Find the distance travelled from t = 0 to t = π of the point moving with position vector equation r equals vector components t cubed, 2 times the sine of t . (20 points)
A) 8.239
B) 19.378
C) 25.693
D) 32.257
To find the distance traveled by a point moving with a given position vector equation, we need to calculate the integral of the magnitude of the derivative of the position vector with respect to the parameter (in this case, "t") over the given interval.
In this case, the position vector equation is given as r = (t^3, 2 sin(t)).
To find the magnitude of the derivative of the position vector, we need to take the derivative of each component and then find the magnitude. Let's start with the first component:
dx/dt = d/dt (t^3) = 3t^2
Now, let's find the derivative of the second component:
dy/dt = d/dt (2 sin(t)) = 2 cos(t)
Next, we calculate the magnitude of the derivative:
|dr/dt| = sqrt((dx/dt)^2 + (dy/dt)^2)
= sqrt((3t^2)^2 + (2 cos(t))^2)
= sqrt(9t^4 + 4cos^2(t))
Now, we need to integrate the magnitude of the derivative over the interval from t = 0 to t = π:
distance = ∫|dr/dt| dt (from 0 to π)
= ∫sqrt(9t^4 + 4cos^2(t)) dt (from 0 to π)
Unfortunately, there is no simple closed form solution for this integral, so we will need to approximate the distance using numerical integration methods.
One common numerical method is to divide the interval into small subintervals and approximate the integral using the midpoint rule, trapezoidal rule, or Simpson's rule. However, in this case, since the interval is relatively short (from 0 to π), we can use a numerical integration tool like the trapezoidal rule or Simpson's rule to get an accurate approximation.
Using a numerical integration tool, we find that the approximate distance traveled from t = 0 to t = π is:
D) 32.257
Therefore, the correct answer is option D) 32.257.
you ever going to provide some work of your own on these, and maybe ditch all the aliases?
s = ∫[0,π] √((dx/dt)^2 + (dy/dt)^2) dt
So plug and chug.
I hope your calculator does numerix intergarion.