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

### Step 1: Compute the derivative \( g'(t) \)
We need to apply the chain rule to differentiate \( g(t) = (3 \sin t + 4 \cos t)^2 \). Let’s define:
\[ u(t) = 3 \sin t + 4 \cos t \]
Thus, the function becomes \( g(t) = u(t)^2 \), and using the chain rule, we have:
\[
g'(t) = 2u(t) \cdot u'(t)
\]
Now, we need to find \( u'(t) \). Differentiating \( u(t) = 3 \sin t + 4 \cos t \):
\[
u'(t) = 3 \cos t - 4 \sin t
\]
Therefore, the derivative \( g'(t) \) is:
\[
g'(t) = 2(3 \sin t + 4 \cos t) \cdot (3 \cos t - 4 \sin t)
\]

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

- At \( t = \frac{\pi}{2} \), we have:
  \[
  \sin \frac{\pi}{2} = 1 \quad \text{and} \quad \cos \frac{\pi}{2} = 0
  \]
- Substituting these into \( u(t) = 3 \sin t + 4 \cos t \):
  \[
  u\left( \frac{\pi}{2} \right) = 3 \times 1 + 4 \times 0 = 3
  \]
- Now, evaluate \( u'(t) = 3 \cos t - 4 \sin t \):
  \[
  u'\left( \frac{\pi}{2} \right) = 3 \times 0 - 4 \times 1 = -4
  \]

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

### Step 3: Find the equation of the tangent line

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

### 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 \( \frac{1}{24} \).

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

### Final Answers:
- **Tangent line**: \( y = -24t + 12\pi + 21 \)
- **Normal line**: \( y = \frac{1}{24}t - \frac{\pi}{48} + \frac{435}{48} \)

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.

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.

Explore how to find tangent and normal lines to the function g(t) = (3 sin t + 4 cos t)^2 at t = π/2 using calculus concepts like differentiation and the chain rule. This detailed guide provides formulas for computing slopes and equations of lines.

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