(1 point) Suppose you start at the point (7,3,8)
 and move 3
 units along the curve:

𝑥(𝑡)=2𝑡+7,𝑦(𝑡)=4𝑡3‾‾√3+3,𝑧(𝑡)=𝑡22+8

in the positive direction. Where are you now?



We are given a point \((7, 3, 8)\) and a parameterized curve described by:

\[
x(t) = 2t + 7, \quad y(t) = 4t^{3/2} + 3, \quad z(t) = \frac{t^2}{2} + 8
\]

We need to find where we end up after moving **3 units** along the curve in the positive direction.

### Step 1: Parametrize the curve.
The velocity vector along the curve is the derivative of the position functions \( x(t), y(t), z(t) \) with respect to \( t \):

\[
\mathbf{v}(t) = \left( \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \right)
\]

So, we differentiate each component:

\[
\frac{dx}{dt} = 2, \quad \frac{dy}{dt} = 6t^{1/2}, \quad \frac{dz}{dt} = t
\]

Thus, the velocity vector is:

\[
\mathbf{v}(t) = \langle 2, 6t^{1/2}, t \rangle
\]

### Step 2: Calculate the speed.
The speed is the magnitude of the velocity vector:

\[
\|\mathbf{v}(t)\| = \sqrt{(2)^2 + (6t^{1/2})^2 + (t)^2} = \sqrt{4 + 36t + t^2}
\]

### Step 3: Arc length formula.
The distance traveled along a curve is given by the arc length formula:

\[
s(t) = \int_0^t \|\mathbf{v}(\tau)\| \, d\tau = \int_0^t \sqrt{4 + 36\tau + \tau^2} \, d\tau
\]

We need to find \( t \) such that the arc length equals 3 units, i.e.,

\[
\int_0^t \sqrt{4 + 36\tau + \tau^2} \, d\tau = 3
\]

Solving this integral directly can be complex, so let me approximate this numerically to find the value of \( t \) corresponding to 3 units of movement.The value of \( t \) that corresponds to moving 3 units along the curve is approximately \( t \approx 0.739 \).

### Step 4: Calculate the new position.
Now, using \( t = 0.739 \), we can find the new position by plugging this value into the parametric equations for \( x(t), y(t), z(t) \):

\[
x(0.739) = 2(0.739) + 7 = 8.478
\]
\[
y(0.739) = 4(0.739)^{3/2} + 3 \approx 4.957
\]
\[
z(0.739) = \frac{(0.739)^2}{2} + 8 \approx 8.273
\]

Thus, after moving 3 units along the curve, your new position is approximately:

\[
(8.478, 4.957, 8.273)
\]

Would you like any further details or explanations?

Here are 5 related questions to expand on this topic:
1. How can you approximate arc length for curves with complex parameterizations?
2. How does the velocity vector help in understanding movement along a curve?
3. What are some techniques for solving integrals involving arc length?
4. How does curvature of a curve affect arc length?
5. What other methods can be used to solve this kind of parametric motion problem?

**Tip**: Numerical methods like integration and root-finding are often used when exact solutions to arc length problems are difficult to obtain.

Suppose you start at the point (7,3,8) and move 3 units along the curve: x(t)=2t+7, y(t)=4t^(3/2)+3, z(t)=t^2/2+8 in the positive direction. Where are you now?

Learn how to determine the new position after moving 3 units along a parametric curve from the point (7, 3, 8). This solution involves parametric equations, vector calculus, and the arc length formula. Suitable for college-level calculus students.

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