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.