Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the slope of the tangent to the curve of intersection of the surface \(2 z=\sqrt{9 x^{2}+9 y^{2}-36}\) and the plane \(y=1\) at the point \(\left(2,1, \frac{3}{2}\right)\).

Step-by-Step Solution

Answer

The slope of the tangent is 3.

Step 1: Interpret the Surface Equation

Given the equation of the surface is \(2z = \sqrt{9x^2 + 9y^2 - 36}\). This describes a surface in three-dimensional space. By rewriting the equation, we can express \(z\) explicitly in terms of \(x\) and \(y\): \(z = \frac{1}{2}\sqrt{9x^2 + 9y^2 - 36}\).

Step 2: Fix the Plane Intersection

The plane intersecting the surface is given by \(y = 1\). Plug \(y = 1\) into the surface equation: \[ z = \frac{1}{2}\sqrt{9x^2 + 9(1)^2 - 36} = \frac{1}{2}\sqrt{9x^2 - 27}. \] This reduces our problem to a curve in the \(xz\)-plane.

Step 3: Simplify and Find Derivative

To evaluate at the specific point \((2, 1, \frac{3}{2})\), plug \(x = 2\) into \(z = \frac{1}{2}\sqrt{9x^2 - 27}\). Simplifying gives \(z = \frac{1}{2}\sqrt{36 - 27} = \frac{3}{2}\), which matches our point.To find the tangent slope, differentiate \(z\) with respect to \(x\):\[ \frac{dz}{dx} = \frac{1}{2} \times \frac{1}{2}(9x^2 - 27)^{-1/2} \cdot 18x = \frac{9x}{2\sqrt{9x^2 - 27}}. \]

Step 4: Evaluate the Derivative at the Given Point

Substitute \(x = 2\) into the derivative expression:\[ \frac{dz}{dx}\bigg|_{x=2} = \frac{9(2)}{2\sqrt{9(2)^2 - 27}} = \frac{18}{2 \cdot 3} = 3. \]Therefore, the slope of the tangent line at the point \((2, 1, \frac{3}{2})\) is 3.

Key Concepts

Surface Intersection

To tackle problems involving surface intersections in calculus, it's crucial to understand how different geometric entities interact in space. In this exercise, we focus on the intersection of a surface and a plane.

The surface is defined by the equation: \(2z = \sqrt{9x^2 + 9y^2 - 36}\). This equation suggests a surface in 3D space. A plane, on the other hand, simplifies to one of the coordinate planes or planes parallel to them, allowing for convenient intersection analysis. Here, the plane is \(y = 1\).

By substituting \(y = 1\) into the surface equation, the problem reduces to a simpler curve in the \(xz\)-plane: \(z = \frac{1}{2}\sqrt{9x^2 - 27}\). This simplification is essential for finding the tangent slope, a concept we'll explore further below. Remember that understanding the geometry of intersecting surfaces and planes helps in visualizing and solving such problems effectively.

Derivatives

Derivatives are foundational in calculus, dealing with rates of change. In the context of our problem, they help in finding the slope of the tangent line to the curve.

The equation for the curve after intersection is simplified to \(z = \frac{1}{2}\sqrt{9x^2 - 27}\). To find how \(z\) changes with \(x\), we differentiate \(z\) with respect to \(x\). This involves applying the chain rule of differentiation to a function involving a square root.

The derivative, \(\frac{dz}{dx}\), captures the slope of the tangent to the curve, given by:

• First, differentiate the outer square root function.
• Apply the chain rule to integrate the square inside the root.
• Simplify to reach \(\frac{9x}{2\sqrt{9x^2 - 27}}\).

This derivative provides the necessary slope when evaluated at specific points.

Tangent Slope

The tangent slope to a curve at a given point reveals the direction and steepness of the curve at that particular point. In this context, we find the tangent slope to the curve formed by the intersection.

Knowing the derivative \(\frac{dz}{dx} = \frac{9x}{2\sqrt{9x^2 - 27}}\), we can compute the slope at any point \((x, y, z)\) on the curve.

To find the specific tangent slope at the point \((2, 1, \frac{3}{2})\):

• Plug \(x = 2\) into the derivative.
• Simplify: \(\frac{9(2)}{2\cdot3} = 3\).

Therefore, the slope of the tangent at this point is 3. This number gives precise insight into how steeply the curve rises at this spot, a critical component in understanding the behavior of the function around a point.

3D Geometry

Understanding 3D geometry allows us to interpret complex calculus problems effortlessly. In this intersection problem, we work within the realm of xyz-space, where each point has three coordinates defining its position.

The surface given by \(2z = \sqrt{9x^2 + 9y^2 - 36}\) creates a shell-like shape. The plane \(y = 1\) slices this shell, leaving a trace, a curve, in the \(xz\)-plane. This curve represents the path along which the tangent slope is calculated.

Visualizing these elements and their interactions in a 3D context can be challenging but rewarding. By viewing the geometrical arrangement, students can gain a better grasp of how derivatives apply spatially, understand the critical role these calculations play in real-world applications, and see the visual representation of analytical solutions to calculus problems.