Calculus Review

[Noob88]

With our AP exam coming up in May, our teacher has given us a list of problems that we SHOULD be able to do for our exam, that we can turn in for a grade if we so wish, but it will not be counted against us if we don't. I've been having a relatively easy time until I got to the word problem ones, which confused me.

This is the first of the problems I'm having trouble with:



You are driving along a highway at a steady 60 mph (88 ft/sec) when you see an accident ahead and slam on the brakes. What constant deceleration is required to stop your car in 242 ft? To find out, do the following:

(A) Solve the initial value problem
Differential equation: d^(2)s/dt^(2) = -k
Initial conditions: ds/dt=88, s=0 when t=0

(B) Find the value of t that makes ds/dt = 0

[C] Find the value of k that makes s = 242 for the value of t you found in (B)

-------------------------------

I did try to work this out:
(A)
integral(-k)= -kt
integral(-kt)= (-kt^2)/2
integral(88)=88t
s=(-kt^2)/2 + 88t

(B)
ds/dt=-kt + 88
0=-kt + 88
kt=88
t=88/k

[C]
This is where I'm stuck and I can't get it to work out the way it's supposed to, can anyone help me out?

[DarkerLine]

The equations you've already worked out are:
t=88/k
s=(-kt^2)/2 + 88t

You also have the condition that s=242. Therefore,
(-kt^2)/2 + 88t=242

Now substitute the equation for t and solve.

[Noob88]

Well I keep trying and it won't come out right... I keep getting:

484k-7744k+7744=0

And the answer is supposed to be 16... I don't know what I'm doing wrong

[DarkerLine]

Possibly, you're substituting wrong. After substituting t=88/k, you should get

-88²/(2k) +88²/k=242

Which does indeed come out to k=16.

[Noob88]

Ah ok substitution error... Ok here is next what I can't figure out:

Show that

y=ln(abs(cos(3)/cos(x)))+5

is the solutiuon to the initial value problem

dy/dx=tan(x), f(3)=5




The problem here... When I take the integral of tan(x), I was under the impression that it was sec^2(x), so I have no idea how to go about doing this.

[DarkerLine]

You're confusing integral and derivative.

[Noob88]

Ok stupid mistake sorry,

Using substitution I got the derivative to be:
y=-ln(cos(x))+C

Now f(3)=5

But plugging in 3 gives me a negative inside ln( which leads to imaginaries... Right? I don't know how to arrive at the conclusion I'm supposed to

[DarkerLine]

When you have a (possible) solution to an equation, the thing that makes the most sense is to try plugging it into the equation. You have a possible solution: y=ln|cos 3/cos x|+5 and an equation: dy/dx=tan x, y(3)=5. So check if it satisfies both conditions of the equation?

What you're doing is solving the equation all over again. This is a valid approach, when it works. But it might not work all the time (complicated differential equations are a whole lot harder to solve than to check) and sometimes, as in this case, you get a solution in a slightly different form.