2.4 Solving Nonlinear Equations

Example 2.8

The number of deer, N, in a preserve t years after being introduced is given by the formula
N(t) = (6.21/(0.035+(0.45^t)))

Part 1

NDeer[t_]:=6.21/(0.035+(0.45^t))

Plot[NDeer[t],{t,0,15}, AxesLabel->{"t years","Number of Deer"}]

Part 2

Plot[{NDeer[t],85},{t,0,15}, AxesLabel->{"t years","Number of Deer"}]

We simply set N(t) = 85 and solve for t.

Part 3

NSolve[NDeer[t]==85,t,Reals]

By looking at the graph in Part 2 and the solve command here, it appears that there should be 85 deer four years after they were introduced into the reserve.

Example 2.9

This problem involves Archimedes Law which involves the amount of water displaced by an object.  In this case, the object is a ball of  diameter 4 ft and known weight of 436 lbs.  A formula that relates these quantities to the amount d  ft of its diameter which will be below the water line is given by:

62.4π (d^2)(2-(d/3)) = 436

We can treat the left side as a function, and try to determine where it crosses the horizontal line y=436.

A[d_]:=62.4*π*(d^2)*(2-(d/3))

Plot[{A[d],436},{d,0,4}]

NSolve[A[d]==436,d,Reals]

So we have three answers here.  Clearly we cannot have a negative answer.  But why can we not have d=5.805 ft?  See problem 11 in the exercises for more questions regarding these three solutions.

Exercise 13

The distance a parachuter falls in feet after t seconds can be modeled by:
d(t) = 12.5((0.2^t)-1)+20t
How long will it take to fall 140 feet?

NSolve[12.5*((0.2^t)-1)+20t==140,t, Reals]

Plot[12.5*((0.2^t)-1)+20t,{t,0,8}]

Using the NSolve command, we see that after 7.63 seconds, the parachuter has fallen 140 feet.

Exercise 14

The temperature (in Fahrenheit) of a coffee cup t minutes after it is poured is given by the formula

c(t) = 125(e^(-0.03t))+75

Part a.

cup[t_]:=125*(E^(-0.03t))+75

Plot[cup[t],{t,0,100},AxesLabel->{"t minutes","Temp (F)"}, PlotRange->{{0,100},{0,200}}]

Part b.

Plot[{cup[t],150},{t,0,100},AxesLabel->{"t minutes","Temp (F)"}, PlotRange->{{0,100},{0,200}}]

NSolve[cup[t]==150,t,Reals]

After 17 minutes, the cup of coffee will be at 150 degrees F.

Part c.

The temperature of the coffee pot is the temperature of the coffee at t=0, which we see to be 200 F.

cup[0]

Part d.

Let’s plug in a large value for time, say 1000 minutes.  We get 75 F, which should be the temperature of the room.  The graph agrees as well.

cup[1000]

Plot[{cup[t],75},{t,0,500},AxesLabel->{"t minutes","Temp (F)"}, PlotRange->{{0,500},{0,200}}]