Suppose you start at the point (10,10,6) and move 8 units along the curve: x(t)=2t+10,y(t)=4sqrt(t^3)/3+10,z(t)=t^2/2+6 in the positive direction. Where are you now? This is very important to my career, please take your time and go through any necessary steps.
( Hint: Your answer may be quite messy. You do not need to simplify.)

Let's walk through this step by step, as it's important to be accurate.

### Step 1: Understand the position on the curve

You are given the parametric equations of a curve:
\[
x(t) = 2t + 10, \quad y(t) = \frac{4\sqrt{t^3}}{3} + 10, \quad z(t) = \frac{t^2}{2} + 6.
\]
The goal is to find the new position on the curve after moving **8 units** from the initial point \((10, 10, 6)\).

### Step 2: Confirm the initial value of \( t \)

The initial point is \( (10, 10, 6) \), which corresponds to the initial parameter \( t \). To find the value of \( t \) for which the position on the curve matches this initial point, we solve each parametric equation:

1. **For \(x(t) = 10\):**
   \[
   2t + 10 = 10 \implies t = 0.
   \]
   
   So, the initial point corresponds to \( t = 0 \).

### Step 3: Compute the arc length to move along the curve

To find the new position after moving 8 units, we need to compute the arc length \( S \) along the curve. The formula for the differential of arc length is:
\[
ds = \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2} \, dt.
\]

Let's compute the necessary derivatives:

1. \(\frac{dx}{dt} = 2\),
2. \(\frac{dy}{dt} = \frac{d}{dt} \left( \frac{4\sqrt{t^3}}{3} + 10 \right) = \frac{2t^{1/2}}{\sqrt{3}}\),
3. \(\frac{dz}{dt} = t\).

Now, we plug these into the arc length formula:
\[
ds = \sqrt{(2)^2 + \left( \frac{2t^{1/2}}{\sqrt{3}} \right)^2 + (t)^2} \, dt.
\]
Simplifying the expression under the square root:
\[
ds = \sqrt{4 + \frac{4t}{3} + t^2} \, dt.
\]

Thus, the arc length \( S \) as a function of \( t \) is:
\[
S(t) = \int_0^t \sqrt{4 + \frac{4t}{3} + t^2} \, dt.
\]

### Step 4: Solve for \( t \) such that the arc length equals 8

To find the new position after moving 8 units, we need to solve:
\[
S(t) = 8.
\]
While this integral doesn't have a simple closed form, we can approach it numerically to find the value of \( t \).

### Step 5: Calculate the new position

Once we find the value of \( t \) such that \( S(t) = 8 \), we can substitute that \( t \) value back into the parametric equations to find the new coordinates \((x(t), y(t), z(t))\).

Let me calculate the value of \( t \) that satisfies this condition.After moving 8 units along the curve, you are at the parameter value \( t \approx 2.802 \). The corresponding coordinates on the curve are:

\[
(x, y, z) \approx (15.604, 16.255, 9.926).
\]

Let me know if you'd like any clarifications or further details!

Here are five related questions to further explore the topic:
1. How is arc length calculated for general parametric curves?
2. Why is the square root necessary in the arc length formula?
3. How do changes in each parametric equation affect the total arc length?
4. What happens to the curvature of a curve as \( t \) increases?
5. How would the process differ if the curve was given in Cartesian form?

**Tip:** Always remember that calculating arc lengths often requires numerical methods unless the integrals simplify nicely.

Suppose you start at the point (10,10,6) and move 8 units along the curve: x(t)=2t+10, y(t)=4sqrt(t^3)/3+10, z(t)=t^2/2+6 in the positive direction. Where are you now?

Learn how to calculate the new position after moving 8 units along a parametric curve given by x(t)=2t+10, y(t)=(4sqrt(t^3))/3+10, and z(t)=t^2/2+6. Explore the arc length formula and detailed calculus steps in this undergraduate-level problem.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation