Rate of Change Related Questions

[Sig00]

Alright, so I submitted a thread a little over a day ago and it never showed up. So here's V2 I guess...

I just have a few problems that I need help doing. I really am looking to understand the solution to the problems and not just get the answer, so any sort of explanations would be greatly appreciated!

Problem #1

An object is launched straight upward from a platform above ground. Its height ish(t) = 10 + v₀t − 4.9t²

where h is in meters and t is in seconds.
It hits the ground with a velocity of−25.0
m/s. Findv₀
to three sig figs.v₀ =

When did it hit the ground? (3 sig figs.)



So, I found the derivative of the equation which comes out to be

h'(t)=v(sub)0-9.8t

So I would assume you would go about this equation somehow by setting the h'(t) equation equal to -25.0 m/s and then plugging in a time, however I don't how you could find the time.



Problem #2

The pressure in a cylinder is given byP(t) = 101 + (k/t)

where P is in kilopascals (WebAssign abbreviation kPa) and t is in minutes.
Assume thatt ≥ 1
and k is constant.
At the instant when pressure is 114 kPa, it is changing at−0.85
kPa/min.
When does this happen?
What is the pressure whent = 1
minute.



So I found the derivative which is

P'(t)=-(k/(t^2))

Then using the data given in the problems I found the following two equations

114=101 + (k/t)
(which comes out to be)

13 kPa = (k/t)
and
-.85 kPa/min = -(k/(t^2))

So I know that those two equations are interrelated but I don't know how to go from there. The only thing I can think of that would make some sense would be setting the equations equal to each other or something along those lines.



Problem #3

The electrical potential in a circuit is given byV(t) = 15 − 15e^−kt

where V is in volts, t is in seconds, and k is constant.
When t = 0 the rate of change of potential is6.9 V/s.
Find the rate of change 3 seconds later.

Find k with correct units. You may find it useful to know the following
FACT: The input to an exponential function must be dimensionless. I.e., it has no units.

k =





So I thought I had been doing this one right.

I found the derivative of the original equation which is

V'(t) = 15ke^(-kt)

using the data given in the problem I got the equation

6.9 = 15k (because you get e^0 which just turns into 1 leaving you with 15*k*1)

so k=.46

When I plug .46 in and solve for 3 seconds later I got 11.2263217, but both of these answers are wrong. Soooo, I'm pretty much at a loss for what to do here as well.





Any help that anybody can provide would be greatly appreciated!!!!!!!

 

[DrPhil]

Alright, so I submitted a thread a little over a day ago and it never showed up. So here's V2 I guess...

I just have a few problems that I need help doing. I really am looking to understand the solution to the problems and not just get the answer, so any sort of explanations would be greatly appreciated!

Problem #1

An object is launched straight upward from a platform above ground. Its height is h(t) = 10 + v₀t − 4.9t²

where h is in meters and t is in seconds.
It hits the ground with a velocity of−25.0
m/s. Findv₀
to three sig figs.v₀ =

When did it hit the ground? (3 sig figs.)



So, I found the derivative of the equation which comes out to be

h'(t)=v(sub)0-9.8t

So I would assume you would go about this equation somehow by setting the h'(t) equation equal to -25.0 m/s and then plugging in a time, however I don't how you could find the time.

Two unknowns, so you need a second equation relating v_0 and t. Lets set h(t)=0:
..............\(\displaystyle 0 = 10 + v_0\ t + 4.9 t^2 \)
combined with
..............\(\displaystyle -25 = v_0\ t - 9.8\ t\)
I would solve the 2nd equation for t, then substitute that expression in the first (quadratic) equation.

Problem #2

The pressure in a cylinder is given by P(t) = 101 + (k/t)

where P is in kilopascals (WebAssign abbreviation kPa) and t is in minutes.
Assume that t ≥ 1 and k is constant.
At the instant when pressure is 114 kPa, it is changing at −0.85 kPa/min.
When does this happen?
What is the pressure when t = 1 minute.

So I found the derivative which is

P'(t)=-(k/(t^2))

Then using the data given in the problems I found the following two equations

114=101 + (k/t)
(which comes out to be)

13 kPa = (k/t)
and
-.85 kPa/min = -(k/(t^2))

So I know that those two equations are interrelated but I don't know how to go from there. The only thing I can think of that would make some sense would be setting the equations equal to each other or something along those lines.

Like problem #1, you have two equations in two unknowns. They aren't linear eauations, so some of your favorite methods won't work. BUT you can almost always use substitution. For instance,
1st eqn. --> k = (13 kPa)*t
2nd eqn. --> -0.85 kPa/min = -(13 kPa)/ t

Problem #3

The electrical potential in a circuit is given byV(t) = 15 − 15e^−kt

where V is in volts, t is in seconds, and k is constant.
When t = 0 the rate of change of potential is 6.9 V/s.
Find the rate of change 3 seconds later.

Find k with correct units. You may find it useful to know the following
FACT: The input to an exponential function must be dimensionless. I.e., it has no units.

k = per second

So I thought I had been doing this one right.

I found the derivative of the original equation which is

V'(t) = 15 k e^(-kt)

using the data given in the problem I got the equation

6.9 V/s = (15 V)k (because you get e^0 which just turns into 1 leaving you with 15*k*1)

so k=0.46 /s

When I plug .46 in and solve for 3 seconds later I got 11.2263217, but both of these answers are wrong. Soooo, I'm pretty much at a loss for what to do here as well.

Did you include the units?

The rate has to get smaller as time passes, so 11 can't be right. The factor will be e^-(k*3s).

 

[Sig00]

Did you include the units?

The rate has to get smaller as time passes, so 11 can't be right. The factor will be e^-(k*3s).

Yes I did include units and it was wrong, so I assume my math went awry somewhere (or was going down the wrong path to start!). Thanks for your help, I'll try to solve the other two problems now in a few minutes.

 

[Sig00]

Okay, I was able to solve problem 2 using substitution, thanks for the help, I didn't even think of that!



As for problem number 1, I got

t = (v_0-9.8)/-25

so when I plug that into the other equation I get

0 = 10+v_0((v_0-9.8)/-25)+4.9((v_0-9.8)/-25)^2

I can't use the quadratic to solve this because v_0 is before the ((v_0-9.8)/-25) and if I distribute out then I would have two different variables I would be solving for. D:



I still haven't been able to make any progress on problem three. Anybody have any ideas?

 

[DrPhil]

Problem #1

An object is launched straight upward from a platform above ground. Its height is h(t) = 10 + v₀t − 4.9t²

where h is in meters and t is in seconds.
It hits the ground with a velocity of−25.0
m/s. Findv₀
to three sig figs.v₀ =

When did it hit the ground? (3 sig figs.)
So, I found the derivative of the equation which comes out to be

h'(t)=v(sub)0-9.8t

So I would assume you would go about this equation somehow by setting the h'(t) equation equal to -25.0 m/s and then plugging in a time, however I don't how you could find the time.

As for problem number 1, I got

t = (v_0-9.8)/-25 X The units don't match up
so when I plug that into the other equation I get

0 = 10+v_0((v_0-9.8)/-25)+4.9((v_0-9.8)/-25)^2

I can't use the quadratic to solve this because v_0 is before the ((v_0-9.8)/-25) and if I distribute out then I would have two different variables I would be solving for. D:

\(\displaystyle -25 \text{ m/s} = v_0 - (9.8 \text{ m/s}^2) t\)

\(\displaystyle t = \dfrac{v_0 + 25 \text{ m/s}}{9.8 \text{ m/s}^2} = \dfrac{v_0}{g} + 2.551\text{ s}\)

\(\displaystyle 0 = 10 + v_0 \left(\dfrac{v_0+25}{g}\right) - \dfrac{g}{2} \left(\dfrac{v_0+25}{g}\right)^2 \)

where \(\displaystyle g = 9.8 \text{ m/s}^2\) is the acceleration of gravity.

When you multiply this out, you will have a quadratic equation in \(\displaystyle v_0\). I suspect one solution will be negative (downward), but the problem states "upward" so you will have to use the positive solution.

For #3: if the given numbers (15 V and 6.9 V/s) are correct as you typed them, then k = 0.46 s^-1.