Math Problem Statement

find an equation of the tangent line to the curve y at the point. y = 4x sin(x) , (π2, 2π)\frac{\pi}{2},\text{ }2\pi)

Solution

To find the equation of the tangent line to the curve y=4xsin(x)y = 4x \sin(x) at the point (π2,2π)\left( \frac{\pi}{2}, 2\pi \right), we need two things:

  1. The derivative of the function yy, which gives the slope of the tangent line.
  2. The point on the curve, which we already have as (π2,2π)\left( \frac{\pi}{2}, 2\pi \right).

Step 1: Find the derivative of y=4xsin(x)y = 4x \sin(x)

To differentiate y=4xsin(x)y = 4x \sin(x), we apply the product rule:

[ \frac{d}{dx}[4x \sin(x)] = \frac{d}{dx}[4x] \cdot \sin(x) + 4x \cdot \frac{d}{dx}[\sin(x)] ]

The derivative of 4x4x is 4, and the derivative of sin(x)\sin(x) is cos(x)\cos(x). So:

dydx=4sin(x)+4xcos(x)\frac{dy}{dx} = 4 \sin(x) + 4x \cos(x)

Step 2: Evaluate the derivative at x=π2x = \frac{\pi}{2}

We substitute x=π2x = \frac{\pi}{2} into the derivative to find the slope of the tangent line at that point.

dydxx=π2=4sin(π2)+4π2cos(π2)\frac{dy}{dx} \Bigg|_{x = \frac{\pi}{2}} = 4 \sin\left( \frac{\pi}{2} \right) + 4 \cdot \frac{\pi}{2} \cdot \cos\left( \frac{\pi}{2} \right)

Since sin(π2)=1\sin\left( \frac{\pi}{2} \right) = 1 and cos(π2)=0\cos\left( \frac{\pi}{2} \right) = 0, we get:

dydxx=π2=41+4π20=4\frac{dy}{dx} \Bigg|_{x = \frac{\pi}{2}} = 4 \cdot 1 + 4 \cdot \frac{\pi}{2} \cdot 0 = 4

So, the slope of the tangent line is m=4m = 4.

Step 3: Use the point-slope form of the line equation

The point-slope form of the equation of a line is:

yy1=m(xx1)y - y_1 = m(x - x_1)

Where mm is the slope and (x1,y1)(x_1, y_1) is the point on the line. Substituting m=4m = 4, x1=π2x_1 = \frac{\pi}{2}, and y1=2πy_1 = 2\pi:

y2π=4(xπ2)y - 2\pi = 4\left(x - \frac{\pi}{2}\right)

Step 4: Simplify the equation

Now, we simplify the equation:

y2π=4x2πy - 2\pi = 4x - 2\pi

Adding 2π2\pi to both sides:

y=4xy = 4x

Thus, the equation of the tangent line is:

y=4xy = 4x

Summary:

The equation of the tangent line to the curve y=4xsin(x)y = 4x \sin(x) at the point (π2,2π)\left( \frac{\pi}{2}, 2\pi \right) is y=4xy = 4x.

Would you like more details on any of the steps? Or do you have other questions?

5 Follow-up Questions:

  1. How do you differentiate products of functions (product rule)?
  2. What does the point-slope form of a line equation look like?
  3. How can I use the derivative to determine if a function is increasing or decreasing?
  4. How do I evaluate trigonometric functions like sin(x)\sin(x) and cos(x)\cos(x) at key points?
  5. How can I find the equation of the tangent line for other functions?

Tip: When using the product rule, remember that it applies when you have two functions multiplied together. The rule is: ddx[uv]=uv+uv\frac{d}{dx}[uv] = u'v + uv', where uu and vv are functions of xx.

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Tangent Lines
Trigonometric Functions

Formulas

Product Rule: d/dx[uv] = u'v + uv'
Point-Slope Form of a Line: y - y1 = m(x - x1)

Theorems

Product Rule for differentiation
Point-Slope Formula for a line

Suitable Grade Level

Grades 11-12