Someone help me im DYING with this (1 Viewer)

hayabusaboston · Apr 25, 2017

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.

InteGrand · Apr 25, 2017

hayabusaboston said:
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.$

hayabusaboston · Apr 25, 2017

integrand said:
$\noindent b) the initial value problem (ivp) is$

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

$\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

hayabusaboston · Apr 25, 2017

for c it asks for impact velocity in terms of h

is my answer correct
I got

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

InteGrand · Apr 25, 2017

hayabusaboston said:
for c it asks for impact velocity in terms of h

is my answer correct
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.

hayabusaboston · Apr 25, 2017

InteGrand said:
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? ^_^

InteGrand · Apr 25, 2017

hayabusaboston said:
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).$

hayabusaboston · Apr 25, 2017

InteGrand said:
$\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)

hayabusaboston · Apr 25, 2017

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

InteGrand · Apr 25, 2017

hayabusaboston said:
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).

hayabusaboston · Apr 25, 2017

InteGrand said:
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?

hayabusaboston · Apr 25, 2017

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)

InteGrand · Apr 25, 2017

hayabusaboston said:
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.

A100p · Apr 29, 2017

how did you identify y for part c?

InteGrand · Apr 29, 2017

A100p said:
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 $y = -1$.$

You can see here for more information and examples: http://tutorial.math.lamar.edu/Classes/DE/EquilibriumSolutions.aspx .

hayabusaboston · May 9, 2017

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 :'(

InteGrand · May 9, 2017

hayabusaboston said:
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/TextMap...r_ODEs/2.6:_Forced_Oscillations_and_Resonance .

hayabusaboston · May 9, 2017

InteGrand said:
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/TextMap...r_ODEs/2.6:_Forced_Oscillations_and_Resonance .

how to find value of w so there is resonance WOT idk m8 lost

InteGrand · May 9, 2017

hayabusaboston said:
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.$

hayabusaboston · May 9, 2017

InteGrand said:
$\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?

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