Problem 70 - Calculus of a Single Variable

Question

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.

Step-by-Step Solution

Answer

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 1: 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 \(x\).

Step 2: 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 \((\frac{4}{9}, \frac{2}{3})\).

Step 3: 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.