https://file.aiquickdraw.com/mathbot/file/1730685944600zbazg877.jpg

To solve this problem, we need to determine the points on the curve \( x = 4t^3 \) and \( y = 2 + 16t - 8t^2 \) where the tangent line has a slope of \( 1 \).

### Solution Steps
1. **Find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \):**
   - Since \( x = 4t^3 \), we differentiate with respect to \( t \):
     \[
     \frac{dx}{dt} = 12t^2
     \]
   - Since \( y = 2 + 16t - 8t^2 \), we differentiate with respect to \( t \):
     \[
     \frac{dy}{dt} = 16 - 16t
     \]

2. **Calculate \( \frac{dy}{dx} \):**
   - The slope of the tangent line \( \frac{dy}{dx} \) is given by:
     \[
     \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{16 - 16t}{12t^2} = \frac{16(1 - t)}{12t^2} = \frac{4(1 - t)}{3t^2}
     \]

3. **Set \( \frac{dy}{dx} = 1 \):**
   - To find the points where the slope of the tangent is \( 1 \), we set \( \frac{dy}{dx} = 1 \):
     \[
     \frac{4(1 - t)}{3t^2} = 1
     \]
   - Multiply both sides by \( 3t^2 \):
     \[
     4(1 - t) = 3t^2
     \]
     \[
     4 - 4t = 3t^2
     \]
   - Rearrange to form a quadratic equation:
     \[
     3t^2 + 4t - 4 = 0
     \]

4. **Solve the quadratic equation for \( t \):**
   - Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a = 3 \), \( b = 4 \), and \( c = -4 \):
     \[
     t = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3}
     \]
     \[
     t = \frac{-4 \pm \sqrt{16 + 48}}{6}
     \]
     \[
     t = \frac{-4 \pm \sqrt{64}}{6}
     \]
     \[
     t = \frac{-4 \pm 8}{6}
     \]
   - This gives two solutions:
     \[
     t = \frac{4}{6} = \frac{2}{3} \quad \text{and} \quad t = \frac{-12}{6} = -2
     \]

5. **Find the points on the curve for these values of \( t \):**
   - For \( t = \frac{2}{3} \):
     - \( x = 4\left(\frac{2}{3}\right)^3 = 4 \cdot \frac{8}{27} = \frac{32}{27} \)
     - \( y = 2 + 16\left(\frac{2}{3}\right) - 8\left(\frac{2}{3}\right)^2 \)
       \[
       = 2 + \frac{32}{3} - \frac{32}{9} = \frac{18 + 96 - 32}{9} = \frac{82}{9}
       \]
     - Point: \( \left( \frac{32}{27}, \frac{82}{9} \right) \)

   - For \( t = -2 \):
     - \( x = 4(-2)^3 = 4 \cdot (-8) = -32 \)
     - \( y = 2 + 16(-2) - 8(-2)^2 \)
       \[
       = 2 - 32 - 32 = -62
       \]
     - Point: \( (-32, -62) \)

### Answer
The points on the curve where the tangent line has a slope of \( 1 \) are:
\[
\left( \frac{32}{27}, \frac{82}{9} \right) \quad \text{and} \quad (-32, -62)
\]

Would you like more details or have any questions?

---

### Related Questions
1. How do we find the tangent line of a parametric curve at a given point?
2. What is the interpretation of \( \frac{dy}{dx} \) for parametric equations?
3. How can we use implicit differentiation to find slopes of tangent lines?
4. What are other methods to determine points of a curve with a given slope?
5. How does changing \( t \) affect the curve and the slope of its tangent?

### Tip
For parametric curves, always start by finding \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) separately before calculating \( \frac{dy}{dx} \). This helps in isolating the slope calculation.

At what points on the given curve x = 4t^3, y = 2 + 16t - 8t^2 does the tangent line have slope 1?

Discover how to find points on the curve x = 4t^3, y = 2 + 16t - 8t^2 where the tangent line has a slope of 1. This solution explores parametric equations, differentiation, and calculating dy/dx. Perfect for students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation