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: Identify the Curve of Intersection

The curve of intersection is where the given surface and plane meet. The surface is described by the equation \(2z = \sqrt{9x^2 + 9y^2 - 36}\), and the plane is given by \(y = 1\). By substituting \(y = 1\) into the surface equation, we simplify the expression to determine the parametric equation along which the surface intersects the plane.

Step 2: Substitute Plane Equation into Surface Equation

Setting \(y = 1\) in the surface equation gives:\[2z = \sqrt{9x^2 + 9(1)^2 - 36} = \sqrt{9x^2 + 9 - 36} = \sqrt{9x^2 - 27}\]Thus, \(2z = \sqrt{9(x^2 - 3)}\). Simplifying further gives the expression for \(z\).

Step 3: Solve for z

Simplify the equation obtained from substitution:\[2z = 3\sqrt{x^2 - 3}\]Solving for \(z\) yields:\[z = \frac{3}{2}\sqrt{x^2 - 3}\]This is the equation of the curve of intersection when \(y = 1\).

Step 4: Differentiate to Find Slope of the Tangent

To find the slope of the tangent at a specific point on the curve, we need to differentiate \(z\) with respect to \(x\). Applying the chain rule:\[\frac{dz}{dx} = \frac{3}{2} \cdot \frac{1}{2\sqrt{x^2 - 3}} \cdot (2x) = \frac{3x}{2\sqrt{x^2 - 3}}\]

Step 5: Evaluate the Derivative at the Given Point

Evaluate \(\frac{dz}{dx}\) at \(x = 2\) to find the slope of the tangent at the point \((2, 1, \frac{3}{2})\):\[\frac{dz}{dx} \bigg|_{x=2} = \frac{3(2)}{2\sqrt{2^2 - 3}} = \frac{6}{2(1)} = 3\]Therefore, the slope of the tangent is 3.

Key Concepts

Curve of Intersection

The curve of intersection refers to the curve formed where two different surfaces meet. Imagine the surface equation representing a 3D shape and another plane cutting through it. The resulting shape where these two meet in space is the curve of intersection. In our exercise, the surfaces are defined by the equation \(2z = \sqrt{9x^2 + 9y^2 - 36}\) and the plane is described by \(y = 1\). By substituting this information into the surface equation, we manage to find a parametric representation of the curve. This simplifies the complex surface into a manageable equation that details how the surface intersects with the plane.

Differentiation

Differentiation is a mathematical technique used to calculate the rate at which a function changes. In simpler terms, it measures how a small change in one variable affects another variable. In the context of curves and surfaces, it helps us find how steep or flat the surface is at any given point. For this exercise, we want to find the slope of the tangent to the curve at a specific point. This involves differentiating the function that describes our curve of intersection, allowing us to identify how quickly the \(z\)-value changes as \(x\) changes.

Chain Rule

The chain rule is a fundamental rule in calculus for differentiating composite functions. When we have a function within a function, the chain rule allows us to differentiate these effectively. In our exercise, we apply the chain rule when differentiating the expression \(z = \frac{3}{2}\sqrt{x^2 - 3}\). Here, \(z\) is a function of \(x^2 - 3\), which itself is a function of \(x\). Applying the chain rule involves first differentiating the outer function and then multiplying by the derivative of the inner function. This step-by-step application of the chain rule provides the rate of change of \(z\) concerning \(x\).

Parametric Equations

Parametric equations help describe a curve in 2D or 3D space using parameters, often simplifying the representation of curves. Instead of writing everything strictly as functions of \(x\) and \(y\), parametric equations use one or more parameters to express these variables. In this problem, after substituting \(y = 1\) into the surface equation, we obtain a parametric equation that relates \(z\) and \(x\) in terms of a single parameter \(x\). This makes the math easier and allows us to treat the original complex 3D problem as a simpler one-dimensional problem by focusing on how \(z\) changes with respect to \(x\) only.