A particle moves along a curve so that its position at time t is given by the position vector <4e^(3t - 3), cos(t - 1)>. What is the particle's speed at time t = 1?
I think it is 12.
dx/dt =12 (e^3t-3)
dy/dt = -sin(t-1)
at t = 1
dx/dt = 12
dy/dt = 0
so yes 12
To find the particle's speed at time t = 1, we need to find the magnitude of its velocity vector at that time.
The velocity vector is the derivative of the position vector with respect to time. So, we need to find the derivative of the given position vector.
Let's find the derivative of each component of the position vector separately:
The derivative of 4e^(3t - 3) with respect to t can be found using the chain rule. The derivative of e^(3t - 3) is 3e^(3t - 3) since the derivative of e^u with respect to u is e^u. So, the derivative of 4e^(3t - 3) is 4 * 3e^(3t - 3), which simplifies to 12e^(3t - 3).
The derivative of cos(t - 1) with respect to t is -sin(t - 1) since the derivative of cos(u) with respect to u is -sin(u).
Now, we have the velocity vector: <12e^(3t - 3), -sin(t - 1)>.
To find the speed at time t = 1, we substitute t = 1 into the velocity vector and take its magnitude.
Substituting t = 1, we have: <12e^(3(1) - 3), -sin(1 - 1)> = <12e^0, -sin(0)> = <12, 0>.
The magnitude of the velocity vector is the square root of the sum of the squares of its components. In this case, the magnitude is sqrt(12^2 + 0^2) = sqrt(144) = 12.
Therefore, the particle's speed at time t = 1 is 12. So, your answer is correct.