Parametric Derivatives

Question

The position of a particle at any time t >= 0 is given by

x(t) = t^2 - 3

y(t) = 2/3t^3

Find the magnitude of the velocity vector at t = 5.

Find the total distance traveled by the particle from t = 0 to t= 5 .

I need help so bad with these

Answer 1

Hi Kalea,

To find the velocity vector, you compute the derivative of the x and y-coordinates. For us this is

v(t) = (x'(t), y'(t))

We can differentiate both x(t) and y(t) with the power rule:

v(t) = (2t, 2t^2).

Since the first question asks you for the magnitude of the velocity vector when t = 5, we first calculate v(5), and then take the magnitude. v(5) = (10, 50). Then the magnitude is

|v(5)| = √ ( 10^2 + 50^2 ) = 50.99 (or √2600 if you want an exact answer!)

The next question asks for distance traveled from t = 0 to t = 5. For this, we're going to need to integrate the magnitude of the velocity. (Distance traveled equals the integral of the magnitude of velocity.) The magnitude is | v(t) | = √((2t)^2 + (2t^2)^2) = 2 √t^2(t^2 + 1) = 2t √(t^2 + 1). To integrate this, we use u-substitution, u = t^2 + 1. Then du = 2t dt, which, happily, is exactly the thing sitting outside the square root. So after our u-substitution, we are integrating √u du. The new bounds of integration found by plugging in t = 0 for u and t = 5 for u. When t = 0, u is 1; when t = 5, u is 26. Keep those new bounds of integration in mind because we're going to plug them into the antiderivative below.

The integral of √u can be found by the power rule. √u = u^(1/2). By the power rule for integration, the integral (antiderivative) is 2/3 * u^(3/2). Evaluating this at the bounds of integration u = 26 and u = 1 gives us:

Distance Traveled = [ (2/3 u^3/2 evaluated at u = 26) - (2/3 u^3/2 evaluated at u = 1) ].

=87.7163