Most likely an easy Integration problem

[apchemstudent]

However, I am 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? I am really lost thanks...

 

[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.

 

[dextercioby]

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

Daniel.

 

[apchemstudent]

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

Daniel.

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

Im really confused as to which one is the correct one... please help

 

[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.

 

[apchemstudent]

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.

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

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

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

is this correct? I am 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.

 

[dextercioby]

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

Daniel.

 

[apchemstudent]

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

Daniel.

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

 

[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.

 

[gnome]

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

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

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

is this correct? I am 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.

Your different answers are just due to errors in your solutions of the d.e. The answer is the same either way.

First way:
\begin{align*}\frac{dv}{2v+32} &amp;= -dt \\<br /> \frac{1}{2}ln|2v+32| &amp;= -t + c \quad \text{note: dv is not the deriv. of 2v}\\ <br /> ln|2v+32| &amp;= -2t + c&#039;\\<br /> 2v+32 &amp;= e^{-2t + c&#039;}\\<br /> &amp;= Ce^{-2t}\\<br /> 2v &amp;= Ce^{-2t} -32\\<br /> v &amp;= \frac{1}{2}(Ce^{-2t} -32)\\<br /> -50 &amp;= \frac{C}{2} -16 \\<br /> C &amp;= -68 \\<br /> v &amp;= \frac{1}{2}(-68e^{-2t} -32)\\<br /> v &amp;= -34e^{-2t} -16 \end{align*}


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

 

[dextercioby]

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

Daniel.

 

[apchemstudent]

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

Daniel.

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

 

[dextercioby]

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

Daniel.