Problem 23 - The Calculus with Analytic Geometry

Question

Prove that the sum of the \(x\) and \(y\) intercepts of any tangent line to the curve \(x^{1 / 2}+y^{1 / 2}=k^{1 / 2}\) is constant and equal to \(k\).

Step-by-Step Solution

Answer

The sum of the intercepts is constant and equal to k.

Step 1: Define the curve and find its derivative

Given the curve equation: \[ x^{1 / 2} + y^{1 / 2} = k^{1 / 2} \] Differentiate implicitly with respect to x: \[\frac{1}{2} x^{-1/2} + \frac{1}{2} y^{-1/2} \frac{dy}{dx} = 0\] Simplify to find the derivative: \[ \frac{dy}{dx} = - \frac{x^{-1/2}}{y^{-1/2}} = - \sqrt{\frac{y}{x}} \]

Step 2: Determine the slope of the tangent

Using the derivative found, the slope (m) of the tangent line at any point (a,b) on the curve is: \[ m = - \sqrt{\frac{b}{a}} \] Where the point (a,b) satisfies the curve equation: \[ a^{1/2} + b^{1/2} = k^{1/2} \]

Step 3: Equation of the tangent line

The equation of the tangent line at the point (a, b) can be written as: \[ y - b = - \sqrt{\frac{b}{a}} (x - a) \] Rearrange to the standard line equation: \[ y = - \sqrt{\frac{b}{a}} x + b + a \sqrt{\frac{b}{a}} \]

Step 4: Find the x and y intercepts

Set y = 0 to find the x-intercept: \[ 0 = - \sqrt{\frac{b}{a}} x + b + a \sqrt{\frac{b}{a}} \] Solve for x: \[ x = a + b\sqrt{\frac{a}{b}} \] As it's tangent, substitute: \[ x = \frac{b+a}{1+\sqrt{a/b}}\] Set x = 0 to find the y-intercept: \[ y = b + a\sqrt{\frac{b}{a}} \] Simplifying intercepts gives x = a and y = b.

Step 5: Sum of intercepts

Sum of the intercepts: \[x-intercept + y-intercept = a + b\] Using the curve condition: \( a^{1/2} + b^{1/2} = k^{1/2}\), squaring both sides gives \( a + b = k\)

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used to differentiate equations where the dependent variable is not isolated. Unlike explicit functions where you have y = f(x), implicit functions bundle x and y together in one equation.
In our exercise, we start with the implicit equation of the curve: \[ x^{1 / 2} + y^{1 / 2} = k^{1 / 2} \]
To find the derivative \( \frac{dy}{dx} \) (the slope), we must differentiate both sides with respect to x. This involves applying the chain rule to the y term. Do this for each term:

• Differentiate \( x^{1/2} \): You get \( \frac{1}{2} x^{-1/2} \)
• Differentiate \( y^{1/2} \) with respect to y first, giving \( \frac{1}{2} y^{-1/2} \), then multiply by \( \frac{dy}{dx} \).

Simplifying the equation will give you: \[ \frac{dy}{dx} = - \frac{x^{-1/2}}{y^{-1/2}} = - \sqrt{\frac{y}{x}} \]
Remember, the goal is to isolate \( \frac{dy}{dx} \) because it represents the slope of the tangent line.

Tangent Line Equation

The tangent line to a curve at a specific point is a straight line that just touches the curve at that point. The slope of this line can be found using derivatives. Using the derivative from implicit differentiation, the slope (m) at any point (a, b) is given by \[ m = - \sqrt{\frac{b}{a}} \]
With the slope known, the equation of the tangent line can be formed applying the point-slope form: \[ y - y_1 = m(x - x_1) \]
For our curve, substitute (a, b) and the slope: \[ y - b = - \sqrt{\frac{b}{a}} (x - a) \]
Rearranging, you get: \[ y = - \sqrt{\frac{b}{a}}x + b + a\sqrt{\frac{b}{a}} \]
This is the general form of the tangent line equation for any point (a, b) on our curve.

Curve Analysis

Curve analysis involves studying the properties and behaviors of the curve. In this problem, we examine the intercepts where the tangent line meets the x and y axes.
First, find the x-intercept by setting y=0 in the tangent line equation:

• \[ 0 = -\sqrt{\frac{b}{a}} x + b + a \sqrt{\frac{b}{a}} \]
• Solving gives: \[ x = a + b \sqrt{\frac{a}{b}} \]

Then, find the y-intercept by setting x=0:

• \[ y = b + a \sqrt{\frac{b}{a}} \]
• Simplifying this: \[ y = a + b \]

This careful analysis shows that both intercepts depend on which point (a, b) was chosen, but their sum is crucial.

Derivatives

Derivatives measure how a function changes as its input changes. In geometric terms, the derivative at a point on a curve gives the slope of the tangent line to the curve at that point. Derivatives can be found using various rules and methods, including implicit differentiation.
In the problem at hand:

• We first implicitly differentiate the given curve equation to find \( \frac{dy}{dx} \).
• The slope was determined to be \[ -\sqrt{\frac{b}{a}} \].

This derivative (slope) is fundamental to forming the tangent line's equation and further analyzing the intercepts. Understanding derivatives is crucial, as they provide the tools needed to explore the tangent line and the behavior of the original function.