# Most likely an easy Integration problem

1. Mar 3, 2005

### apchemstudent

However, im stuck on this problem

Let v(t) be the velocity, in feet per second, of a skydiver at time t seconds, t>=0. After her parachute opens, her velocity satisfies the differential equation dv/dt = -2v -32, with initial condition v(0) = -50.

A) Find an expression for v in terms of t, where t is measured in seconds.

well I have problems right from the start...
dv/(-2v-32) = dt
integrate both sides --> ln(-2v-32) = -2t
I solve for v and i got v = e^(-2t)/-2 - 16 + C

C = -33.5

However, i could also have done it like this

dv/(-2(v+16)) = dt
bring the -2 over and integrate it to get --> ln(v+16) = -2t
solve for v = e^(-2t) -16 + C

C = -35

Why are they different?!!!! i must be doing something wrong...

b) Terminal velocity is defined as lim t->infinity v(t). Find the terminal velocity of the skydiver to the nearest foot per second.

well for the first equation i got
i get -49.5 feet per second, since the 1/infinity = 0

while the second one is -51 feet per second.... which is most likely not correct.

c) It is safe to land when her speed is 20 feet per second. At what time t does she reach this speed?

well for both cases i got t = negative seconds. I really don't think this is right. Can some one please help me figure out my mistake? Im really lost thanks...

2. Mar 3, 2005

### StatusX

In the first part, the constant C has to be added right after the integration. It ends up in the exponent and becomes a constant factor.

3. Mar 3, 2005

### dextercioby

I think they meant the absolute value for the velocity...Negative time would be when he's still in the plane...

Daniel.

4. Mar 3, 2005

### apchemstudent

the negative velocity will work if the equation is like this

dv/(-2v - 32) = dt

integrate both sides --> ln(-2v-32)/-2 = t + c

-2v - 32 = e^(-2t - 2c)

v = e^(-2t - 2c) /-2 - 16

However the velocity will not work if it is like this
ln(v+16) = -2t + c
v = e^(-2t + c) - 16

5. Mar 3, 2005

### dextercioby

No,there's one way to do this problem:
$$\int_{-50}^{v} \frac{dv}{-2v-32}=\int_{0}^{t}dt$$

Solve this and tell me what u get...

Daniel.

6. Mar 3, 2005

### apchemstudent

alright, i get ln(-2v-32)/-2 = t + c

so (-2v-32) = e^(-2t - 2c)

v = e^(-2t - 2c)/-2 -16

is this correct? im just wondering, why is it that if you bring out the -2 before you integrate it, you get a different answer?
I don't want to plug in 50 yet... i just want to find out why the equation is different as compared to my previous posts.

7. Mar 3, 2005

### dextercioby

You did't do what i said,right...?Suit yourself.As for this issue,i believe StatusX gave the explanation...

Daniel.

8. Mar 3, 2005

### apchemstudent

Well first of all, i don't understand the integration values completely yet (50 - 0 ?) We've only just started this as you see :tongue2: ... Well i did what Status said, but i still don't get the same answer... Can anyone explain this?

9. Mar 3, 2005

### dextercioby

That's integrating with "corresponding limits".At t=0,the v=-50 ([ms^{-1}]),at t=t,the v=v([ms^{-1}])...There's no big deal.It's much more simple than integrating indefinitely & then imposing conditions on the antiderivative...

Daniel.

10. Mar 3, 2005

### gnome

First way:
\begin{align*}\frac{dv}{2v+32} &= -dt \\ \frac{1}{2}ln|2v+32| &= -t + c \quad \text{note: dv is not the deriv. of 2v}\\ ln|2v+32| &= -2t + c'\\ 2v+32 &= e^{-2t + c'}\\ &= Ce^{-2t}\\ 2v &= Ce^{-2t} -32\\ v &= \frac{1}{2}(Ce^{-2t} -32)\\ -50 &= \frac{C}{2} -16 \\ C &= -68 \\ v &= \frac{1}{2}(-68e^{-2t} -32)\\ v &= -34e^{-2t} -16 \end{align*}

Second way:
\begin{align*}\frac{dv}{v+16} &= -2dt \\ ln|v+16| &= -2t + c\\ v+16 &= e^{-2t + c}\\ &= Ce^{-2t}\\ v &= Ce^{-2t} -16\\ -50 &= C -16 \\ C &= -34 \\ v &= -34e^{-2t} -16 \end{align*}

11. Mar 4, 2005

### dextercioby

I believe and hope it's all clear and u realize where you went wrong...

Daniel.

12. Mar 4, 2005

### apchemstudent

thank's gnome and dexter.... it makes sense now. I didn't know that e^c can be just used as C.

13. Mar 4, 2005

### dextercioby

That's a standard trick...Once you get used to this type of integration,it will become obvious...

Daniel.