Length of Line - CLEP Calculus

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Question

Consider a function with first-derivative:

$\frac{dy}{dx}=\tan{x}$ .

Which integral can calculate the length along this curve between and $x=2\pi$ ?

Answer

To determine the length of a curve between two points, we evaluate the integral

$L = \int_0^{2\pi}{\sqrt{1+\left(\frac{dy}{dx} \right )^2}}dx$

There are many reasons this works, but we'll give an informal explanation here:

If we divide this curve into three line-segments, we can see that they become more and more similar to the original curve. By adding up all the little hypotenuses, we can arrive at the length of the curve. Note, that

$\sqrt{dx^2 + dy^2} = \sqrt{1+\left(\frac{dy}{dx} \right )^2}dx$

If we think of the integration symbol as a sum of infinitely small parts, this gets us the formula for length:

$L = \int_a^b{\sqrt{1+\left(\frac{dy}{dx} \right )^2}}dx$ .

Returning to the problem, we plug the derivative into the length formula:

$L=\int_{0}^{2\pi}{\sqrt{1+\tan^2{x}}}\quad dx$

substitute:

$L=\int_{0}^{2\pi}{\sqrt{\left( \frac{\cos{x}}{\cos{x}} \right)^2+\left(\frac{\sin{x}}{\cos{x}} \right )^2}}\quad dx$

Simplify:

$L=\int_{0}^{2\pi}{\sqrt{\frac{\cos^2{x}}{\cos^2{x}}+\frac{\sin^2{x}}{\cos^2{x}}}}\quad dx$

$L=\int_{0}^{2\pi}{\sqrt{\frac{\cos^2{x}+\sin^2{x}}{\cos^2{x}}}}\quad dx$

By trigonometric identities we get:

$L=\int_{0}^{2\pi}{\sqrt{\frac{1}{\cos^2{x}}}}\quad dx$

$L=\int_{0}^{2\pi}{\frac{1}{\cos{x}}}\quad dx$

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Question

Find the length of the line segment between points A and B:

Answer

The distance between two points can be easily found using the Distance Formula:

$D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

Applying the points we are given to this formula results in:

$D=\sqrt{((-5)-2)^2+(10-(-7))^2}=\sqrt{(-7)^2+(17)^2}=\sqrt{338}$

This is one of the answer choices.

Compare your answer with the correct one above

Question

Consider a function with first-derivative:

$\frac{dy}{dx}=\tan{x}$ .

Which integral can calculate the length along this curve between and $x=2\pi$ ?

Answer

To determine the length of a curve between two points, we evaluate the integral

$L = \int_0^{2\pi}{\sqrt{1+\left(\frac{dy}{dx} \right )^2}}dx$

There are many reasons this works, but we'll give an informal explanation here:

If we divide this curve into three line-segments, we can see that they become more and more similar to the original curve. By adding up all the little hypotenuses, we can arrive at the length of the curve. Note, that

$\sqrt{dx^2 + dy^2} = \sqrt{1+\left(\frac{dy}{dx} \right )^2}dx$

If we think of the integration symbol as a sum of infinitely small parts, this gets us the formula for length:

$L = \int_a^b{\sqrt{1+\left(\frac{dy}{dx} \right )^2}}dx$ .

Returning to the problem, we plug the derivative into the length formula:

$L=\int_{0}^{2\pi}{\sqrt{1+\tan^2{x}}}\quad dx$

substitute:

$L=\int_{0}^{2\pi}{\sqrt{\left( \frac{\cos{x}}{\cos{x}} \right)^2+\left(\frac{\sin{x}}{\cos{x}} \right )^2}}\quad dx$

Simplify:

$L=\int_{0}^{2\pi}{\sqrt{\frac{\cos^2{x}}{\cos^2{x}}+\frac{\sin^2{x}}{\cos^2{x}}}}\quad dx$

$L=\int_{0}^{2\pi}{\sqrt{\frac{\cos^2{x}+\sin^2{x}}{\cos^2{x}}}}\quad dx$

By trigonometric identities we get:

$L=\int_{0}^{2\pi}{\sqrt{\frac{1}{\cos^2{x}}}}\quad dx$

$L=\int_{0}^{2\pi}{\frac{1}{\cos{x}}}\quad dx$

Compare your answer with the correct one above

Question

Find the length of the line segment between points A and B:

Answer

The distance between two points can be easily found using the Distance Formula:

$D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

Applying the points we are given to this formula results in:

$D=\sqrt{((-5)-2)^2+(10-(-7))^2}=\sqrt{(-7)^2+(17)^2}=\sqrt{338}$

This is one of the answer choices.

Compare your answer with the correct one above

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