Chapter 2, Problem 1RE

Precalculus or Calculus In Exercises 1 and 2, decide whether the problem can be solved using precalculus or whether calculus is required. If the problem can be solved using precalculus, solve it. If the problem seems to require calculus, explain your reasoning and use a graphical or numerical approach to estimate the solution.

Find the distance between the points (1, 1) and (3, 9) along the curve $y=x^2$.

To determine

To calculate: The distance between the points $(1,1)$ and $(3,9)$ along the curve $y=x^2$.

Answer to Problem 1RE

Solution: The distance between the points $(1,1)$ and $(3,9)$ along the curve $y=x^2$ is $\approx \text{8}\text{.26815}$.

Explanation of Solution

Given:

The curve $y=x^2$ and the points $(1,1)$ and $(3,9)$

Formula used:

$L(x_1,x_2)={\displaystyle \underset{x_1}{\overset{x_2}{\int }}\sqrt{1+{\left[f^{\prime }(x)\right]}^2}}dx$ where $L(x_1,x_2)$ is the path between the points.

Calculation:

To solve this problem we need to use calculus as the path between the points, $\left(1,1\right)$ and $(3,9)$ is nonlinear.

Using the formula, $L(x_1,x_2)={\displaystyle \underset{x_1}{\overset{x_2}{\int }}\sqrt{1+{\left[f^{\prime }(x)\right]}^2}}dx$, where $L(x_1,x_2)$ is the pathbetween the points $(1,1)$ and $(3,9)$ along the curve $y=x^2$.

Here, $f(x)=x^2$ and $f^{\prime }(x)=2x$.

Thus, $L(x_1,x_2)={\displaystyle \underset{1}{\overset{3}{\int }}\sqrt{1+4x^2}}dx\approx \text{8}\text{.26815}$

The distance between the points $(1,1)$ and $(3,9)$ along the curve $y=x^2$ is $\approx \text{8}\text{.26815}$.