# Thread: Someone help me im DYING with this

1. ## Someone help me im DYING with this

A parachutist weighing m=75kg jumps from an airplane at an altitude of h meteres with initial vertical velocity v0=0. Let v(t) be his instantaneous vertical velocity at time t with the downwards direction taken as positive. Suppose that he deploys the parachute immediately after jumping from the airplane. The air resistance is assumed to be r(t)=15v^2(t). Let g=9.8ms^-2 be the acceleration due to gravity. We neglect side wind. The equation of motion is
mdv/dt=F

a) Show that the instantaneous vertical velocity satisfies dv(t)/dt + v^2(t)/5 = 9.8
b) solve the initial value problem and determine the instantaneous vertical velocity v(t)
c) determine the impact velocity before landing. Express it in terms of h.

I CANT GET B AND C MAN :'( been trying whole of last night and this morning.

2. ## Re: Someone help me im DYING with this

Originally Posted by hayabusaboston
A parachutist weighing m=75kg jumps from an airplane at an altitude of h meteres with initial vertical velocity v0=0. Let v(t) be his instantaneous vertical velocity at time t with the downwards direction taken as positive. Suppose that he deploys the parachute immediately after jumping from the airplane. The air resistance is assumed to be r(t)=15v^2(t). Let g=9.8ms^-2 be the acceleration due to gravity. We neglect side wind. The equation of motion is
mdv/dt=F

a) Show that the instantaneous vertical velocity satisfies dv(t)/dt + v^2(t)/5 = 9.8
b) solve the initial value problem and determine the instantaneous vertical velocity v(t)
c) determine the impact velocity before landing. Express it in terms of h.

I CANT GET B AND C MAN :'( been trying whole of last night and this morning.
$\noindent b) The initial value problem (IVP) is$

$\frac{\mathrm{d}v}{\mathrm{d}t} = g-kv^{2}, \quad v(0) = 0,$

$\noindent where k = \frac{1}{5}.$

$\noindent Hence we have$

\begin{align*}\int_{0}^{v} \frac{\mathrm{d}\nu}{g - k\nu^{2}} &= \int_{0}^{t} \mathrm{d}\tau \quad (\text{since }v(0) = 0)\\ \Rightarrow \int_{0}^{v} \frac{\mathrm{d}\nu}{g - \left(\sqrt{k}\nu\right)^{2}} &= t.\end{align*}

$\noindent The integral on the left-hand side evaluates to \frac{1}{\sqrt{k}}\times \frac{1}{\sqrt{g}}\left[\tanh^{-1}\left(\frac{\sqrt{k}\nu}{\sqrt{g}}\right)\right]_{\nu = 0}^{\nu = v} = \frac{1}{\sqrt{kg}}\tanh^{-1}\left(\frac{\sqrt{k}v}{\sqrt{g}}\right). So this is equal to t, and solving for v yields$

$v(t) = \sqrt{\frac{g}{k}}\tanh\left(\sqrt{kg}t\right).$

$\noindent Remember that \tanh \xi = \frac{e^{\xi} - e^{-\xi}}{e^{\xi} + e^{-\xi}}.$

$\noindent c) The differential equation can be written as$

$\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{2}v ^{2}\right) + kv^{2} = 9.8 \iff \frac{\mathrm{d}y}{\mathrm{d}x} + 2ky = 9.8,$

$\noindent where y \equiv \frac{1}{2}v^{2}, using the fact that \frac{\mathrm{d}v}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{2}v^{ 2}\right). This is a linear first-order differential equation for y as a function of x, which you should know how to solve. Upon solving this, and utilising initial conditions, you will have an equation linking x and v, from which you should be able to answer this question.$

3. ## Re: Someone help me im DYING with this

Originally Posted by integrand
$\noindent b) the initial value problem (ivp) is$

$\frac{\mathrm{d}v}{\mathrm{d}t} = g-kv^{2}, \quad v(0) = 0,$

$\noindent where k = \frac{1}{5}.$

$\noindent hence we have$

\begin{align*}\int_{0}^{v} \frac{\mathrm{d}\nu}{g - k\nu^{2}} &= \int_{0}^{t} \mathrm{d}\tau \quad (\text{since }v(0) = 0)\\ \rightarrow \int_{0}^{v} \frac{\mathrm{d}\nu}{g - \left(\sqrt{k}\nu\right)^{2}} &= t.\end{align*}

$\noindent the integral on the left-hand side evaluates to \frac{1}{\sqrt{k}}\times \frac{1}{\sqrt{g}}\left[\tanh^{-1}\left(\frac{\sqrt{k}\nu}{\sqrt{g}}\right)\right]_{\nu = 0}^{\nu = v} = \frac{1}{\sqrt{kg}}\tanh^{-1}\left(\frac{\sqrt{k}v}{\sqrt{g}}\right). So this is equal to t, and solving for v yields$

$v(t) = \sqrt{\frac{g}{k}}\tanh\left(\sqrt{kg}t\right).$

$\noindent remember that \tanh \xi = \frac{e^{\xi} - e^{-\xi}}{e^{\xi} + e^{-\xi}}.$

$\noindent c) the differential equation can be written as$

$\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{2}v ^{2}\right) + kv^{2} = 9.8 \iff \frac{\mathrm{d}y}{\mathrm{d}x} + 2ky = 9.8,$

$\noindent where y \equiv \frac{1}{2}v^{2}, using the fact that \frac{\mathrm{d}v}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{1}{2}v^{ 2}\right). This is a linear first-order differential equation for y as a function of x, which you should know how to solve. Upon solving this, and utilising initial conditions, you will have an equation linking x and v, from which you should be able to answer this question.$
thankyouuu <3 <3 <3

4. ## Re: Someone help me im DYING with this

for c it asks for impact velocity in terms of h

I got

v(h)=sqrt(g/k) tanh(arccos(e^(h.sqrt(k/g)))

5. ## Re: Someone help me im DYING with this

Originally Posted by hayabusaboston
for c it asks for impact velocity in terms of h

I got

v(h)=sqrt(g/k) tanh(arccos(e^(h.sqrt(k/g)))
Your proposed answer isn't well-defined. For any h > 0, e^(h.sqrt(k/g)) > 1, so the arccos of this doesn't exist in the real numbers.

6. ## Re: Someone help me im DYING with this

Originally Posted by InteGrand
Your proposed answer isn't well-defined. For any h > 0, e^(h.sqrt(k/g)) > 1, so the arccos of this doesn't exist in the real numbers.
could you pls provide an assist? ^_^

7. ## Re: Someone help me im DYING with this

Originally Posted by hayabusaboston
could you pls provide an assist? ^_^
$\noindent Did you manage to solve the ODE for y as a function of x? Once we have this solution (noting the initial condition of y = 0 when x = h), we will get an equation relating y\equiv \frac{1}{2}v^{2} to x, and will involve h. To get the impact speed, set x = 0 and solve for v (taking the positive root).$

8. ## Re: Someone help me im DYING with this

Originally Posted by InteGrand
$\noindent Did you manage to solve the ODE for y as a function of x? Once we have this solution (noting the initial condition of y = 0 when x = h), we will get an equation relating y\equiv \frac{1}{2}v^{2} to x, and will involve h. To get the impact speed, set x = 0 and solve for v (taking the positive root).$
oops I did another way, I found time taken to travel h metres, called it b, then found the velocity at time b, so v(b)

I did integral from o to b of v(t)

9. ## Re: Someone help me im DYING with this

can I rewrite the arccosh(x) as ln(x+sqrt(x^2-1))?

10. ## Re: Someone help me im DYING with this

Originally Posted by hayabusaboston
can I rewrite the arccosh(x) as ln(x+sqrt(x^2-1))?
In your original answer, did you mean cosh-1 (I'm guessing you did)? You wrote arccos (didn't put an "h" on the end).

11. ## Re: Someone help me im DYING with this

Originally Posted by InteGrand
In your original answer, did you mean cosh-1 (I'm guessing you did)? You wrote arccos (didn't put an "h" on the end).
yea arccosh. So its right in that case?

12. ## Re: Someone help me im DYING with this

next question if anyone can provide guidance would b appreciated

Consider the differential equation for y(t) (t>=0)

dy/dt=sech(y)-2e/(e^2+1)

a) find the equilibrium solutions (is it 1 and -1?)
b)draw a phase plot of dy/dt versus y and label all intercepts (upside down parabola style at -1/1 intercepts right?)
c) Determine whether the equilibrium solutions are stable or not (is -1 unstable, and 1 stable?)
d) draw rough sketches of y(t) against t for several initial values y(0). Choose these initial values so they are spread across the regions defined by the equilibrium solutions and make sure that you illustrate all of the main shapes of population versus time graphs that can occur. You should draw a minimum of 4 such curves (NO IDEA PLS HALP)

13. ## Re: Someone help me im DYING with this

Originally Posted by hayabusaboston
next question if anyone can provide guidance would b appreciated

Consider the differential equation for y(t) (t>=0)

dy/dt=sech(y)-2e/(e^2+1)

a) find the equilibrium solutions (is it 1 and -1?)
b)draw a phase plot of dy/dt versus y and label all intercepts (upside down parabola style at -1/1 intercepts right?)
c) Determine whether the equilibrium solutions are stable or not (is -1 unstable, and 1 stable?)
d) draw rough sketches of y(t) against t for several initial values y(0). Choose these initial values so they are spread across the regions defined by the equilibrium solutions and make sure that you illustrate all of the main shapes of population versus time graphs that can occur. You should draw a minimum of 4 such curves (NO IDEA PLS HALP)
a) Yes.
b) Pretty much.
c) Correct!
d) Have a play around with this: https://www.geogebra.org/m/W7dAdgqc (type in sech(y) - sech(1) for the f'(x,y) in the top left).

Basically you need to sketch the direction field and draw some solution curves passing through various different points on the y-axis (corresponding to different values of y(0)). You should be able to see that depending on where you choose your y(0) value (y-intercept) in relation to the two bands of stable solutions, the solution curve that results will behave differently. If you have y(0) > 0, the solution will approach 1 as t -> oo. If we have -1 < y(0) < 1, the solution will approach 1 as t -> oo. If we have y(0) < -1, the solution will go to -oo as t -> oo.

In that applet I linked, you can tick all the boxes for Solution A-D in the bottom left to get various solution curves, and you can drag the points they are forced to go through to different places to see how the resulting solution curve changes.

14. ## Re: Someone help me im DYING with this

how did you identify y for part c?

15. ## Re: Someone help me im DYING with this

Originally Posted by A100p
how did you identify y for part c?
Just find the roots of the RHS function of the DE y' = f(y).

$\noindent In more detail, equilibrium solutions to an autonomous differential equation y' = f(y) are just constant solutions of the form y = y_{0} where y_{0} satisfies f(y_{0}) = 0 (it is easy to see that these are solutions). In other words, to find the equilibrium solutions, just find the zeros of the function on the RHS. In this thread's particular example, the DE is y' = \mathrm{sech}(y) - \mathrm{sech}(1), noting that \frac{2e}{e^{2} + 1} = \frac{2}{e + e^{-1}} = \mathrm{sech}(1). So the function f(y) here is \mathrm{sech}(y) - \mathrm{sech}(1), whose roots are \pm 1. Hence the equilibrium solutions are y = 1 and y = -1.$

16. ## Re: Someone help me im DYING with this

InteGrand if you can help with this would be amazing.

1. Find the general solution to the differential equation y''+2y'+3y=1+t^2+e^-t

is y(t)=Ae^-t.cos(sqrt2)t+Be^-t.sin(sqrt2)t+(e^-t)/2+(t^2)/3-(4t/9)+17/27?

and 2.Consider an electric circuit

L
-------<<------
i i
[V] i
i i
i i
-------II--------
C
With an inductance of L=0.5 Henry and a capacitor with a capacitance of C=0.1 farad connected in series with a voltage source of V(t)=50sin(wt) Volts. Initially the charge on the capacitor is 2 Coulombs and there is no current in the circuit.

a) Show that the differential equation satisfied by the charge q(t) in this circuit is q''+20q=100sin(wt)
b) Determine the value(s) Wres for which resonance occurs
c) Solve the initial value problem to find the charge qNR(t) as a function of time in the case where there is NO resonance.
d) Solve the initial value problem to find the charge qR(t) as a function of time for ONE of the values w(res) where there IS resonance (You can choose which wres value to consider.

Question 2 idk :'(

17. ## Re: Someone help me im DYING with this

Originally Posted by hayabusaboston
InteGrand if you can help with this would be amazing.

1. Find the general solution to the differential equation y''+2y'+3y=1+t^2+e^-t

is y(t)=Ae^-t.cos(sqrt2)t+Be^-t.sin(sqrt2)t+(e^-t)/2+(t^2)/3-(4t/9)+17/27?

and 2.Consider an electric circuit

L
-------<<------
i i
[V] i
i i
i i
-------II--------
C
With an inductance of L=0.5 Henry and a capacitor with a capacitance of C=0.1 farad connected in series with a voltage source of V(t)=50sin(wt) Volts. Initially the charge on the capacitor is 2 Coulombs and there is no current in the circuit.

a) Show that the differential equation satisfied by the charge q(t) in this circuit is q''+20q=100sin(wt)
b) Determine the value(s) Wres for which resonance occurs
c) Solve the initial value problem to find the charge qNR(t) as a function of time in the case where there is NO resonance.
d) Solve the initial value problem to find the charge qR(t) as a function of time for ONE of the values w(res) where there IS resonance (You can choose which wres value to consider.

Question 2 idk :'(
For Q1) you basically find a particular solution and add it on to the solution to the homogenous equation to get the overall solution. I don't want to do the calculations now but I assume you know the steps.

And for Q2), you should probably review your resonance theory (and did you manage to set up the ODE? Did you try Kirchhoff's Laws?). Here's some info and examples on resonance: https://math.libretexts.org/TextMaps..._and_Resonance .

18. ## Re: Someone help me im DYING with this

Originally Posted by InteGrand
For Q1) you basically find a particular solution and add it on to the solution to the homogenous equation to get the overall solution. I don't want to do the calculations now but I assume you know the steps.

And for Q2), you should probably review your resonance theory (and did you manage to set up the ODE? Did you try Kirchhoff's Laws?). Here's some info and examples on resonance: https://math.libretexts.org/TextMaps..._and_Resonance .
how to find value of w so there is resonance WOT idk m8 lost

19. ## Re: Someone help me im DYING with this

Originally Posted by hayabusaboston
how to find value of w so there is resonance WOT idk m8 lost
$\noindent Remember, there will be resonance iff the value of \omega is equal to the natural frequency'' \omega_{0} of the system. The ODE mx'' + kx = f(t) (where f(t) is the external force, or input) has natural frequency \omega_{0} = \sqrt{\frac{k}{m}}, if k,m are positive constants.$

20. ## Re: Someone help me im DYING with this

Originally Posted by InteGrand
$\noindent Remember, there will be resonance iff the value of \omega is equal to the natural frequency'' \omega_{0} of the system. The ODE mx'' + kx = f(t) (where f(t) is the external force, or input) has natural frequency \omega_{0} = \sqrt{\frac{k}{m}}, if k,m are positive constants.$
so if sin(wt) is equal to the general solution values of sin and cos?

characteristic equation results in 20i,-20i so y(t)=Acos20t+Bsin20t, if times by t will be equal with sinwt so

20 is it? later questions say "pick a value where resonance occurs" isnt there only the one?

21. ## Re: Someone help me im DYING with this

$\noindent Here's an example (different numbers to your Q.). Suppose we are given the differential equation$

$2x'' + 0.5x = \cos \left(\omega t\right).$

$\noindent Here m = 2 and k = 0.5, so the natural frequency'' of the system is \omega_{0} = \sqrt{\frac{0.5}{2}} = \sqrt{\frac{1}{4}} = \frac{1}{2}. So there'll be resonance if \omega = \frac{1}{2}, and no resonance otherwise (instead, a beats'' phenomenon occurs -- see the link I provided above).$

22. ## Re: Someone help me im DYING with this

Originally Posted by hayabusaboston
so if sin(wt) is equal to the general solution values of sin and cos?

characteristic equation results in 20i,-20i so y(t)=Acos20t+Bsin20t, if times by t will be equal with sinwt so

20 is it? later questions say "pick a value where resonance occurs" isnt there only the one?
Yeah there's only one positive value, but you could also use -20 and that'd still give resonance. Usually omega is thought of as positive I guess. Using a negative omega here just makes the RHS of your ODE become negative sine, since sine is odd. This means your ''particular solution'' becomes negative of the old one.

23. ## Re: Someone help me im DYING with this

Originally Posted by InteGrand
Yeah there's only one positive value, but you could also use -20 and that'd still give resonance. Usually omega is thought of as positive I guess. Using a negative omega here just makes the RHS of your ODE become negative sine, since sine is odd. This means your ''particular solution'' becomes negative of the old one.
for c)do you just get rid of w and say sint then do initial value problem?

for d) do you just solve the initial problem with w=20?

24. ## Re: Someone help me im DYING with this

Originally Posted by hayabusaboston
for c)do you just get rid of w and say sint then do initial value problem?

for d) do you just solve the initial problem with w=20?
$\noindent For c), you'd do it by solving it with a general \omega where \omega \neq \omega_{0}, where \omega_{0} is the natural frequency, which is the \textbf{square root} of 20. I.e. solve q'' + 20 q = 100 \sin \omega t, where \omega \neq \sqrt{20} (with the given initial values). For d), you'd solve the IVP with \omega = \sqrt{20}. As you'll find, the form of the particular solution is different in each case.$

25. ## Re: Someone help me im DYING with this

Originally Posted by InteGrand
$\noindent For c), you'd do it by solving it with a general \omega where \omega \neq \omega_{0}, where \omega_{0} is the natural frequency, which is the \textbf{square root} of 20. I.e. solve q'' + 20 q = 100 \sin \omega t, where \omega \neq \sqrt{20} (with the given initial values). For d), you'd solve the IVP with \omega = \sqrt{20}. As you'll find, the form of the particular solution is different in each case.$
wot how do u do that? solve w not equal to w0?

Page 1 of 3 123 Last

There are currently 1 users browsing this thread. (0 members and 1 guests)

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•