(2 points) Consider the parametric curve defined by the image of r⃗ (t)=⟨−6t2,t3−5t⟩
.




(c) Find d2y/dx2
 when t=2.5
.

d2ydx2∣∣∣t=2.5=

To find \( \frac{d^2y}{dx^2} \) for the parametric curve defined by the vector function \( \mathbf{r}(t) = \langle -6t^2, t^3 - 5t \rangle \) at \( t = 2.5 \), we proceed as follows:

### Step 1: Find \( \frac{dy}{dx} \)
From parametric equations, the slope of the curve is given by:

\[
\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}
\]

The components of \( \mathbf{r}(t) \) are:
- \( x(t) = -6t^2 \)
- \( y(t) = t^3 - 5t \)

Now, compute the derivatives:
- \( \frac{dx}{dt} = \frac{d}{dt}(-6t^2) = -12t \)
- \( \frac{dy}{dt} = \frac{d}{dt}(t^3 - 5t) = 3t^2 - 5 \)

So, we have:

\[
\frac{dy}{dx} = \frac{3t^2 - 5}{-12t}
\]

### Step 2: Find \( \frac{d^2y}{dx^2} \)
Next, we compute \( \frac{d^2y}{dx^2} \), which is the derivative of \( \frac{dy}{dx} \) with respect to \( x \). Using the chain rule, we get:

\[
\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{\frac{d}{dt} \left( \frac{dy}{dx} \right)}{\frac{dx}{dt}}
\]

First, find \( \frac{d}{dt} \left( \frac{dy}{dx} \right) \):
\[
\frac{d}{dt} \left( \frac{dy}{dx} \right) = \frac{d}{dt} \left( \frac{3t^2 - 5}{-12t} \right)
\]

We apply the quotient rule to differentiate:
\[
\frac{d}{dt} \left( \frac{3t^2 - 5}{-12t} \right) = \frac{(-12t)(6t) - (-12)(3t^2 - 5)}{(-12t)^2}
\]

Simplifying the numerator:
\[
(-12t)(6t) = -72t^2, \quad (-12)(3t^2 - 5) = -36t^2 + 60
\]

So the numerator becomes:
\[
-72t^2 - (-36t^2 + 60) = -72t^2 + 36t^2 - 60 = -36t^2 - 60
\]

Thus, we have:
\[
\frac{d}{dt} \left( \frac{dy}{dx} \right) = \frac{-36t^2 - 60}{144t^2}
\]

Simplifying:
\[
\frac{d}{dt} \left( \frac{dy}{dx} \right) = \frac{-36(t^2 + 5/3)}{144t^2} = \frac{-t^2 - 5/3}{4t^2}
\]

### Step 3: Evaluate at \( t = 2.5 \)
Now, we substitute \( t = 2.5 \) into the expression:

1. \( \frac{dx}{dt} \) at \( t = 2.5 \):
   \[
   \frac{dx}{dt} = -12(2.5) = -30
   \]

2. \( \frac{d}{dt} \left( \frac{dy}{dx} \right) \) at \( t = 2.5 \):
   \[
   \frac{d}{dt} \left( \frac{dy}{dx} \right) = \frac{-(2.5)^2 - 5/3}{4(2.5)^2} = \frac{-6.25 - 5/3}{4(6.25)}
   \]
   Simplifying further:
   \[
   = \frac{-6.25 - 1.6667}{25} = \frac{-7.9167}{25} \approx -0.3167
   \]

Finally, we compute \( \frac{d^2y}{dx^2} \):
\[
\frac{d^2y}{dx^2} = \frac{-0.3167}{-30} \approx 0.01056
\]

Thus, the second derivative \( \frac{d^2y}{dx^2} \) at \( t = 2.5 \) is approximately:

\[
\boxed{0.0106}
\]

Let me know if you want details on any steps!

---

### Related Questions:
1. How would you compute \( \frac{dy}{dx} \) for a general parametric curve?
2. What is the geometric interpretation of \( \frac{d^2y}{dx^2} \) on a curve?
3. Can \( \frac{d^2y}{dx^2} \) be negative? What does it represent in that case?
4. How does the chain rule apply to parametric curves in higher dimensions?
5. Can this method be extended to curves in three dimensions?

### Tip:
When dealing with parametric curves, always start by computing the first derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) to ensure you have the proper components for further calculations.

Consider the parametric curve defined by the image of r⃗ (t)=⟨−6t^2,t^3−5t⟩. (c) Find d²y/dx² when t=2.5.

Learn how to compute the second derivative (d²y/dx²) for a parametric curve defined by r(t) = ⟨-6t², t³-5t⟩ at t = 2.5. This detailed step-by-step solution uses key calculus concepts like parametric equations, derivatives, and the quotient and chain rules. Suitable for advanced high school students in Grades 11-12.

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