Chapter 5 - The Integral - 5.3 The Indefinite Integral - Exercises - Page 253: 52

$z=2t^{-\frac{1}{2}}$

$\frac{dz}{dt}=t^{-\frac{3}{2}};$ $z(4)=-1$ $dz=t^{-\frac{3}{2}}dt$ $\int dz=\int t^{-\frac{3}{2}}dt$ $z=-2t^{-\frac{1}{2}}+C$ Substitute in the initial condition to solve for $C$ $-1=-2(4)^{-\frac{1}{2}}+C$ $C=0$ $z=2t^{-\frac{1}{2}}$