Problem 55 - Calculus Single Variable

Question

Let \(a\) be a positive constant. Let \(T\) be any tangent line to $$ \sqrt{x}+\sqrt{y}=a $$ Compute the sum of the intercepts of \(T\) on the coordinate axes. (The answer does not depend on the choice of \(T .)\)

Step-by-Step Solution

Answer

The sum of the intercepts is always \( a^2 \), independent of \( T \).

Step 1: Define the Function

The given curve is \( \sqrt{x} + \sqrt{y} = a \). Let's solve it for \( y \) as a function of \( x \). For any point \((x_0, y_0)\) on the curve, we get \( \sqrt{x_0} + \sqrt{y_0} = a \). Solving for \( y_0 \), we have \( \sqrt{y_0} = a - \sqrt{x_0} \). Thus, \( y_0 = (a - \sqrt{x_0})^2 \).

Step 2: Find the Derivative

To find the slope of the tangent line \( T \), we differentiate implicitly. Start by differentiating both sides of \( \sqrt{x} + \sqrt{y} = a \) with respect to \( x \). We have \( \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = 0 \). Solving for \( \frac{dy}{dx} \), we get \( \frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}} \).

Step 3: Find the Equation of the Tangent Line

The slope of the tangent line at point \((x_0, y_0)\) is \( m = -\frac{\sqrt{y_0}}{\sqrt{x_0}} \). The equation of the tangent line can be written using the point-slope form: \( y - y_0 = -\frac{\sqrt{y_0}}{\sqrt{x_0}} (x - x_0) \).

Step 4: Find X-Intercept and Y-Intercept

Set \( y = 0 \) in the equation \( y - y_0 = -\frac{\sqrt{y_0}}{\sqrt{x_0}} (x - x_0) \) to find the x-intercept: \( 0 - y_0 = -\frac{\sqrt{y_0}}{\sqrt{x_0}} (x - x_0) \). Solving, \( x = x_0 + \frac{y_0 \sqrt{x_0}}{\sqrt{y_0}} \). Set \( x = 0 \) to find the y-intercept: \( y - y_0 = -\frac{\sqrt{y_0}}{\sqrt{x_0}} (0 - x_0) \). Solving, \( y = y_0 + \frac{x_0 \sqrt{y_0}}{\sqrt{x_0}} \).

Step 5: Sum the Intercepts

Now sum the x-intercept \( x_0 + \frac{y_0 \sqrt{x_0}}{\sqrt{y_0}} \) and the y-intercept \( y_0 + \frac{x_0 \sqrt{y_0}}{\sqrt{x_0}} \). This simplifies to \( x_0 + a\sqrt{x_0} + y_0 + a\sqrt{y_0} \), which simplifies further to exactly \( a^2 \), demonstrating the independence of the sum from the point \((x_0, y_0)\).

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used to find the derivative of a function when it is not explicitly solved for one of the variables. In many cases, the equation contains both

• x
• y

which are intricately connected. By using implicit differentiation, we can differentiate both sides of an equation with respect to x and then solve for \( \frac{dy}{dx} \). For the equation \( \sqrt{x} + \sqrt{y} = a \), we start by differentiating:

• For \( \sqrt{x} \), the derivative is \( \frac{1}{2\sqrt{x}} \).
• For \( \sqrt{y} \), the derivative is \( \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} \).

Combining both, we get: \( \frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \cdot \frac{dy}{dx} = 0 \). Solving for \( \frac{dy}{dx} \), we find \( \frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}} \). This derivative tells us how y changes with x along the curve.

Intercepts

Intercepts are points where a graph intersects the axes. The x-intercept occurs where the graph crosses the x-axis (y = 0), and the y-intercept is where it crosses the y-axis (x = 0). In the context of tangent lines for the curve \( \sqrt{x} + \sqrt{y} = a \), finding these intercepts helps us understand how the tangent line interacts with the coordinate axes. For the tangent line equation \[ y - y_0 = -\frac{\sqrt{y_0}}{\sqrt{x_0}}(x - x_0) \]

• Set \( y = 0 \) to find the x-intercept.
• Set \( x = 0 \) to find the y-intercept.

The intercepts calculated from our steps are indicative of definite numbers that describe the relationship between the point

• x0
• y0

and the curve's tangent.

Derivative

A derivative represents the rate at which one quantity changes with respect to another. In calculus, derivatives are fundamental in understanding how functions behave. For this exercise, the derivative of the curve \( \sqrt{x} + \sqrt{y} = a \) provides the slope of the tangent line at any point \((x_0, y_0)\). When we obtain \[ \frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}} \]we are essentially finding how steep or flat the line is as it touches the curve at that point. Differentiation is pivotal, especially when dealing with equations that are not neatly laid out for solving traditional slopes. That's where implicit differentiation shines by enabling us to obtain derivatives even when variables are tangled.

Point-Slope Form

The point-slope form is a way to express the equation of a line using a known point and the slope. It is written as \[ y - y_1 = m(x - x_1) \]where

• \(m\) is the slope
• \((x_1, y_1)\) is a point on the line

For this exercise, once we determined the slope \( m = -\frac{\sqrt{y_0}}{\sqrt{x_0}} \), we can write the equation for the tangent using any chosen point on the curve. This form provides a straightforward way to describe a line's equation without needing the y-intercept explicitly. It is especially useful in situations like this exercise, where our objective was to find how the tangent line expressed by point-slope impacts rather than just graphing the line.