Homework, 2.8: related rates
Calculus I
Andy Long

#19

A[b_,h_]:=b*h/2
eq=D[A[b[t],h[t]],t]
Solve[A[b,10]==100,b]
Solve[eq==2,b'[t]]

#20

(* The hypothenuse h is changing at the rate of 1m/s: dh/dt=1m/s.
If the height is 1 m, we want to know how fast x is changing (dx/dt) when x=8. 
h^2=1+x^2, hence
2*h*dh/dt=2*x*dx/dt
So dx/dt=h/x*dy/dt=Sqrt[1+8^2]/8*1m/s
*)
xprime=Sqrt[1+8^2]/8*1
N[%]

#27

(* The volume V is changing at the rate of 30ft^3/min: dV/dt=30.
The cone has volume V(t)=Pi/3h^3, and diameter and h are equal (so r=h/2) in this case.
Hence V(t)=Pi/12h(t)^3. 
we want to know how fast h is changing (dh/dt) when h=10ft. 
dV/dt=Pi*h^2/4*dh/dt
dh/dt=1/(Pi*h^2)*dV/dt=4/(Pi*10^2)*30
*)
hprime=4/(Pi*10^2)*30
N[%]

#28

Solve[D[Tan[th[t]]==100/x[t],t],th'[t]]
-100Cos[Pi/6]^2*8/30000

#40

(* The angle theta is changing at the rate of 4rev/min: dT/dt=8PI/min.
We want to relate the angle and x. 3Tan[T]=x.
So dx/dt=3Sec^2[T]dT/dt, or 
since Tan[T]=1/3 at the point of interest, Sec^2[T]=1+Tan^2[T]=10/9. So
dx/dt=3Sec^2[T]dT/dt=3*10/9*8Pi (km/min)
*)
xprime=3*10/9*8Pi
N[%]