Problem 29 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that the curves \(y=\sqrt{2} \sin x\) and \(y=\sqrt{2} \cos x\) intersect at right angles at a certain point with \(0

Step-by-Step Solution

Answer

The curves intersect at right angles at \(x = \frac{\pi}{4}\).

Step 1: Find Intersection Points

To find the intersection points of the curves, equate the equations for the curves: \(\sqrt{2} \sin x = \sqrt{2} \cos x\). Simplify this to \(\tan x = 1\). Solving for \(x\) gives \(x = \frac{\pi}{4}\), which is in the given interval \(0

Step 2: Compute Derivatives

Find the derivatives of each curve to determine their slopes at the intersection point. The derivative of \(y = \sqrt{2} \sin x\) is \(y' = \sqrt{2} \cos x\). The derivative of \(y = \sqrt{2} \cos x\) is \(y' = -\sqrt{2} \sin x\).

Step 3: Evaluate Slopes at Intersection

At \(x = \frac{\pi}{4}\), evaluate the derivatives. For \(y' = \sqrt{2} \cos x\), substitute \(x = \frac{\pi}{4}\) to get \(y' = \sqrt{2} \frac{1}{\sqrt{2}} = 1\). For \(y' = -\sqrt{2} \sin x\), substitute \(x = \frac{\pi}{4}\) to get \(y' = -\sqrt{2} \frac{1}{\sqrt{2}} = -1\).

Step 4: Determine Perpendicularity

Two curves intersect at right angles if the product of their slopes is \(-1\). At \(x = \frac{\pi}{4}\), the product of the slopes is \(1 \times (-1) = -1\). Therefore, the curves intersect at right angles.

Key Concepts

Trigonometric Functions

Trigonometric functions are essential in describing periodic phenomena or patterns, such as waves and circular motion. In our exercise, the curves provided use two fundamental trigonometric functions: sine and cosine:

• For the first curve, we have the function: \( y = \sqrt{2} \sin x \).
• For the second curve: \( y = \sqrt{2} \cos x \).

These functions oscillate between their respective maximum and minimum values. Both have a period of \( 2\pi \), meaning they repeat their values every \( 2\pi \) units, although in our task we focus on an interval from \( 0 \) to \( \frac{\pi}{2} \).
The intersection occurs when these trigonometric functions essentially equate, leading to a special scenario where \( \tan x = 1 \) is fulfilled, indicating a classic property where sine equals cosine at particular points.

Derivative Computation

Derivative computation allows us to find the rate of change or slope of the curve at any point. Think of it like assessing how steep a hill is at a specific location.
For the given curves:

• The derivative of \( y = \sqrt{2} \sin x \) is given by \( \frac{d}{dx}(\sqrt{2} \sin x) = \sqrt{2} \cos x \).
• The derivative of \( y = \sqrt{2} \cos x \) is \( \frac{d}{dx}(\sqrt{2} \cos x) = -\sqrt{2} \sin x \).

These derivatives tell us how rapidly the functions are increasing or decreasing at any given point.
Understanding the derivatives is crucial to evaluate the behavior of one function compared to another, especially when determining angles at intersections.

Slope Evaluation

Slope evaluation is the method of determining the steepness or inclination of a line at a specific point on a curve using its derivative.
By computing the derivative at an intersection point, we evaluate how each curve behaves there:

• For \( y' = \sqrt{2} \cos x \), substituting \( x = \frac{\pi}{4} \) results in \( y' = 1 \).
• For \( y' = -\sqrt{2} \sin x \), substituting \( x = \frac{\pi}{4} \) results in \( y' = -1 \).

This indicates one line has a positive slope whereas the other has a negative one at the intersection point, emphasizing contrasting directions.
Such evaluations are vital for geometric interpretations, such as determining the angle of intersection.

Intersection Points Calculation

Finding intersection points involves setting two equations equal to determine where the curves meet. This step solves where the behavior of both is identical in value.
For the curves \( y = \sqrt{2} \sin x \) and \( y = \sqrt{2} \cos x \), equating them yields:

• \( \sqrt{2} \sin x = \sqrt{2} \cos x \).
• Simplifying gives \( \tan x = 1 \), leading to \( x = \frac{\pi}{4} \).

This solution is within the given interval, indicating a valid intersection at \( x = \frac{\pi}{4} \).
Calculating these points precisely uncovers where to apply other mathematical analyses, such as derivatives and slope assessments, to fully grasp the geometrical insights of the curves' intersection.

Previous

Problem 28

Next

Problem 29

Other exercises in this chapter

Problem 28

The rate of change of electric charge with respect to time is called current. Suppose that \(\frac{1}{3} t^{3}+t\) coulombs of charge flow through a wire in \(t

View solution

Problem 28

In Problems 21-28, find the indicated derivative. $$ \frac{d y}{d t}, \text { where } y=\left[\sin t \tan \left(t^{2}+1\right)\right] $$

View solution

Problem 29

If \(s=\frac{1}{2} t^{4}-5 t^{3}+12 t^{2}\), find the velocity of the moving object when its acceleration is zero.

View solution

Problem 29

Find the indicated derivative. \(D_{x} e^{\sqrt{x+2}}\)

View solution

Key Concepts

Trigonometric Functions

Trigonometric functions are essential in describing periodic phenomena or patterns, such as waves and circular motion. In our exercise, the curves provided use two fundamental trigonometric functions: sine and cosine:

• For the first curve, we have the function: \( y = \sqrt{2} \sin x \).
• For the second curve: \( y = \sqrt{2} \cos x \).

These functions oscillate between their respective maximum and minimum values. Both have a period of \( 2\pi \), meaning they repeat their values every \( 2\pi \) units, although in our task we focus on an interval from \( 0 \) to \( \frac{\pi}{2} \).
The intersection occurs when these trigonometric functions essentially equate, leading to a special scenario where \( \tan x = 1 \) is fulfilled, indicating a classic property where sine equals cosine at particular points.

Derivative Computation

Derivative computation allows us to find the rate of change or slope of the curve at any point. Think of it like assessing how steep a hill is at a specific location.
For the given curves:

• The derivative of \( y = \sqrt{2} \sin x \) is given by \( \frac{d}{dx}(\sqrt{2} \sin x) = \sqrt{2} \cos x \).
• The derivative of \( y = \sqrt{2} \cos x \) is \( \frac{d}{dx}(\sqrt{2} \cos x) = -\sqrt{2} \sin x \).

These derivatives tell us how rapidly the functions are increasing or decreasing at any given point.
Understanding the derivatives is crucial to evaluate the behavior of one function compared to another, especially when determining angles at intersections.

Slope Evaluation

Slope evaluation is the method of determining the steepness or inclination of a line at a specific point on a curve using its derivative.
By computing the derivative at an intersection point, we evaluate how each curve behaves there:

• For \( y' = \sqrt{2} \cos x \), substituting \( x = \frac{\pi}{4} \) results in \( y' = 1 \).
• For \( y' = -\sqrt{2} \sin x \), substituting \( x = \frac{\pi}{4} \) results in \( y' = -1 \).

This indicates one line has a positive slope whereas the other has a negative one at the intersection point, emphasizing contrasting directions.
Such evaluations are vital for geometric interpretations, such as determining the angle of intersection.

Intersection Points Calculation

Finding intersection points involves setting two equations equal to determine where the curves meet. This step solves where the behavior of both is identical in value.
For the curves \( y = \sqrt{2} \sin x \) and \( y = \sqrt{2} \cos x \), equating them yields:

• \( \sqrt{2} \sin x = \sqrt{2} \cos x \).
• Simplifying gives \( \tan x = 1 \), leading to \( x = \frac{\pi}{4} \).

This solution is within the given interval, indicating a valid intersection at \( x = \frac{\pi}{4} \).
Calculating these points precisely uncovers where to apply other mathematical analyses, such as derivatives and slope assessments, to fully grasp the geometrical insights of the curves' intersection.