### Instruction: A tennis ball dipped in red paint rolls around on the coordinate plane, so that it is at
\[(x,y) = (3t^2 - 9t - 5, t^2 - 3t + 2)\]at time $t,$ where $0 \le t \le 4.$  Find the length of the paint trail left by the tennis ball.### Response: I need to find the length of the curve described by the parametric equations $x(t) = 3t^2 - 9t - 5$ and $y(t) = t^2 - 3t + 2$ for $0 \le t \le 4.$
One way to do that is to use the formula for the arc length of a parametric curve, which is \[\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt,\] where $a$ and $b$ are the endpoints of the interval.
So I need to find the derivatives of $x(t)$ and $y(t)$ with respect to $t.$
Using the power rule, I get \[\frac{dx}{dt} = 6t - 9\] and \[\frac{dy}{dt} = 2t - 3.\]
Squaring these and adding them, I get \[\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = (6t - 9)^2 + (2t - 3)^2 = 40t^2 - 120t + 90.\]
Now I need to take the square root of this expression and integrate it from $t = 0$ to $t = 4.$
I notice that the expression under the square root is a perfect square: \[40t^2 - 120t + 90 = 10(2t - 3)^2.\]
So I can simplify the square root as \[\sqrt{40t^2 - 120t + 90} = \sqrt{10}(2t - 3).\]
Now I have a simpler integral to evaluate: \[\int_0^4 \sqrt{10}(2t - 3) \, dt.\]
I can use the power rule again to find the antiderivative of the integrand: \[\int \sqrt{10}(2t - 3) \, dt = \sqrt{10}(t^2 - 3t) + C.\]
To find the definite integral, I need to plug in the endpoints of the interval and subtract: \[\int_0^4 \sqrt{10}(2t - 3) \, dt = \sqrt{10}((4^2 - 3 \cdot 4) - (0^2 - 3 \cdot 0)) = \sqrt{10}(4 - 0) = 4\sqrt{10}.\]