What is the velocity vector of a particle traveling to the right along the hyperbola y=x−1 with constant speed 1 when the particle's location is (4, 14).
(Use symbolic notation and fractions where needed. Give your answer in the form of comma separated list of i, j components.)
y = x-1
dy/dx = slope = Hey ! this is not a hyperbola but a straight line = 1
the particle is never at 4,14
To find the velocity vector of a particle, we need to differentiate its position vector with respect to time. However, in this case, we are not given the parameterization of the particle's position, but rather its position along the hyperbola y = x - 1. We need to establish a connection between the particle's position on the hyperbola and the parameter that represents time.
Let's consider the hyperbola equation y = x - 1. We can rewrite it as x = y + 1 to express x in terms of y. Now, we can introduce a parameter t that represents time and consider x and y as functions of t, denoted as x(t) and y(t).
x(t) = y(t) + 1
y(t) = x(t) - 1
Next, we need to find an expression for t in terms of x or y. We can start with the given position of the particle being (4, 14) on the hyperbola.
For x = 4, we can solve the equation y = x - 1 to find the corresponding y-coordinate:
y = 4 - 1 = 3
Therefore, at t = ?, the particle is at (4, 14), where x = 4 and y = 3.
Having established the relationship between the particle's position and time, we can proceed to find the velocity vector. Differentiating the position vector (x, y) with respect to time t will give us the velocity vector (v_x, v_y).
v_x = d(x)/dt
v_y = d(y)/dt
Differentiating x(t) = y(t) + 1 with respect to t, we get:
v_x = d(y(t) + 1)/dt
= dy(t)/dt
Similarly, differentiating y(t) = x(t) - 1 with respect to t, we get:
v_y = d(x(t) - 1)/dt
= dx(t)/dt
As we have established our positions x(t) and y(t) as y(t) = x(t) - 1, we can substitute y(t) into the expression for v_x:
v_x = dy(t)/dt
= dy(x(t) - 1)/dt
Since we are given that the particle is moving with constant speed 1, the magnitude of the velocity vector is always 1. Therefore, the velocity vector is a unit vector in the direction of the velocity. We can normalize (v_x, v_y) by dividing by its magnitude to get the unit vector.
To summarize, the steps to find the velocity vector (v_x, v_y) are as follows:
1. Rewrite the hyperbola equation as x = y + 1 to express x in terms of y.
2. Introduce a parameter t to represent time and consider x(t) and y(t) as functions of t.
3. Find the value of t at which the particle is located at (4, 14) on the hyperbola.
4. Differentiate x(t) = y(t) + 1 with respect to t to find v_x.
5. Differentiate y(t) = x(t) - 1 with respect to t to find v_y.
6. Normalize the velocity vector (v_x, v_y) by dividing by its magnitude to get the unit vector.
Unfortunately, with the given information, it is not possible to determine the exact value of (v_x, v_y). However, using the steps outlined above, you should be able to find the velocity vector once you have the parameterization of the particle's position.