Math Problem Statement

calculate the equation of the tangent line to f(t)=3t^2 × cos(t) at t=pi

Solution

To find the equation of the tangent line to the function f(t)=3t2cos(t)f(t) = 3t^2 \cos(t) at t=πt = \pi, we need to follow these steps:

  1. Calculate f(π)f(\pi) to find the point where the tangent touches the curve.
  2. Differentiate f(t)f(t) with respect to tt to get f(t)f'(t), the slope of the tangent line.
  3. Evaluate f(π)f'(\pi) to find the slope at t=πt = \pi.
  4. Use the point-slope form of the tangent line: yf(π)=f(π)(tπ)y - f(\pi) = f'(\pi) (t - \pi).

Step 1: Calculate f(π)f(\pi)

f(π)=3π2cos(π)=3π2(1)=3π2f(\pi) = 3\pi^2 \cos(\pi) = 3\pi^2 \cdot (-1) = -3\pi^2 So, the point on the curve at t=πt = \pi is (π,3π2)(\pi, -3\pi^2).

Step 2: Differentiate f(t)f(t) to find f(t)f'(t)

Since f(t)=3t2cos(t)f(t) = 3t^2 \cos(t), we apply the product rule: f(t)=ddt(3t2)cos(t)+3t2ddt(cos(t))f'(t) = \frac{d}{dt}(3t^2) \cdot \cos(t) + 3t^2 \cdot \frac{d}{dt}(\cos(t)) =(6t)cos(t)+3t2(sin(t))= (6t) \cos(t) + 3t^2 \cdot (-\sin(t)) =6tcos(t)3t2sin(t)= 6t \cos(t) - 3t^2 \sin(t)

Step 3: Evaluate f(π)f'(\pi)

Substitute t=πt = \pi into f(t)f'(t): f(π)=6πcos(π)3π2sin(π)f'(\pi) = 6\pi \cos(\pi) - 3\pi^2 \sin(\pi) Since cos(π)=1\cos(\pi) = -1 and sin(π)=0\sin(\pi) = 0: f(π)=6π(1)3π20=6πf'(\pi) = 6\pi \cdot (-1) - 3\pi^2 \cdot 0 = -6\pi The slope of the tangent line at t=πt = \pi is 6π-6\pi.

Step 4: Write the equation of the tangent line

Using the point-slope form yy1=m(tt1)y - y_1 = m(t - t_1), where m=6πm = -6\pi, t1=πt_1 = \pi, and y1=3π2y_1 = -3\pi^2: y+3π2=6π(tπ)y + 3\pi^2 = -6\pi (t - \pi) Simplifying, y=6πt+6π23π2y = -6\pi t + 6\pi^2 - 3\pi^2 y=6πt+3π2y = -6\pi t + 3\pi^2

Final Answer

The equation of the tangent line to f(t)=3t2cos(t)f(t) = 3t^2 \cos(t) at t=πt = \pi is: y=6πt+3π2y = -6\pi t + 3\pi^2

Would you like further details or have any questions?

Related Questions

  1. How do we find the tangent line of f(t)=t3sin(t)f(t) = t^3 \sin(t) at t=π/2t = \pi/2?
  2. What is the derivative of g(t)=etcos(t)g(t) = e^t \cos(t) using the product rule?
  3. How do we find the tangent line of a trigonometric function at a specific point?
  4. What is the geometric interpretation of the slope of the tangent line?
  5. How does the product rule apply when differentiating h(t)=t2sin(t2)h(t) = t^2 \sin(t^2)?

Tip: Remember that the product rule states: if u(t)u(t) and v(t)v(t) are functions of tt, then ddt[u(t)v(t)]=u(t)v(t)+u(t)v(t)\frac{d}{dt} [u(t)v(t)] = u'(t)v(t) + u(t)v'(t).

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Trigonometric Functions
Tangent Line

Formulas

Product Rule: (uv)' = u'v + uv'
Point-slope form of a line: y - y1 = m(x - x1)

Theorems

Product Rule for Differentiation

Suitable Grade Level

Grades 11-12