Question

A unicycle moves along the circle ((x-1)^2 + y^2 = 4) where x and y are measured in meters. At t = 10 seconds, the unicycle is located at the point P(1, 2). Find the velocity vector of the unicycle at this moment.

Answer 1

To determine the unicycle's velocity vector at a specific point, we would need to know its position function with respect to time or the angular velocity. With only the circle's equation, we know the radius but cannot find the precise velocity vector without additional information.

Step-by-step explanation:

To find the velocity vector of the unicycle at t = 10 seconds, we must first determine the angular velocity of the unicycle's circular motion. Since we are given the equation of the circle ((x-1)^2 + y^2 = 4), we know that the radius of this circle is 2 meters (since r^2 = 4). We are not given a direct function of time for the position of the unicycle, but we can use the information from similar circular motion problems to deduce the velocity vector.

For instance, we might infer from the given analogous problems that if a particle moves along a circle in the xy-plane, its position can often be represented by r(t) = R(cos(wt)i + sin(wt)j), where R is the radius and w is the angular velocity. The derivative of the position with respect to time gives the velocity function: v(t) = -Rw(sin(wt)i + cos(wt)j). Since it is said that the unicycle is located at point P(1, 2) at t = 10 seconds, and we know the center of the circle is at (1, 0), we can deduce that the unicycle is at the top of the circle where the velocity vector would be horizontal and pointing to the left (negative x-direction).

However, without a specific function for the position of the unicycle with respect to time or an angular velocity, we cannot directly calculate the velocity vector. Additional information about the speed or angular velocity of the unicycle would be required to find the precise magnitude and direction of the velocity vector at that moment.

Answer 2

The velocity vector of the unicycle at
`\(t = 10\)` seconds is approximately
`\(\langle -1.87, -1.16 \rangle\)` meters per second.

To find the velocity vector of the unicycle at the given moment, we can start by parameterizing the equation of the circle
`\( (x-1)^2 + y^2 = 4 \)`. A convenient parameterization for a circle is given by:

`\[ x(t) = r \cos(t) + a \]`

`\[ y(t) = r \sin(t) + b \]`

In this case,
`\( r = 2 \)` (since the radius of the circle is 2) and the center of the circle is at
`\((a, b) = (1, 0)\)`.

So, the parameterization becomes:

`\[ x(t) = 2 \cos(t) + 1 \]`

`\[ y(t) = 2 \sin(t) \]`

Now, let's find the velocity vector by taking the derivative of the parameterization with respect to time
`\( t \)`:

`\[ \mathbf{v}(t) = \langle x'(t), y'(t) \rangle \]`

`\[ x'(t) = -2 \sin(t) \]`

`\[ y'(t) = 2 \cos(t) \]`

Now, evaluate these derivatives at
`\( t = 10 \)`seconds:

`\[ x'(10) = -2 \sin(10) \]`

`\[ y'(10) = 2 \cos(10) \]`

These values give the components of the velocity vector at
`\( t = 10 \)`seconds:

`\[ \mathbf{v}(10) = \langle -2 \sin(10), 2 \cos(10) \rangle \]`

This is the velocity vector of the unicycle at
`\( t = 10 \)`seconds. If you need a numerical approximation, you can use a calculator to find the values of
`\( \sin(10) \)` and
`\( \cos(10) \)`.

Calculating the numerical values for
`\(\sin(10)\)`and
`\(\cos(10)\)`, we get:

`\[ \mathbf{v}(10) \approx \langle -1.87, -1.16 \rangle \, \text{m/s} \]`