Problem 28 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the slope of the tangent to the curve of intersection of the cylinder \(4 z=5 \sqrt{16-x^{2}}\) and the plane \(y=3\) at the point \((2,3,5 \sqrt{3} / 2)\)

Step-by-Step Solution

Answer

The slope is \(-\frac{5\sqrt{3}}{12}\).

Step 1: Identify the intersection curve

To find the slope of the tangent to the curve of intersection, first consider the given equations: the cylinder equation is \(4z = 5 \sqrt{16-x^2}\) and the plane equation is \(y = 3\). The curve of intersection lies above the surface formed by fixing \(y = 3\) in both the cylinder and the plane, effectively reducing the problem to examining the behavior of \(x\) and \(z\).

Step 2: Substitute and simplify

Substitute \(y = 3\) in the cylinder equation, it remains unchanged for variable \(z\) as it depends only on \(x\). The equation for \(z\) simplifies to \(z = \frac{5}{4} \sqrt{16-x^2}\).

Step 3: Differentiate to find slope

To find the slope of the tangent line at \((2, 3, 5\sqrt{3}/2)\), differentiate the equation \(z = \frac{5}{4} \sqrt{16-x^2}\) with respect to \(x\). Use the chain rule to find \(\frac{dz}{dx} = \frac{d}{dx} \left( \frac{5}{4} \sqrt{16-x^2} \right)\).

Step 4: Apply chain rule

Differentiate: \(\frac{d}{dx} \left(\frac{5}{4} \sqrt{16-x^2} \right) = \frac{5}{4} \cdot \frac{1}{2\sqrt{16-x^2}} \cdot (-2x) = \frac{-5x}{4\sqrt{16-x^2}}\).

Step 5: Evaluate derivative at given point

Plug \(x = 2\) into \(\frac{dz}{dx}\): \(\frac{dz}{dx} \bigg|_{x=2} = \frac{-5(2)}{4\sqrt{16-2^2}} = \frac{-10}{4\sqrt{12}} = \frac{-10}{8\sqrt{3}} = -\frac{5}{4\sqrt{3}}\).

Step 6: Rationalize the denominator

Multiply the numerator and the denominator by \(\sqrt{3}\) to rationalize: \(-\frac{5}{4\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = -\frac{5\sqrt{3}}{12}\). Therefore, the slope of the tangent line is \(-\frac{5\sqrt{3}}{12}\).

Key Concepts

Tangent Line

A tangent line to a curve is a straight line that just "touches" the curve at a particular point, matching its slope at that point. The tangent line represents the instantaneous rate of change of the curve, effectively summarizing how steeply the curve rises or falls at the point of intersection.

• When the tangent line is drawn, it gives a linear approximation of the curve near the specific point.
• The slope of the tangent line is the main point of interest, especially when analyzing curves and surfaces.

For our given problem, when intersecting the cylinder and the plane, finding the tangent line involves differentiating the equation of the intersection curve.

Cylinder

A cylinder in mathematics is a 3D geometric surface where a pair of parallel lines extends through every point of a closed plane curve, typically a circle. Cylinders play an essential role in calculus, as they often interact with other geometric forms like planes.

• In the context of our problem, the cylinder is described by the equation: \[ 4z = 5 \sqrt{16-x^2} \]
• This description indicates a circular cylinder since it involves a square root of a term involving \(x^2\).

The goal is to understand how the cylinder intersects with a plane to produce a new curve.

Plane Intersection

The intersection between a plane and another geometric shape, such as a cylinder, results in a curve. This curve is a critical entity, as it provides various properties of the joint surfaces and is often the focus of calculus problems.

• The plane for this scenario is described by the equation: \( y = 3 \).
• By 'slicing' the cylinder with this plane, we can examine the curve where these two objects meet.

The challenge then becomes to describe this curve mathematically and to explore the properties, such as its tangent at a given point.

Derivative

The derivative in calculus symbolizes the rate of change of a function. It's a crucial concept when determining slopes and tangent lines. When we differentiate a function, we get another function that gives us the slope of the tangent at any point.

• To determine the slope of the tangent line at the curve's intersection, we differentiated the equation \( z = \frac{5}{4} \sqrt{16-x^2} \).
• This involved using the chain rule, due to the presence of a composite function (a function within a square root).

Once we find the derivative, we can substitute specific values to determine the slope at certain points, such as \( x = 2 \) in this problem. This discusses how the slope is calculated as \(-\frac{5\sqrt{3}}{12} \).