SOLUTION: A point travels as described by the following parametric equations x=10t+10cos3.14t, y=20t+10sin3.14t and z=30t, where x,y,z, are in meters, t in seconds, all angles in radians. Th

Question 1199547: A point travels as described by the following parametric equations x=10t+10cos3.14t, y=20t+10sin3.14t and z=30t, where x,y,z, are in meters, t in seconds, all angles in radians. The vector locating the body at any time is r=ix+jy+kz. Determine the magnitude of the velocity of the body in meters per second at time t = 0.75 second.
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959)   (Show Source): You can put this solution on YOUR website!
To determine the magnitude of the velocity of the body at \( t = 0.75 \) seconds, we follow these steps:
---
### **Step 1: Velocity vector**
The velocity vector \( \mathbf{v} \) is the time derivative of the position vector \( \mathbf{r} \):
\[
\mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} + z(t) \mathbf{k}
\]
\[
\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt} = \frac{dx(t)}{dt} \mathbf{i} + \frac{dy(t)}{dt} \mathbf{j} + \frac{dz(t)}{dt} \mathbf{k}
\]
---
### **Step 2: Compute derivatives**
The given parametric equations are:
\[
x = 10t + 10\cos(\pi t), \quad y = 20t + 10\sin(\pi t), \quad z = 30t
\]
1. Derivative of \( x \):
\[
\frac{dx}{dt} = 10 - 10\pi \sin(\pi t)
\]
2. Derivative of \( y \):
\[
\frac{dy}{dt} = 20 + 10\pi \cos(\pi t)
\]
3. Derivative of \( z \):
\[
\frac{dz}{dt} = 30
\]
Thus:
\[
\mathbf{v}(t) = \left( 10 - 10\pi \sin(\pi t) \right) \mathbf{i} + \left( 20 + 10\pi \cos(\pi t) \right) \mathbf{j} + 30 \mathbf{k}
\]
---
### **Step 3: Compute magnitude of velocity**
The magnitude of \( \mathbf{v} \) is:
\[
|\mathbf{v}| = \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2}
\]
At \( t = 0.75 \):
1. Compute \( \sin(\pi t) \) and \( \cos(\pi t) \):
\[
t = 0.75 \quad \Rightarrow \quad \pi t = 0.75\pi
\]
\[
\sin(0.75\pi) = \cos(0.25\pi) = \frac{\sqrt{2}}{2}, \quad \cos(0.75\pi) = -\sin(0.25\pi) = -\frac{\sqrt{2}}{2}
\]
2. Substitute into derivatives:
\[
\frac{dx}{dt} = 10 - 10\pi \cdot \frac{\sqrt{2}}{2} = 10 - 5\pi\sqrt{2}
\]
\[
\frac{dy}{dt} = 20 + 10\pi \cdot \left(-\frac{\sqrt{2}}{2}\right) = 20 - 5\pi\sqrt{2}
\]
\[
\frac{dz}{dt} = 30
\]
3. Compute magnitude:
\[
|\mathbf{v}| = \sqrt{\left( 10 - 5\pi\sqrt{2} \right)^2 + \left( 20 - 5\pi\sqrt{2} \right)^2 + 30^2}
\]
Expand each term:
\[
\left( 10 - 5\pi\sqrt{2} \right)^2 = 100 - 100\pi\sqrt{2} + 50\pi^2
\]
\[
\left( 20 - 5\pi\sqrt{2} \right)^2 = 400 - 200\pi\sqrt{2} + 50\pi^2
\]
\[
30^2 = 900
\]
Add them together:
\[
|\mathbf{v}|^2 = \left( 100 + 400 + 900 \right) - \left( 100\pi\sqrt{2} + 200\pi\sqrt{2} \right) + \left( 50\pi^2 + 50\pi^2 \right)
\]
\[
|\mathbf{v}|^2 = 1400 - 300\pi\sqrt{2} + 100\pi^2
\]
Finally:
\[
|\mathbf{v}| = \sqrt{1400 - 300\pi\sqrt{2} + 100\pi^2}
\]
---
### **Step 4: Numerical calculation**
Using \( \pi \approx 3.1416 \):
\[
|\mathbf{v}| \approx \sqrt{1400 - 300(3.1416)\sqrt{2} + 100(3.1416)^2}
\]
\[
|\mathbf{v}| \approx \sqrt{1400 - 1331.29 + 986.96} \approx \sqrt{1055.67} \approx 32.50 \, \text{m/s}
\]
---
### **Final Answer**:
The magnitude of the velocity at \( t = 0.75 \) seconds is approximately:
\[
\boxed{32.50 \, \text{m/s}}
\]
Answer by ikleyn(52778)   (Show Source): You can put this solution on YOUR website!
.
A point travels as described by the following parametric equations x = 10t+10cos(3.14t), y = 20t+10sin(3.14t) and z=30t,
where x,y,z, are in meters, t in seconds, all angles in radians. The vector locating the body at any time is r = ix+jy+kz.
Determine the magnitude of the velocity of the body in meters per second at time t = 0.75 second.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        In opposite to the long unreadable solution by @CPhill,
        I will give the solution in simple readable human form.

The components of velocity at t = 0.75 seconds are x'(0.75) = = = -12.20315293 m/s; y'(0.75) = = = -2.203152929 m/s; z'(0.75) = 30 m/s. The magnitude of the velocity is |V| = = = 32.46 m/s. ANSWER

Solved.

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