Problem 32 - Advanced Engineering Mathematics

Question

Show that the sum of the \(x-, y-\), and \(z\) -intercepts of every tangent plane to the graph of \(\sqrt{x}+\sqrt{y}+\sqrt{z}=\sqrt{a}\) \(a>0\), is the number \(a\).

Step-by-Step Solution

Answer

The sum of the intercepts is \( a \).

Step 1: Identify the curve and equation of the tangent plane

The given surface is defined by \( \sqrt{x} + \sqrt{y} + \sqrt{z} = \sqrt{a} \). A tangent plane to a surface at a point is determined using the gradient vector of the surface equations. We need to find partial derivatives to form the equation of the tangent plane.

Step 2: Calculate the partial derivatives of the surface

To find the tangent plane at a point \((x_0, y_0, z_0)\), compute the partial derivatives: \( \frac{\partial}{\partial x} = \frac{1}{2\sqrt{x}} \), \( \frac{\partial}{\partial y} = \frac{1}{2\sqrt{y}} \), \( \frac{\partial}{\partial z} = \frac{1}{2\sqrt{z}} \). Evaluate these partials at the point \((x_0, y_0, z_0)\) to get specific values.

Step 3: Write the equation of the tangent plane

Using point \((x_0, y_0, z_0)\) on the surface, the equation of the tangent plane is given by: \[ \left( \frac{1}{2\sqrt{x_0}} \right)(x-x_0) + \left( \frac{1}{2\sqrt{y_0}} \right)(y-y_0) + \left( \frac{1}{2\sqrt{z_0}} \right)(z-z_0) = 0. \] Simplify this equation to standard form: \[ \frac{x}{2\sqrt{x_0}} + \frac{y}{2\sqrt{y_0}} + \frac{z}{2\sqrt{z_0}} = C, \] where \( C \) is a constant determined by simplification.

Step 4: Determine the intercepts of the tangent plane

To find the \(x\)-intercept, set \(y = 0\) and \(z = 0\) and solve for \(x\). Similarly, solve for \(y\)-intercept by setting \(x = 0\) and \(z = 0\), and for \(z\)-intercept by setting \(x = 0\) and \(y = 0\). The intercepts are \( x = 2\sqrt{a}\sqrt{x_0}, \; y = 2\sqrt{a}\sqrt{y_0}, \; z = 2\sqrt{a}\sqrt{z_0} \).

Step 5: Sum up the intercepts

Sum up the expressions for the intercepts: \[ 2\sqrt{x_0} + 2\sqrt{y_0} + 2\sqrt{z_0} = 2(\sqrt{x_0} + \sqrt{y_0} + \sqrt{z_0}). \] Since \( \sqrt{x_0} + \sqrt{y_0} + \sqrt{z_0} = \sqrt{a} \), substitute to get \( 2\sqrt{a} = a \), confirming that the sum of the intercepts is \( a \).

Key Concepts

Partial Derivatives

In calculus, partial derivatives are a fundamental tool for understanding how functions change. When dealing with functions of multiple variables, like in our equation \( \sqrt{x} + \sqrt{y} + \sqrt{z} = \sqrt{a} \), we use partial derivatives to analyze the change with respect to one variable while keeping others constant. For instance, the partial derivative \( \frac{\partial}{\partial x} \) measures how the function changes as \( x \) changes and \( y \) and \( z \) stay fixed.

For our function, the partial derivatives are calculated as follows:

• The partial derivative with respect to x is \( \frac{1}{2\sqrt{x}} \).
• For y, it's \( \frac{1}{2\sqrt{y}} \).
• And for z, you get \( \frac{1}{2\sqrt{z}} \).

Understanding these partial derivatives is crucial as they lead us to determine the tangent plane, which is a locally linear approximation of the surface near a point.

Gradient Vector

The gradient vector is a powerful concept in calculus, particularly in the study of multivariable functions. It combines all the partial derivatives of a function, giving us direction towards the steepest ascent. For a function \( f(x, y, z) \), the gradient vector is \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \).

In the context of our surface \( \sqrt{x} + \sqrt{y} + \sqrt{z} = \sqrt{a} \), each component of this gradient points in the direction where the function increases most rapidly for each variable:

• \( \frac{1}{2\sqrt{x}} \) for x,
• \( \frac{1}{2\sqrt{y}} \) for y,
• and \( \frac{1}{2\sqrt{z}} \) for z.

The importance of the gradient vector lies in its orthogonality to the tangent plane, which helps us define the plane's equation. This is crucial when determining plane equations to understand geometrical aspects like intercepts.

Intercepts

In geometry, an intercept refers to where a line or plane crosses the axes of a coordinate system. The concept of intercepts offers substantial insight into the geometry of a plane. For a tangent plane to intersect the x, y, and z-axes in three-dimensional space, solving and understanding intercepts becomes critical.

In our given surface equation, to find each intercept, we set the other two variables to zero. Here’s how you find each axis intercept for the tangent plane:

• X-intercept: Set \( y = 0 \) and \( z = 0 \), solve for \( x \) which gives \( x = 2\sqrt{a} \sqrt{x_0} \).
• Y-intercept: For \( x = 0 \) and \( z = 0 \), solve for \( y \), obtaining \( y = 2\sqrt{a} \sqrt{y_0} \).
• Z-intercept: With \( x = 0 \) and \( y = 0 \), solve for \( z \), leading to \( z = 2\sqrt{a} \sqrt{z_0} \).

Adding these intercepts gives the result \( 2(\sqrt{x_0} + \sqrt{y_0} + \sqrt{z_0}) \), which if using the property \( \sqrt{x_0} + \sqrt{y_0} + \sqrt{z_0} = \sqrt{a} \), confirms that their total is \( a \). This elegant solution highlights how intercepts can simplify understanding complex spatial structures and their properties.