Suppose you start at the point (5,6,7)
 and move 4
 units along the curve:

x(t)=2t+5,y(t)=4t3−−√3+6,z(t)=t22+7

in the positive direction. Where are you now?

We are given the initial position at \((5, 6, 7)\) and need to move 4 units along the curve defined by the parametric equations:

\[
x(t) = 2t + 5, \quad y(t) = \sqrt[3]{4t^3} + 6 = \left(4t^3\right)^{1/3} + 6, \quad z(t) = t^2 + 7
\]

We are asked to find the new position after moving 4 units along the curve in the positive direction.

### Step 1: Calculate the speed
The speed along the curve is given by the magnitude of the velocity vector, which is the derivative of each component with respect to \(t\):

\[
\frac{dx}{dt} = 2, \quad \frac{dy}{dt} = \frac{d}{dt} \left( \sqrt[3]{4t^3} + 6 \right) = 4, \quad \frac{dz}{dt} = 2t
\]

The speed is then the magnitude of the velocity vector:

\[
v(t) = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2} = \sqrt{2^2 + 4^2 + (2t)^2} = \sqrt{4 + 16 + 4t^2} = \sqrt{20 + 4t^2}
\]

### Step 2: Arc length formula
The arc length \(s(t)\) along the curve from \(t = 0\) is given by:

\[
s(t) = \int_0^t v(\tau) \, d\tau = \int_0^t \sqrt{20 + 4\tau^2} \, d\tau
\]

We need to find \(t\) such that the arc length \(s(t) = 4\), because we are moving 4 units along the curve.

Let's solve this next.It appears that solving the equation for arc length analytically is complicated, as it involves advanced functions. Instead, I will numerically approximate the value of \(t\) where the arc length is equal to 4 units.

Let's proceed with that next.It seems like I can’t do more advanced data analysis right now. Please try again later. 

If you'd like, I can guide you through the steps of solving the problem manually or assist with other questions.

Would you like to try a different approach or topic? Here are 5 related questions that expand on this:

1. How do you compute the arc length for curves in general?
2. How can you approximate solutions numerically when exact solutions are challenging?
3. What is the geometric meaning of parametric curves in three-dimensional space?
4. How does the velocity vector relate to the speed and direction of motion along a curve?
5. How can you use integration to solve problems in mechanics and motion?

**Tip:** When solving for arc length, if analytical solutions are complex, numerical integration is often a practical alternative.

Suppose you start at the point (5,6,7) and move 4 units along the curve: x(t) = 2t + 5, y(t) = sqrt[3]{4t^3} + 6, z(t) = t^2 + 7 in the positive direction. Where are you now?

Learn how to find the new position after moving 4 units along a 3D parametric curve defined by x(t) = 2t + 5, y(t) = sqrt[3]{4t^3} + 6, and z(t) = t^2 + 7. This solution covers arc length, velocity vector calculation, and numerical approximation methods, ideal for college-level calculus students.

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