A particle moves along the curve x = t^5 +15, y= 2t^2 -10 and z=3t+50 where t is the time. Find the components of its velocity and acceleration at t=1 in the direction i+ 2j +2k.

To solve the problem, we'll first calculate the components of velocity and acceleration in the \(x\), \(y\), and \(z\) directions. Then, we'll determine the velocity and acceleration at \(t = 1\). Finally, we'll project these vectors onto the direction vector \( \mathbf{i} + 2\mathbf{j} + 2\mathbf{k} \).

### Step 1: Find the Velocity Components
The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j} + z(t)\mathbf{k} \) with respect to time \( t \):

Given:
- \( x(t) = t^5 + 15 \)
- \( y(t) = 2t^2 - 10 \)
- \( z(t) = 3t + 50 \)

Compute the derivatives:
- \( \frac{dx}{dt} = v_x(t) = 5t^4 \)
- \( \frac{dy}{dt} = v_y(t) = 4t \)
- \( \frac{dz}{dt} = v_z(t) = 3 \)

So, the velocity vector is:
\[
\mathbf{v}(t) = 5t^4 \mathbf{i} + 4t \mathbf{j} + 3 \mathbf{k}
\]

At \( t = 1 \):
\[
\mathbf{v}(1) = 5(1)^4 \mathbf{i} + 4(1) \mathbf{j} + 3 \mathbf{k} = 5\mathbf{i} + 4\mathbf{j} + 3\mathbf{k}
\]

### Step 2: Find the Acceleration Components
The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \):

Compute the derivatives:
- \( \frac{d^2x}{dt^2} = a_x(t) = 20t^3 \)
- \( \frac{d^2y}{dt^2} = a_y(t) = 4 \)
- \( \frac{d^2z}{dt^2} = a_z(t) = 0 \)

So, the acceleration vector is:
\[
\mathbf{a}(t) = 20t^3 \mathbf{i} + 4 \mathbf{j} + 0 \mathbf{k}
\]

At \( t = 1 \):
\[
\mathbf{a}(1) = 20(1)^3 \mathbf{i} + 4 \mathbf{j} + 0 \mathbf{k} = 20\mathbf{i} + 4\mathbf{j} + 0\mathbf{k}
\]

### Step 3: Direction Vector and Normalization
Given direction vector:
\[
\mathbf{d} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k}
\]

Normalize the direction vector:
\[
|\mathbf{d}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3
\]
\[
\hat{\mathbf{d}} = \frac{1}{3} (\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})
\]

### Step 4: Project Velocity and Acceleration onto the Direction Vector
- **Velocity projection**:
\[
\text{Velocity in direction of } \mathbf{d} = \mathbf{v}(1) \cdot \hat{\mathbf{d}} = (5\mathbf{i} + 4\mathbf{j} + 3\mathbf{k}) \cdot \frac{1}{3}(\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})
\]
\[
= \frac{1}{3} (5(1) + 4(2) + 3(2)) = \frac{1}{3} (5 + 8 + 6) = \frac{19}{3}
\]

- **Acceleration projection**:
\[
\text{Acceleration in direction of } \mathbf{d} = \mathbf{a}(1) \cdot \hat{\mathbf{d}} = (20\mathbf{i} + 4\mathbf{j}) \cdot \frac{1}{3} (\mathbf{i} + 2\mathbf{j} + 2\mathbf{k})
\]
\[
= \frac{1}{3} (20(1) + 4(2) + 0(2)) = \frac{1}{3} (20 + 8) = \frac{28}{3}
\]

### Final Results:
- **Velocity component in the direction \( \mathbf{i} + 2\mathbf{j} + 2\mathbf{k} \) at \( t = 1 \)**: \( \frac{19}{3} \) units.
- **Acceleration component in the direction \( \mathbf{i} + 2\mathbf{j} + 2\mathbf{k} \) at \( t = 1 \)**: \( \frac{28}{3} \) units.

Would you like more details or have any questions? Here are some related questions to expand your understanding:

1. How do you calculate the magnitude of a vector?
2. What is the physical significance of projecting a vector onto another vector?
3. How does the velocity vector change as the parameter \( t \) increases?
4. What is the geometric interpretation of a direction vector?
5. How can you use vectors to describe motion in three-dimensional space?

**Tip:** When dealing with vector projections, always ensure the direction vector is normalized to maintain accurate component calculations.

Explore the velocity and acceleration vectors of a particle moving along a parametric curve x = t^5 + 15, y = 2t^2 - 10, z = 3t + 50. Understand how to calculate velocity and acceleration components at t = 1 and project them onto a specified direction vector. Suitable for advanced high school students and calculus enthusiasts.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation