose arc length is poorly approximated by four line segments.

Part (b) An equation of a function $f\left(x\right)=e^x$ on $\left[a,b\right]$ that gives rise to an arc length integral that we do not know how to solve.

Part (c) An equation of a surface of revolution defined by revolving a function $f\left(x\right)=\ln x$ on [a, b] around the x-axis that gives rise to a surface area integral that we do not know how to solve. 

Part (a) Step 1. Given Information.

It is given that the function arc length is approximated by four line segments.

Part (a) Step 2. Construct example.

Let the graph of a function$f\left(x\right)=\sin x$on the interval$\left[0,4\mathrm{\pi }\right]$and the four-line segments are$[0,\pi ], [\pi ,2\pi ], [2\pi ,3\pi ], [3\pi ,4\pi ].$

The graph is 

Thus, the approximation of the arc length by the line segments is $4\mathrm{\pi }$ and the actual arc length is $15.075.$

Therefore, the arc length is poorly approximated by the four-line segments.

Part (b) Step 1. Construct example.

Let the function be $f\left(x\right)=e^x$ on the interval $\left[a,b\right].$ So, the arc length in the integral is $I={\int }_a^b\sqrt{1+e^{2x}}dx.$ The integral we get can't be solved by the integration.

Part (c) Step 1. Construct example.

Let the function be $f\left(x\right)=\ln x$ on the interval $\left[a,b\right].$

So, the surface area as a definite integral is $S=2\mathrm{\pi }{\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{lnx}\sqrt{1+{\left(\frac{1}{\mathrm{x}}\right)}^2}\mathrm{dx}.$

The integral we get can't be solved by the integration.