A curve is defined by the parametric equations \[x=2^t\] and \[y=8^t\].

**What is \[\dfrac{d^2y}{dx^2}\] in terms of \[t\] ?**

To find the second derivative \(\frac{d^2y}{dx^2}\) in terms of \(t\), we will follow these steps:

### Step 1: Find \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\)

Given the parametric equations:
\[
x = 2^t \quad \text{and} \quad y = 8^t
\]

First, compute \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\):

\[
\frac{dx}{dt} = \frac{d}{dt}(2^t) = 2^t \cdot \ln(2)
\]
\[
\frac{dy}{dt} = \frac{d}{dt}(8^t) = 8^t \cdot \ln(8)
\]

### Step 2: Find \(\frac{dy}{dx}\)

\(\frac{dy}{dx}\) is given by:
\[
\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{8^t \cdot \ln(8)}{2^t \cdot \ln(2)}
\]

Now, simplify this expression. Notice that \(8^t = (2^3)^t = (2^t)^3 = (2^t)^3 = 2^{3t}\):

\[
\frac{dy}{dx} = \frac{2^{3t} \cdot \ln(8)}{2^t \cdot \ln(2)} = 2^{3t - t} \cdot \frac{\ln(8)}{\ln(2)} = 2^{2t} \cdot \frac{\ln(8)}{\ln(2)}
\]

Since \(\ln(8) = \ln(2^3) = 3\ln(2)\), the expression simplifies to:
\[
\frac{dy}{dx} = 2^{2t} \cdot 3 = 3 \cdot 2^{2t}
\]

### Step 3: Find \(\frac{d^2y}{dx^2}\)

To find \(\frac{d^2y}{dx^2}\), differentiate \(\frac{dy}{dx}\) with respect to \(x\):
\[
\frac{d^2y}{dx^2} = \frac{d}{dx}\left(3 \cdot 2^{2t}\right)
\]

Since \(\frac{d}{dx}\) means we need to differentiate with respect to \(t\) and then divide by \(\frac{dx}{dt}\), we first find the derivative of \(3 \cdot 2^{2t}\) with respect to \(t\):

\[
\frac{d}{dt}\left(3 \cdot 2^{2t}\right) = 3 \cdot 2^{2t} \cdot 2 \ln(2) = 6 \cdot 2^{2t} \ln(2)
\]

Now, divide this by \(\frac{dx}{dt}\):
\[
\frac{d^2y}{dx^2} = \frac{6 \cdot 2^{2t} \ln(2)}{2^t \ln(2)} = 6 \cdot 2^{2t - t} = 6 \cdot 2^t
\]

### Final Answer:
\[
\frac{d^2y}{dx^2} = \boxed{6 \cdot 2^t}
\]

Would you like any further explanation or have other questions?

Here are some related questions:
1. How would you find \(\frac{d^2y}{dx^2}\) for different parametric equations?
2. What is the geometric interpretation of \(\frac{d^2y}{dx^2}\)?
3. How does the value of \(t\) affect the curvature of the curve?
4. What would \(\frac{dy}{dx}\) look like if \(x\) and \(y\) were linear functions of \(t\)?
5. Can you express \(\frac{d^2y}{dx^2}\) without using the parameter \(t\)?

**Tip:** When dealing with parametric equations, always be careful with the chain rule, especially when differentiating with respect to one parameter and then converting to another.

Explore how to find the second derivative d^2y/dx^2 in terms of t for the parametric equations x = 2^t and y = 8^t. This detailed solution covers differentiation of exponential functions and application of the chain rule.

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