A particle moves in the x-y plane for t > 0 so that x(t) = t2 + 6t and y(t) = e-t. At time t = 1, the particle is moving (20 points)
A) to the left and up
B) to the right and up
C) to the left and down
D) to the right and down
You meant:
x(t) = t^2 + 6t and y(t) = e-t
then x'(t) = 2t and y'(t) = -e^t
x'(1) = 2 and y'(1) = -e
x'(1) is positve and y'(1) is negative.
What does not mean?
To determine the direction of motion of the particle at time t = 1, we need to analyze the derivatives of its x and y coordinates with respect to time. That is, we will find the velocity vector of the particle at t = 1.
Given:
x(t) = t^2 + 6t
y(t) = e^-t
The velocity vector v(t) of the particle is given by:
v(t) = (dx/dt) * i + (dy/dt) * j
Taking the derivatives of x(t) and y(t) with respect to t:
dx/dt = 2t + 6
dy/dt = -e^-t
Now, we can calculate the velocity vector at t = 1:
v(1) = (2(1) + 6) * i + (-e^-1) * j
= 8 * i - e^-1 * j
From the velocity vector, we can determine the direction of motion. The i-component of the velocity vector represents the motion in the x direction, and the j-component represents the motion in the y direction.
Since the i-component of the velocity is positive (8), the particle is moving to the right. And since the j-component is negative (-e^-1), the particle is moving downward.
Therefore, the correct answer is:
D) to the right and down