integration problem

Quote:

Originally Posted by mastermin346

Given that $\frac{dp}{dt}=kt^2$ where k is constant.Given also $p=\frac{11}{3}$ and $\frac{dp}{dt}=-2$ when $t=-2.$ Express $p$ in term of $t$

$\frac{dp}{dt}=kt^2\ \Rightarrow\ p=\frac{k}{3}t^3+C$

$\frac{dp}{dt}=-2\ for\ t=-2\ \Rightarrow\ k(-2)^2=-2\ \Rightarrow\ 4k=-2$

From that you find k and substitute that value of k to find C.

$\frac{11}{3}=\left(-\frac{1}{2}\right)\left(\frac{1}{3}\right)(-2)^3+C$