Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that the surfaces \(z=x^{2} y\) and \(y=\frac{1}{4} x^{2}+\frac{3}{4}\) intersect at \((1,1,1)\) and have perpendicular tangent planes there.

Step-by-Step Solution

Answer

The surfaces intersect at (1,1,1) and have perpendicular tangent planes there.

Step 1: Verify Intersection Point

We start by checking if the given point \((1,1,1)\) lies on both surfaces. Substitute \((x, y, z) = (1, 1, 1)\) into the first equation: \[ z = x^2 y \]\[ 1 = 1^2 \times 1 \]\[ 1 = 1 \] The point satisfies the first surface. Now, substitute into the second equation: \[ y = \frac{1}{4}x^2 + \frac{3}{4} \]\[ 1 = \frac{1}{4}\times1^2 + \frac{3}{4} \]\[ 1 = 1 \] The point satisfies the second surface as well. Thus, the point \((1,1,1)\) is indeed on both surfaces.

Step 2: Find Tangent Plane to the First Surface

Next, calculate the gradient of the first surface \(z = x^2 y\) at the point \((1,1,1)\). The gradient is given by\[ abla f = \left( \frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}, \frac{\partial z}{\partial z} \right) \]Compute the partial derivatives:\[ \frac{\partial z}{\partial x} = 2xy \Rightarrow 2 \times 1 \times 1 = 2 \]\[ \frac{\partial z}{\partial y} = x^2 \Rightarrow 1^2 = 1 \]\[ \frac{\partial z}{\partial z} = -1 \]The gradient vector at this point is \((2, 1, -1)\), which gives the normal vector to the tangent plane.

Step 3: Find Tangent Plane to the Second Surface

Calculate the gradient of the second surface \(y = \frac{1}{4}x^2 + \frac{3}{4}\). By moving all terms to one side, it becomes \(F(x, y) = \frac{1}{4}x^2 + \frac{3}{4} - y = 0\). The gradient is:\[ abla F = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} \right) \]Calculate the partial derivatives:\[ \frac{\partial F}{\partial x} = \frac{1}{2}x \Rightarrow \frac{1}{2}\times1 = \frac{1}{2} \]\[ \frac{\partial F}{\partial y} = -1 \]\[ \frac{\partial F}{\partial z} = 0 \]The gradient vector at \((1, 1, 1)\) is \((\frac{1}{2}, -1, 0)\), which represents the normal vector to the tangent plane.

Step 4: Check Perpendicularity of Tangent Planes

To check if two planes are perpendicular, the dot product of their normal vectors should be zero. Calculate the dot product:\[ (2, 1, -1) \cdot \left(\frac{1}{2}, -1, 0\right) = 2 \times \frac{1}{2} + 1 \times (-1) + (-1) \times 0 \]\[ = 1 - 1 + 0 = 0 \]The dot product is zero, confirming the tangent planes are perpendicular at the point \((1,1,1)\).

Key Concepts

Tangent Planes

A tangent plane serves as the best flat-surface approximation to a given surface at a specific point. When dealing with surfaces in three-dimensional space, being able to define a tangent plane is crucial for understanding the local behavior of the surface at specific points.
In simple terms, imagine laying a flat piece of paper on a smooth object like a balloon. Where the paper touches the balloon, that's akin to the tangent plane 'touching' the surface without cutting through it.

• For the surface described by an equation like \(z = x^2 y\), the tangent plane at a point \((x_0, y_0, z_0)\) can be found by using partial derivatives of the function with respect to \(x\) and \(y\).
• These derivatives inform us about the slope of the surface in various directions. By securing the normal vector (a vector that stands perpendicular to the tangent plane), the plane can be mathematically described.

This mathematical description helps us in comparing different surfaces like checking for perpendicularity or testing solutions to differential equations at specific points.

Partial Derivatives

Partial derivatives are tools that help us understand how a function changes as we tweak one variable while holding others constant. This is crucial in multivariable calculus, where functions often depend on more than one variable.
When dealing with surfaces like \(z = x^2 y\), partial derivatives tell us how much \(z\) changes as \(x\) or \(y\) changes while the other stays constant.

• The partial derivative with respect to \(x\), written \(\frac{\partial z}{\partial x}\), measures the rate of change of \(z\) if only \(x\) varies.
• Similarly, \(\frac{\partial z}{\partial y}\) measures the rate of change if only \(y\) varies.

In the context of tangent planes, partial derivatives are used to calculate the gradient vector, which then gives us the plane's normal vector and helps define the plane itself. Understanding these derivatives is essential for anyone venturing into calculus and understanding geometric space.

Gradient Vectors

A gradient vector is a powerful tool in the calculus toolbox, providing essential information about the behavior and direction of functions of several variables.
The gradient vector emerges from the derivatives of a multivariable function. It is a vector composed of all the partial derivatives of a function. For a function \(f(x, y, z)\), the gradient \(abla f\) is given by:\[abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)\]

• The gradient vector points in the direction of the steepest ascent of the function.
• For functions defined on surfaces, the gradient at a given point is orthogonal (perpendicular) to the tangent plane at that point.

In the context of the original problem, the gradients derived from both functions \(z = x^2 y\) and \(y = \frac{1}{4} x^2 + \frac{3}{4}\) yield the normal vectors to their respective tangent planes. Checking these vectors' perpendicularity helps determine if the surfaces intersect perpendicularly.