Math Problem Statement

If x ∈ [0, 2π], find the x-coordinate of the curve y = e^x cos x if the slope of its tangent is parallel to the x-axis.

Solution

To solve this problem, let's break it down systematically.


The problem asks to find the xx-coordinate of the curve y=excosxy = e^x \cos x where the slope of the tangent is parallel to the xx-axis. This condition implies that the slope is 00, i.e., dydx=0\frac{dy}{dx} = 0.

Step 1: Find the derivative dydx\frac{dy}{dx}

The function is: y=excosxy = e^x \cos x

Using the product rule for derivatives (uv)=uv+uv(uv)' = u'v + uv', let:

  • u=exu = e^x, so u=exu' = e^x,
  • v=cosxv = \cos x, so v=sinxv' = -\sin x.

Now, dydx=ddx(excosx)=ex(sinx)+ex(cosx).\frac{dy}{dx} = \frac{d}{dx}(e^x \cos x) = e^x(-\sin x) + e^x(\cos x).

Simplify: dydx=ex(cosxsinx).\frac{dy}{dx} = e^x(\cos x - \sin x).


Step 2: Set the derivative equal to zero

The slope of the tangent is parallel to the xx-axis when: dydx=0.\frac{dy}{dx} = 0.

From the equation: ex(cosxsinx)=0.e^x (\cos x - \sin x) = 0.

Since ex0e^x \neq 0 for any xx, the equation simplifies to: cosxsinx=0.\cos x - \sin x = 0.


Step 3: Solve cosxsinx=0\cos x - \sin x = 0

Rearrange: cosx=sinx.\cos x = \sin x.

Divide both sides by cosx\cos x (valid where cosx0\cos x \neq 0): 1=tanx.1 = \tan x.

Thus: tanx=1.\tan x = 1.

The general solution for tanx=1\tan x = 1 is: x=π4+nπ,nZ.x = \frac{\pi}{4} + n\pi, \quad n \in \mathbb{Z}.


Step 4: Restrict the solution to [0,2π][0, 2\pi]

Within the interval [0,2π][0, 2\pi], the solutions are: x=π4,x=5π4.x = \frac{\pi}{4}, \quad x = \frac{5\pi}{4}.


Final Answer:

The xx-coordinates where the slope of the tangent is parallel to the xx-axis are: x=π4,x=5π4.x = \frac{\pi}{4}, \quad x = \frac{5\pi}{4}.


Would you like a deeper explanation or a visual illustration of the solution?

Here are 5 related questions to expand your understanding:

  1. What is the significance of the derivative being zero in terms of the curve's behavior?
  2. Can we generalize this method for any function y=f(x)g(x)y = f(x)g(x)?
  3. What happens if y=exsinxy = e^x \sin x instead? How does the solution change?
  4. Why is tanx=1\tan x = 1 periodic, and how is the period determined?
  5. How can you verify the solution graphically?

Tip: Always ensure to apply the product rule carefully when differentiating products of functions.

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Trigonometric Equations

Formulas

Product Rule: (uv)' = u'v + uv'
Trigonometric Identity: tan(x) = sin(x) / cos(x)

Theorems

Conditions for horizontal tangents (dy/dx = 0)

Suitable Grade Level

Grade 12 or early undergraduate