/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 70 Find the length of the curve \(y... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 70

Find the length of the curve \(y^{2}=x^{3}\) from the origin to the point where the tangent makes an angle of \(45^{\circ}\) with the \(x\) -axis.

Short Answer

Expert verified
The length of the curve \(y^{2}=x^{3}\) from the origin to the point where the tangent makes an angle of \(45^{\circ}\) with the \(x\) -axis is found by evaluating the integral \(\int_{0}^{\frac{4}{9}}\sqrt{1+\frac{9x}{4}}dx\).

Step by step solution

01

Calculate Derivative

Differentiate the equation \(y^{2}=x^{3}\) with respect to \(x\) to find \(y'(x)\). Rewriting equation to \(y = \sqrt{x^3}\), we find \(y'(x) = \frac{3x^2}{2\sqrt{x^3}} = \frac{3}{2}\sqrt{x}\). This is slope of the curve at any point \(x\).
02

Set up equation for tangent making \(45^{\circ}\) with x-axis

The slope of the curve at the point where the tangent makes an angle of \(45^{\circ}\) with the x-axis can be found using the tangent of the angle. Since \(\tan(45^{\circ}) = 1\), we equate \(y'(x) = 1\), leading to \(\frac{3}{2}\sqrt{x} = 1\). Solving this gives \(x = \frac{4}{9}\) and \(y = \sqrt{(\frac{4}{9})^3} = \frac{2}{3}\). So, the point of tangency is \((\frac{4}{9}, \frac{2}{3})\).
03

Calculate arc length

Finally, apply the arc length formula \(\int_{a}^{b}\sqrt{1+[y'(x)]^2}dx\). Here, \(a=0\) and \(b=\frac{4}{9}\). Substituting \(y'(x)\) derived above gives \(\int_{0}^{\frac{4}{9}}\sqrt{1+(\frac{3}{2}\sqrt{x})^2}dx\). Simplify under radical gives \(\int_{0}^{\frac{4}{9}}\sqrt{1+\frac{9x}{4}}dx\) which needs to be evaluated to find the length of the curve.

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Most popular questions from this chapter

Lateral Surface Area of a Cone A right circular cone is generated by revolving the region bounded by \(y=h x / r,\) \(y=h,\) and \(x=0\) about the \(y\) -axis. Verify that the lateral surface area of the cone is \(S=\pi r \sqrt{r^{2}+h^{2}}\)

Finding Arc Length In Exercises \(17-26,\) (a) sketch the graph of the function, highlighting the part indicated by the given interval, (b) find a definite integral that represents the arc length of the curve over the indicated interval and observe that the integral cannot be evaluated with the techniques studied so far, and (c) use the integration capabilities of a graphing utility to approximate the are length. $$ y=\frac{1}{x+1}, \quad 0 \leq x \leq 1 $$

Verifying a Formula (a) Given a circular sector with radius \(L\) and central angle \(\theta\) (see figure), show that the area of the sector is given by $$S=\frac{1}{2} L^{2} \theta .$$ (b) By joining the straight-line edges of the sector in part (a), a right circular cone is formed (see figure) and the lateral surface area of the cone is the same as the area of the sector. Show that the area is \(S=\pi r L,\) where \(r\) is the radius of the base of the cone. (Hint: The arc length of the sector equals the circumference of the base of the cone.) (c) Use the result of part (b) to verify that the formula for thelateral surface area of the frustum of a cone with slant height \(L\) and radii \(r_{1}\) and \(r_{2}\) (see figure) is \(S=\pi\left(r_{1}+r_{2}\right) L .\) (Note: This formula was used to develop the integral for finding the surface area of a surface of revolution.)

Arc Length and Area Let \(C\) be the curve given by \(f(x)=\cosh x\) for \(0 \leq x \leq t,\) where \(t>0 .\) Show that the arc length of \(C\) is equal to the area bounded by \(C\) and the \(x\) -axis. Identify another curve on the interval \(0 \leq x \leq t\) with this property.

Finding the Area of a Surface of Revolution In Exercises \(37-42,\) set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the \(x\) -axis. $$ y=2 \sqrt{x} $$
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