Math Problem Statement

write the eq of a line tangent to y=tan(2x) at x=pi/12

Solution

To find the equation of the line tangent to the function y=tan(2x)y = \tan(2x) at x=π12x = \frac{\pi}{12}, follow these steps:

1. Find the derivative of y=tan(2x)y = \tan(2x)

The derivative of y=tan(2x)y = \tan(2x) gives the slope of the tangent line. Using the chain rule, we differentiate tan(2x)\tan(2x):

dydx=ddx[tan(2x)]=2sec2(2x)\frac{dy}{dx} = \frac{d}{dx}[\tan(2x)] = 2 \sec^2(2x)

2. Evaluate the slope at x=π12x = \frac{\pi}{12}

Substitute x=π12x = \frac{\pi}{12} into the derivative to find the slope of the tangent line at this point:

dydxx=π12=2sec2(2×π12)=2sec2(π6)\frac{dy}{dx} \Bigg|_{x=\frac{\pi}{12}} = 2 \sec^2\left( 2 \times \frac{\pi}{12} \right) = 2 \sec^2\left(\frac{\pi}{6}\right)

We know sec(π6)=2\sec\left(\frac{\pi}{6}\right) = 2, so:

dydxx=π12=2×22=8\frac{dy}{dx} \Bigg|_{x=\frac{\pi}{12}} = 2 \times 2^2 = 8

The slope of the tangent line is 8.

3. Find the y-coordinate at x=π12x = \frac{\pi}{12}

Substitute x=π12x = \frac{\pi}{12} into the original function y=tan(2x)y = \tan(2x):

y(π12)=tan(2×π12)=tan(π6)y\left(\frac{\pi}{12}\right) = \tan\left(2 \times \frac{\pi}{12}\right) = \tan\left(\frac{\pi}{6}\right)

We know tan(π6)=13\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}, so the y-coordinate at x=π12x = \frac{\pi}{12} is 13\frac{1}{\sqrt{3}}.

4. Use the point-slope form to write the equation of the tangent line

The point-slope form 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 curve. We have m=8m = 8, x1=π12x_1 = \frac{\pi}{12}, and y1=13y_1 = \frac{1}{\sqrt{3}}. Substituting these values:

y13=8(xπ12)y - \frac{1}{\sqrt{3}} = 8 \left( x - \frac{\pi}{12} \right)

This is the equation of the tangent line.

Would you like more details or have any questions?

Here are some related questions you might want to explore:

  1. How do you find the normal line to a curve?
  2. What happens to the tangent line as the slope changes rapidly, like near asymptotes?
  3. Can we extend this method to find tangent lines for other trigonometric functions?
  4. How does the chain rule apply to differentiating composite functions in other contexts?
  5. How does the derivative of a trigonometric function change with different amplitudes and frequencies?

Tip: When dealing with trigonometric functions, it's often helpful to review the values of sine, cosine, and tangent at key angles, such as π6,π4,\frac{\pi}{6}, \frac{\pi}{4}, and π3\frac{\pi}{3}.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Calculus
Derivatives
Tangent Line

Formulas

Derivative of y = tan(2x): dy/dx = 2 sec^2(2x)
Point-slope form: y - y1 = m(x - x1)

Theorems

Chain Rule

Suitable Grade Level

Grades 11-12