Math Problem Statement

Find the tangent line and normal line

Find the tangent line and normal line to the function \( g(t) = (4\sin t + 4\cos t)^2 \) at \( t = \frac{\pi}{2} \).

Solution

To find the tangent line and normal line to the function g(t)=(4sint+4cost)2g(t) = (4\sin t + 4\cos t)^2 at t=π2t = \frac{\pi}{2}, we need to follow these steps:

  1. Find g(t)g(t) and g(t)g'(t) at t=π2t = \frac{\pi}{2}:

    • Evaluate g(t)g(t) at t=π2t = \frac{\pi}{2}.
    • Compute the derivative g(t)g'(t) and evaluate it at t=π2t = \frac{\pi}{2} for the slope of the tangent line.
  2. Equation of the Tangent Line:

    • Use the point-slope form of the line equation: yg(t0)=g(t0)(tt0)y - g(t_0) = g'(t_0)(t - t_0), where t0=π2t_0 = \frac{\pi}{2}.
  3. Equation of the Normal Line:

    • The slope of the normal line is the negative reciprocal of the slope of the tangent line.

Step 1: Evaluate g(t)g(t) at t=π2t = \frac{\pi}{2}

Given: g(t)=(4sint+4cost)2g(t) = (4\sin t + 4\cos t)^2

At t=π2t = \frac{\pi}{2}: sin(π2)=1,cos(π2)=0\sin\left(\frac{\pi}{2}\right) = 1, \quad \cos\left(\frac{\pi}{2}\right) = 0

Thus, g(π2)=(41+40)2=16g\left(\frac{\pi}{2}\right) = (4 \cdot 1 + 4 \cdot 0)^2 = 16

Step 2: Compute g(t)g'(t)

Use the chain rule: g(t)=(4sint+4cost)2g(t) = (4\sin t + 4\cos t)^2 Let u=4sint+4costu = 4\sin t + 4\cos t, then g(t)=u2g(t) = u^2.

The derivative is: g(t)=2ududtg'(t) = 2u \cdot \frac{du}{dt}

Now, find dudt\frac{du}{dt}: u=4sint+4cost    dudt=4cost4sintu = 4\sin t + 4\cos t \implies \frac{du}{dt} = 4\cos t - 4\sin t

Thus, g(t)=2(4sint+4cost)(4cost4sint)g'(t) = 2(4\sin t + 4\cos t)(4\cos t - 4\sin t)

Evaluate g(t)g'(t) at t=π2t = \frac{\pi}{2}: g(π2)=2(41+40)(4041)=2(4)(4)=32g'\left(\frac{\pi}{2}\right) = 2(4\cdot 1 + 4\cdot 0)(4\cdot 0 - 4\cdot 1) = 2(4)(-4) = -32

Step 3: Equation of the Tangent Line

Using the point (π2,16)\left(\frac{\pi}{2}, 16\right) and slope m=32m = -32: y16=32(tπ2)y - 16 = -32(t - \frac{\pi}{2}) So, the tangent line equation is: y=32t+16+32π2y = -32t + 16 + 32\frac{\pi}{2} y=32t+16+16πy = -32t + 16 + 16\pi

Step 4: Equation of the Normal Line

The slope of the normal line is the negative reciprocal of 32-32, which is 132\frac{1}{32}.

Thus, the normal line equation is: y16=132(tπ2)y - 16 = \frac{1}{32}(t - \frac{\pi}{2}) So, y=132tπ64+16y = \frac{1}{32}t - \frac{\pi}{64} + 16

Summary

  • Tangent line equation: y=32t+16+16πy = -32t + 16 + 16\pi
  • Normal line equation: y=132tπ64+16y = \frac{1}{32}t - \frac{\pi}{64} + 16

Would you like further details or have any questions?

Further Exploration

  1. How do we derive the equation of a tangent line in parametric form?
  2. What is the significance of the negative reciprocal in finding the normal line?
  3. How can the derivative g(t)g'(t) be interpreted geometrically?
  4. What happens to the tangent and normal lines if we choose a different point tt?
  5. Can we generalize this approach to three-dimensional functions?

Tip

Always check if the function simplifies before differentiating to avoid complex calculations.

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Math Problem Analysis

Mathematical Concepts

Calculus
Trigonometric Functions
Derivative
Tangent Line
Normal Line

Formulas

Point-slope form of the line equation: \( y - y_1 = m(x - x_1) \)
Chain rule: \( (f(g(t)))' = f'(g(t)) \cdot g'(t) \)

Theorems

Chain Rule for Differentiation

Suitable Grade Level

Grades 11-12