Math Problem Statement

Find the equations of the tangent and normal lines to the graph of the function at the given point. g(t) = (3 sin t + 4 cos t)^2 at t = π/2.

Solution

We are tasked with finding the tangent and normal lines to the graph of the function g(t)=(3sint+4cost)2g(t) = (3 \sin t + 4 \cos t)^2 at t=π2t = \frac{\pi}{2}.

Step 1: Compute the derivative g(t)g'(t)

We need to apply the chain rule to differentiate g(t)=(3sint+4cost)2g(t) = (3 \sin t + 4 \cos t)^2. Let’s define: u(t)=3sint+4costu(t) = 3 \sin t + 4 \cos t Thus, the function becomes g(t)=u(t)2g(t) = u(t)^2, and using the chain rule, we have: g(t)=2u(t)u(t)g'(t) = 2u(t) \cdot u'(t) Now, we need to find u(t)u'(t). Differentiating u(t)=3sint+4costu(t) = 3 \sin t + 4 \cos t: u(t)=3cost4sintu'(t) = 3 \cos t - 4 \sin t Therefore, the derivative g(t)g'(t) is: g(t)=2(3sint+4cost)(3cost4sint)g'(t) = 2(3 \sin t + 4 \cos t) \cdot (3 \cos t - 4 \sin t)

Step 2: Evaluate g(t)g(t) and g(t)g'(t) at t=π2t = \frac{\pi}{2}

  • At t=π2t = \frac{\pi}{2}, we have: sinπ2=1andcosπ2=0\sin \frac{\pi}{2} = 1 \quad \text{and} \quad \cos \frac{\pi}{2} = 0
  • Substituting these into u(t)=3sint+4costu(t) = 3 \sin t + 4 \cos t: u(π2)=3×1+4×0=3u\left( \frac{\pi}{2} \right) = 3 \times 1 + 4 \times 0 = 3
  • Now, evaluate u(t)=3cost4sintu'(t) = 3 \cos t - 4 \sin t: u(π2)=3×04×1=4u'\left( \frac{\pi}{2} \right) = 3 \times 0 - 4 \times 1 = -4

Now substitute into g(t)g'(t): g(π2)=2(3)(4)=24g'\left( \frac{\pi}{2} \right) = 2(3) \cdot (-4) = -24

Step 3: Find the equation of the tangent line

The tangent line at t=π2t = \frac{\pi}{2} has the form: yg(π2)=g(π2)(tπ2)y - g\left( \frac{\pi}{2} \right) = g'\left( \frac{\pi}{2} \right) (t - \frac{\pi}{2}) We already know g(π2)=24g'\left( \frac{\pi}{2} \right) = -24. Now, find g(π2)g\left( \frac{\pi}{2} \right): g(π2)=(3)2=9g\left( \frac{\pi}{2} \right) = (3)^2 = 9 Thus, the equation of the tangent line is: y9=24(tπ2)y - 9 = -24 \left( t - \frac{\pi}{2} \right) Simplifying: y=24t+12π+21y = -24t + 12\pi + 21

Step 4: Find the equation of the normal line

The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of the tangent line's slope, which is 124\frac{1}{24}.

Thus, the equation of the normal line is: y9=124(tπ2)y - 9 = \frac{1}{24} \left( t - \frac{\pi}{2} \right) Simplifying: y=124tπ48+43548y = \frac{1}{24}t - \frac{\pi}{48} + \frac{435}{48}

Final Answers:

  • Tangent line: y=24t+12π+21y = -24t + 12\pi + 21
  • Normal line: y=124tπ48+43548y = \frac{1}{24}t - \frac{\pi}{48} + \frac{435}{48}

Would you like more details, or do you have any questions?

Here are 5 related questions:

  1. How do you find the slope of a normal line in general?
  2. What is the geometric meaning of a tangent line?
  3. How does the chain rule apply in multivariable calculus?
  4. What are other techniques for differentiating products of functions?
  5. How do parametric curves affect tangent and normal line computations?

Tip: The negative reciprocal is key to finding the slope of a normal line perpendicular to a tangent.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Calculus
Differentiation
Slope of a line

Formulas

Chain rule: g'(t) = 2u(t) * u'(t)
Derivative of sine and cosine: u'(t) = 3 cos t - 4 sin t
Tangent line equation: y - g(t0) = g'(t0) (t - t0)
Normal line equation: y - g(t0) = -(1/g'(t0)) (t - t0)

Theorems

Chain Rule

Suitable Grade Level

College/University