9.4 Defining and Differentiating Vector-Valued Functions

Because parametric functions are associated with time, they are also generally used to calculate motion and velocity, and the College Board usually uses parametrics in this context. When we deal with parametrics in the context of motion, we express them as vector-valued functions. Vector-valued functions aren’t graphed with the points x and y like we are used to seeing. Instead, each “point” on a vector-valued function is determined by a position vector (a vector that starts at the origin) that exists in the direction of the point. 🧍

One of the key concepts students learn is the derivative of a function. When dealing with Cartesian functions, such as y = f(x), the derivative of the function gives us the slope of the tangent line at any point on the curve. This slope is often represented as dy/dx, and it gives us a measure of how much y is changing as x changes.

When we move to the realm of vector-valued functions, we are still looking at how much a variable is changing, but now we are looking at how much the position of a particle is changing in both the x and y direction. In this case, the position vector, represented by r(t), tells us the position of a particle at any given time t. The velocity vector, represented by v(t), tells us how fast the particle is moving in the x and y direction at any given time t. And the acceleration vector, represented by a(t), tells us how the velocity of the particle is changing at any given time t.

Just like Cartesian functions, if we take the derivative of the position vector, we would get the velocity vector, and if we take the derivative of the velocity vector, we would get the acceleration vector. These vectors are related, and together they give us a complete picture of how the position of a particle is changing over time.

When we were taking the derivative of a parametric function to find dy/dx, we were trying to find the slope of the tangent line that was determined by both the x and y functions of the curve. However, when we are looking at vector-valued functions, we aren’t looking at the curve itself; we are looking at how much our particle is moving in the direction of x and how much it is moving in the direction of y. This means that when we are taking derivatives of vector-valued functions, we take the derivative of the components separately. 🐞

A particle is moving in a two-dimensional plane, and its position as a function of time, t, is given by the vector-valued function r(t) = <t^3 - 9t, 3t^2 - 6t>. The particle's velocity at any time t is given by the derivative of the position vector with respect to time, and its acceleration at any time t is given by the derivative of the velocity vector with respect to time. ✈️

(1) Find the velocity vector-valued function, v(t), of the particle.

(2) Find the acceleration vector-valued function, a(t), of the particle.

(3) At what time(s) is the velocity of the particle equal to zero?

(4) At what time(s) is the acceleration of the particle equal to zero?

(5) At what time(s) is the particle moving in the direction of increasing x?

(6) At what time(s) is the particle moving in the direction of increasing y?

(7) At what time(s) is the particle moving in the direction of decreasing x?

(8) At what time(s) is the particle moving in the direction of decreasing y?

(9) At what time(s) is the particle moving at its maximum speed?

(10) At what time(s) is the particle accelerating in the direction of increasing x?

(11) At what time(s) is the particle accelerating in the direction of increasing y?

(12) At what time(s) is the particle accelerating in the direction of decreasing x?

(13) At what time(s) is the particle accelerating in the direction of decreasing y?

(14) At what time(s) is the particle accelerating at its maximum rate?

(15) At what time(s) is the particle slowing down at its maximum rate?

(16) At what time(s) is the particle experiencing uniform acceleration?

(17) At what time(s) is the particle experiencing non-uniform acceleration?

(1) To find the velocity vector-valued function, v(t), we take the derivative of the position vector-valued function, r(t), with respect to time. So, v(t) = r'(t) = <3t^2 - 9, 6t - 6>.

(2) To find the acceleration vector-valued function, a(t), we take the derivative of the velocity vector-valued function, v(t), with respect to time. So, a(t) = v'(t) = <6t, 6>.

(3) The velocity of the particle is equal to zero when the x and y components of the velocity vector are both zero. This means that 3t^2 - 9 = 0 and 6t - 6 = 0. Solving for t, we get t = 3/sqrt(3) and t = 1.

(4) The acceleration of the particle is equal to zero when the x and y components of the acceleration vector are both zero. This means that 6t = 0. Solving for t, we get t = 0.

(5) The particle is moving in the direction of increasing x when the x component of the velocity vector is positive. This means that 3t^2 - 9 > 0. Solving for t, we get t < -sqrt(3) or t > sqrt(3).

(6) The particle is moving in the direction of increasing y when the y component of the velocity vector is positive. This means that 6t - 6 > 0. Solving for t, we get t > 1.

(7) The particle is moving in the direction of decreasing x when the x component of the velocity vector is negative. This means that 3t^2 - 9 < 0. Solving for t, we get -sqrt(3) < t < sqrt(3).

(8) The particle is moving in the direction of decreasing y when the y component of the velocity vector is negative. This means that 6t - 6 < 0. Solving for t, we get t < 1.

(9) The particle is moving at its maximum speed when the magnitude of the velocity vector is a maximum. To find the maximum speed, we can take the derivative of the magnitude of the velocity vector with respect to time and set it equal to zero. This gives us (3t^2 - 9)^2 + (6t - 6)^2 = 0. Solving this equation gives no real solution.

(10) The particle is accelerating in the direction of increasing x when the x component of the acceleration vector is positive. This means that 6t > 0. Solving for t, we get t > 0.

(11) The particle is accelerating in the direction of increasing y when the y component of the acceleration vector is positive. This means that 6 > 0, which is always true.

(12) The particle is accelerating in the direction of decreasing x when the x component of the acceleration vector is negative. This means that 6t < 0. Solving for t, we get t < 0.

(13) The particle is not accelerating in the direction of decreasing y.

(14) The particle is accelerating at its maximum rate when the magnitude of the acceleration vector is a maximum. To find the maximum acceleration, we can take the derivative of the magnitude of the acceleration vector with respect to time and set it equal to zero. This gives us 12t = 0. Solving for t gives t = 0.

(15) The particle is slowing down at its maximum rate when the magnitude of the acceleration vector is a minimum. To find the minimum acceleration, we can take the derivative of the magnitude of the acceleration vector with respect to time and set it equal to zero. The acceleration vector is constant, and thus it's magnitude is constant and equal to 6, so the particle is not slowing down at any point.

(16) The particle is experiencing uniform acceleration when the acceleration vector is constant. The acceleration vector is constant, and thus the particle is experiencing uniform acceleration.

(17) The particle is not experiencing non-uniform acceleration since the acceleration vector is constant.