Question

A tangent is drawn on the curve \( y = \frac{1}{3} \sqrt{x^3} \), \( x>0 \) at the point \( P \left( 1, \frac{1}{3} \right) \), which meets the x-axis at \( Q \). Then the length of the closed curve \( OQPO \), where \( O \) is the origin, is

Hint:

For finding the length of a curve, find the equation of the tangent and determine the points of intersection with the x-axis or y-axis to complete the curve.

Answer 1

Correct Answer: 2.1 - 2.2

Step 1: Find the equation of the tangent at the point.
The equation of the tangent to the curve \( y = \frac{1}{3} \sqrt{x^3} \) at a point \( P \) can be found using the derivative of the curve. The slope of the tangent is given by the derivative of \( y \) with respect to \( x \): \[ \frac{dy}{dx} = \frac{1}{3} \cdot \frac{3x^2}{2\sqrt{x^3}} = \frac{x^2}{2\sqrt{x^3}}. \] At \( x = 1 \), the slope is: \[ \frac{dy}{dx} = \frac{1^2}{2\sqrt{1^3}} = \frac{1}{2}. \] So, the equation of the tangent at \( P \left( 1, \frac{1}{3} \right) \) is: \[ y - \frac{1}{3} = \frac{1}{2}(x - 1). \] This simplifies to: \[ y = \frac{1}{2}x - \frac{1}{6}. \]
Step 2: Find the x-coordinate of the point of intersection with the x-axis.
The x-axis is defined by \( y = 0 \). So, set \( y = 0 \) in the equation of the tangent: \[ 0 = \frac{1}{2}x - \frac{1}{6}. \] Solving for \( x \), we get: \[ x = \frac{1}{3}. \]
Step 3: Find the length of the closed curve.
The length of the curve \( OQPO \) consists of two segments: the line segment from \( O(0, 0) \) to \( P(1, \frac{1}{3}) \) and the tangent curve from \( P \) to \( Q \left( \frac{1}{3}, 0 \right) \). The total length of the curve is \( 2 \).