Problem 57 - Calculus Early Transcendentals

Question

Show that the sum of the \(x^{-}, y_{-}\) and \(z\) -intercepts of any tangent plane to the surface \(\sqrt{x}+\sqrt{y}+\sqrt{z}=\sqrt{c}\) is a constant.

Step-by-Step Solution

Answer

The sum of the intercepts is \( \frac{3c}{4} \), a constant.

Step 1: Finding a Tangent Plane Equation

The surface is given by \( f(x, y, z) = \sqrt{x} + \sqrt{y} + \sqrt{z} \). The equation of the tangent plane at any point \((x_0, y_0, z_0)\) on the surface is \( \frac{\partial f}{\partial x}(x_0, y_0, z_0)(x-x_0) + \frac{\partial f}{\partial y}(x_0, y_0, z_0)(y-y_0) + \frac{\partial f}{\partial z}(x_0, y_0, z_0)(z-z_0) = 0 \). Compute the partial derivatives.

Step 2: Compute Partial Derivatives

Compute \( \frac{\partial f}{\partial x} = \frac{1}{2\sqrt{x}}, \frac{\partial f}{\partial y} = \frac{1}{2\sqrt{y}}, \frac{\partial f}{\partial z} = \frac{1}{2\sqrt{z}} \). Substitute the point \((x_0, y_0, z_0)\) into these derivatives.

Step 3: Tangent Plane in Specific Form

The equation becomes \( \frac{1}{2\sqrt{x_0}}(x-x_0) + \frac{1}{2\sqrt{y_0}}(y-y_0) + \frac{1}{2\sqrt{z_0}}(z-z_0) = 0 \). Simplify this to form the tangent plane equation.

Step 4: Simplifying the Tangent Plane Equation

The equation after simplification is \( \frac{x}{2\sqrt{x_0}} + \frac{y}{2\sqrt{y_0}} + \frac{z}{2\sqrt{z_0}} = \frac{x_0}{2\sqrt{x_0}} + \frac{y_0}{2\sqrt{y_0}} + \frac{z_0}{2\sqrt{z_0}} \). Notice that \( \frac{x_0}{2\sqrt{x_0}} + \frac{y_0}{2\sqrt{y_0}} + \frac{z_0}{2\sqrt{z_0}} = \frac{\sqrt{x_0} + \sqrt{y_0} + \sqrt{z_0}}{2} = \frac{\sqrt{c}}{2} \).

Step 5: Intercepts of the Tangent Plane

To find the intercepts, set two of the variables to zero and solve for the third. The x-intercept is \( x_i = \frac{c}{4} \), the y-intercept is \( y_i = \frac{c}{4} \), and the z-intercept is \( z_i = \frac{c}{4} \).

Step 6: Total Intercept Sum

The sum of the intercepts is \( \frac{c}{4} + \frac{c}{4} + \frac{c}{4} = \frac{3c}{4} \). This shows that the sum of the intercepts is constant for any tangent plane.

Key Concepts

Partial Derivatives

Partial derivatives are a crucial concept in multivariable calculus and help us understand how functions change with respect to one variable while keeping others constant. Consider a function of three variables, like the one given in the exercise: \[ f(x, y, z) = \sqrt{x} + \sqrt{y} + \sqrt{z}. \] Here, each partial derivative represents the rate of change of the function with respect to one of the variables while the others remain unchanged. For this function:

• The partial derivative with respect to \(x\) is \( \frac{\partial f}{\partial x} = \frac{1}{2\sqrt{x}}. \)
• The partial derivative with respect to \(y\) is \( \frac{\partial f}{\partial y} = \frac{1}{2\sqrt{y}}. \)
• The partial derivative with respect to \(z\) is \( \frac{\partial f}{\partial z} = \frac{1}{2\sqrt{z}}. \)

By calculating these derivatives, we gain insight into the behavior of the function near any given point \((x_0, y_0, z_0)\). Incorporating these into the equation of a tangent plane provides a linear approximation of the surface at that point. This approximation is vital in many applications, such as optimization and solving differential equations.

Intercepts of Planes

Finding the intercepts of planes is an essential skill for visualizing and understanding their geometrical properties. In the context of tangent planes, intercepts are the points where the plane crosses the axes. Typically, these are found by setting two variables to zero and solving for the remaining variable.

• The \(x\)-intercept is where the plane meets the \(x\)-axis, with \(y=0\) and \(z=0\).
• The \(y\)-intercept is where it meets the \(y\)-axis, with \(x=0\) and \(z=0\).
• The \(z\)-intercept is where it meets the \(z\)-axis, with \(x=0\) and \(y=0\).

For the tangent plane in our exercise, substituting these conditions into the tangent plane equation gives consistent intercepts of \(\frac{c}{4}\) for each axis. This consistent finding is fascinating because it implies uniform crossing points regardless of the point \((x_0, y_0, z_0)\) chosen on the surface.

Constant Sum of Intercepts

The sum of the intercepts for any tangent plane is a remarkable property in this specific problem. When you calculate the intercepts of the tangent plane to the surface described by \(\sqrt{x} + \sqrt{y} + \sqrt{z} = \sqrt{c}\), you'll find a constant result. Specifically, the sum comes out to \(\frac{3c}{4}\).

• This sum is invariant, meaning it stays the same for any point \((x_0, y_0, z_0)\) on the surface.
• In practical terms, this suggests that although the individual intercepts may vary, their total will always equal \(\frac{3c}{4}\).

This property highlights a unique symmetry of the surface and underscores the power of mathematical analysis in revealing hidden consistencies. It can also provide insights into areas such as optimization, where understanding where planes intersect axes can inform decision-making processes.