Query 09 redo

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course Mth 279

7/9 12:16 am

Query 09#$&*

course Mth 279

6/30 9:51 pm

Query 09 Differential Equations*********************************************

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Question:  3.6.4.  A 3000 lb car is to be slowed from 220 mph to 50 mph in 4 seconds.  Assume a drag force proportional to speed.  What is the value of k, and how far will the car travel while being slowed?

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Your solution: 

 I feel like I don't have enough values for variables

v' + k/m v = -g

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The car is not falling, so g is not relevant. Your equation is simply

v ' + k / m v = 0.

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You need to solve this equation before you substitute anything.

The equation is first-order linear and homogeneous.

It is also separable.

It can be solved by either or, preferably (for the extra insight it provides), by both of those methods.

Having obtained your solution you can plug in the numbers for the symbols, and use the conditions you have correctly stated:

v(0) = 220 mph

v(4) = 50 mph

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Maybe I'm drawing a blank, but if we have mv' = -kv -mg, where g = 0

mv' = -kv

shouldn't you divide by -kv instead of adding it to the other side?

2x = 2 simplifies to x = 1 and not 2x - 2 = 0 ???

mv' + kv = 0

v' + k/m v = 0

e^(k/m t) v' + k/m e^(k/m t) v = 0

integrate

e^(k/m t) v = C

v = Ce^(-k/m t)

when t = 0, v = 220

220 = C

v = 220e^(-k/3000 t)

solving for k, v = 50 when t = 4

50 = 220 e^(-k/3000 * 4)

ln (5/22) = -k/3000 *4

-3000/4 ln(5/22) = k

v = 220 e^(( -3000/4 ln(5/22) )/3000 *t)

However, I don't know how to now incorporate x.

I know that v = 0 when the car has slowed to a stop. Knowing that I'll be able to solve for time.

??? how do you get distance?

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To find the displacement you just integrate v with respect to t.

v(t) is a simple exponential function (once you simplify all your constants) and straightfoward to integrate.

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v' is the change in v over the change in t. Since the car slowed 170 mph (249.333 feet per second) I can substitute

249.3333/4 + k/3000 v = -3.2

However, I have no way to solve for k. The only way to solve for the distance would be to take the integral of the velocity

62.333 + k/3000 v = -3.2

v = -177400/k

integrating

x = -177400/k * t + C

But that just adds another variable I can't solve for

 

v(0) = 220 mph

v(4) = 50 mph

I think there should be a t somewhere in my equation for v(t)

 

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Given Solution: 

 

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Question:  3.6.6.  A vertical projectile of mass m has initial velocity v_0 and drag force of magnitude k v.  How long after being fired will it reach its maximum height?

If the projectile has mass .12 grams and after being fires straight upward at 80 meters / second reaches its maximum height after 2.5 seconds, then what is the value of k?

 

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Your solution: 

 

 How do I know whether the drag force is v or v^2???

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You are told that the drag force has magnitude k v.

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  &&&& will we always be told what it is??? &&&

 

 m dv/dt = -mg - kv

v' + k/m v = -g

dv/dt * dt/dx = dv/dx

dv/dx + k/m v = -g

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dv/dt * dt/dx = dv/dx is true, but this doesn't say that dv/dt = dv/dx.

YOu have replaced v ' = dv/dt with dv/dx, with no change to the rest of the equation.

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&&& how do I put x into the equation ??? &&&

when the velocity = 0, the projectile will be at maximum height

dv/dx + k/m (0) = -g

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Your initial conditions must be substituted after the equations is solved, not before. The equation works for functions of t, not for the constant values of those functions at some specific value of t.

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dv/dx = -g

dv = -gdx

integrate

v = -gx + C

v = -gx + v_0

I don't really know how to solve for k, time *and* distance at the same time. Or separately for that matter.

Does the fact that the ball is thrown upwards make this problem different from the first one???

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By using -g you have declared the positive direction to be up. so the initial velocity must be positive.

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You need to return to the equation

 m dv/dt = -mg - kv

and solve the equation for v(t). The equation is separable and the integration requires just straightforward u substitution.

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dv/dt + k/m v = -g

what do you mean by separable??? as in, I can move the dt to the other side to solve?? how would I solve the left hand side?

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You are solving the equation as first-order linear nonhomogeneous, which is fine. You'll get the same solution as if you treated it as separable.

The equation can be rearranged to the form

dv / (-k/m v - g) = dt.

A straightforward u substitution u = - k / m v - g makes it easy to integration the left-hand side, that the right-hand sides integrates to t + c.

The rest follows in the usual manner.

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e^(k/m t) v' + k/m e^(k/m t ) v = -g e^(k/m t)

integrate

e^(k/m t) v = -mg/k e^(k/m t) + C

v = -mg/k + Ce^(-k/m t)

It will reach max height when v = 0

However, I don't know how to put x into the equation.

plugging in values to solve for k

when t = 0, v = 80

80 = -.12(9.8)/k + C

C = 80 + .12(9.8)/k

0 = -.12(9.8)/k + Ce^(k/.12 (2.5) )

I should be able to solve for C and k.

0 = -.12(9.8)/k + [80 + .12(9.8)/k]*e^(k/.12 (2.5) )

I'm having trouble moving the k's to one side.

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You can't algebraically solve that equation for k.

Newton's Method can work on this equation, as can trial and error.

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Given Solution: 

 

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Self-critique (if necessary):

 

 

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Question:  3.6.10.  A 82 kg skydiver falls freely for 10 seconds then opens his chute.  He reaches the ground 4 seconds later.  Assume air resistance is proportional to speed, and assume that with this chute a 90 kg would reach a terminal velocity of 5 m / s. 

At what altitude was the parachute opened?

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Your solution: 

 In the equation mv' +kv = -mg, which term is terminal velocity??? Or how to I solve for that??? 

 mv' is the same as mass * acceleration, which is force. I don't know what the kv is. Then the -mg is the force of gravity pointing downwards.

 So kv is the air resistance. And when I solve for an equation of v(t), the terminal velocity is v(t) as it approaches infinity???

Not too sure how knowing that is helpful.

Two part problem?

Part one: falling freely with air resistance slowing him down

Part two: falling with chute, both chute and air resistance slowing him down.

Part one:

v' + k/m v = -g

dt = 10 s

m = 82 kg

v' + k/82 v = -9.8

I still don't know how to solve for k ???

In the book example, it kind of just disappears out of the equation.

Part two:

dt = 4 s

terminal velocity = 5 m/s. Is that before or as he's hitting the ground????

How did his mass change to 90 kg??? 

v' + k/90 v = -9.8

I still don't know k or how to find height, or time.

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The first thing you have to do is solve the equation

v' + k/82 v = -9.8

Then you can formulate the given conditions and find the constants, which will be k and the integration constant you get when solving.

This equation is first-order linear nonhomogeneous, and it is also separable. If you can, I recommend solving it both ways and reconciling the two solutions.

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  &&&& how did you get the equation

v' + k/82 v = -9.8 ??? Shouldn't the parachute weight be included in his weight?

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The chute is included in the 90 kg.

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v' + k/(82+90) v = -9.8

e^(k/82 t) v' + k/82 e^(k/82 t) v = -9.8 e^(k/82 t)

integrate

e^(k/82 t) v = -9.8*82/k e^(k/82 t) + C

v = -9.8*82/k + Ce^(-k/82 t)

This is one way to solve ( I think). What is the other way?

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The other way is to use separation of variables.

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Was I correct in assuming this is a two part problem?

How does the terminal velocity help solve the equation?

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At terminal velocity v_term the drag force is equal to the weight, so

k v_term = weight.

You are given v_term and weight, so you can find k.

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@&

Your solutions are often a little out of order. On at least a couple of problems you've tried to impose the given conditions before solving the equation.

In most cases you do have a correct equation. Your first step will be to solve the equation, then to impose the conditions in order to evaluate the constants.

&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).

Be sure to include the entire document, including my notes.

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&#Your work looks good. See my notes. Let me know if you have any questions. &#