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MelodicRose(OD)

Calc 2 Struggles...

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So the first one seems like an arc-length problem. The easiest way to solve that, if I'm recalling arc lengths correctly, is to like this:

L = integral(ds)

where ds is the same ds we saw in the parabola. That is, ds**2 = dx**2 + dy**2.

The second problem is a work problem that involves integrating using discs. Work = distance * weight. Weight = volume * density. You'll also want to integrate along the y-axis again because the parabola opens upwards and you're treating it as a tank.

I figured out the work problem now, found an example of it in the book. Still at a loss for the arc length hyperbolic one, any chance you can muster through another walk through? x-x

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So if you remember that picture I made yesterday, we had that right-triangle of dx, dy and ds (the hypotenuse). dx and dy are the "infinitely small" changes to x and y as you move down the line. ds is the "infinitely small" length of the line covered by (x+dx, y+dy).

Arc Length is the length of a curve. So if you were to pull the curve taut so that it's a straight line and measure that, that's the arc length. So since ds is length of line covered by dx and dy then we want to sum up all of the ds's to get the arc length.

Thus: Arc Length = integral(ds).

Now, from geometry, we know that if c is the hypotenuse and a,b are the other sides then: c**2 = a**2 + b**2. Thus:

ds**2 = dx**2 + dy**2

From this we can solve for ds in terms of either dy or dx, based on what our axis of integration is.

If our axis of integration is the x-axis (dx) then:

ds = sqrt(dx**2 + dy**2)

ds = sqrt( dx**2 * (1 + dy**2/dx**2) )

ds = sqrt(dx**2) * sqrt(1 + (dy/dx)**2)

ds = sqrt(1 + (dy/dx)**2) * dx

or

ds = sqrt(1 + (y')**2) * dx

or

ds = sqrt(1 + (f'(x))**2) * dx

If our axis of integration is the y-axis (dy) then:

ds = sqrt(dx**2 + dy**2)

ds = sqrt( (dx**2/dy**2 + 1) * dy**2 )

ds = sqrt( (dx/dy)**2 + 1) * sqrt(dy**2)

ds = sqrt( 1 + (dx/dy)**2 ) * dy

or

ds = sqrt( 1 + (x')**2) * dy

or

ds = sqrt( 1 + (h'(y))**2 ) * dy

For this problem your integral will be definite since the towers are a set distance apart.

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So if you remember that picture I made yesterday, we had that right-triangle of dx, dy and ds (the hypotenuse). dx and dy are the "infinitely small" changes to x and y as you move down the line. ds is the "infinitely small" length of the line covered by (x+dx, y+dy).

Arc Length is the length of a curve. So if you were to pull the curve taut so that it's a straight line and measure that, that's the arc length. So since ds is length of line covered by dx and dy then we want to sum up all of the ds's to get the arc length.

Thus: Arc Length = integral(ds).

Now, from geometry, we know that if c is the hypotenuse and a,b are the other sides then: c**2 = a**2 + b**2. Thus:

ds**2 = dx**2 + dy**2

From this we can solve for ds in terms of either dy or dx, based on what our axis of integration is.

If our axis of integration is the x-axis (dx) then:

ds = sqrt(dx**2 + dy**2)

ds = sqrt( dx**2 * (1 + dy**2/dx**2) )

ds = sqrt(dx**2) * sqrt(1 + (dy/dx)**2)

ds = sqrt(1 + (dy/dx)**2) * dx

or

ds = sqrt(1 + (y')**2) * dx

or

ds = sqrt(1 + (f'(x))**2) * dx

If our axis of integration is the y-axis (dy) then:

ds = sqrt(dx**2 + dy**2)

ds = sqrt( (dx**2/dy**2 + 1) * dy**2 )

ds = sqrt( (dx/dy)**2 + 1) * sqrt(dy**2)

ds = sqrt( 1 + (dx/dy)**2 ) * dy

or

ds = sqrt( 1 + (x')**2) * dy

or

ds = sqrt( 1 + (h'(y))**2 ) * dy

For this problem your integral will be definite since the towers are a set distance apart.

I understood that part..

So to set up the integral would it be from 0 to 1280? I don't know how to do the derivative of that y= function is my issue. I know the arclength formula, I just don't know how to do it with the hyperbolic function..

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Ah, okay.

One thing I need to ask about the hyperbola function, are you sure the function you linked is correct?

The +/- makes it actually 2 functions, which is fine for hyperbolas. However, both functions are actually just lines, neither one is curved.

In the case of the "-" you get a horizontal line.

In the case of the "+" you get a line with a y-intercept of -499.98 and a slope of 15/(e**500).

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https://gyazo.com/06d66cc6c28593c21f29590a8fc27831

This is literally what the problem says on my paper. It's under the "Hyperbolic Functions" section of the project, so I'm at a loss xD

I probably mistyped something somewhere .. zzz

Edited by MelodicRose(OD)

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I, uh, don't know how to say this...but that function is a straight horizontal line...

y = 0.03 * ( -16666 + 250 * ( x/(e**500) + (-x)/(e**500) ) )

y = 0.03 * ( -16666 + 250 * (x - x) / (e**500) )

y = 0.03 * ( -16666 + 250 * (0) / (e**500) )

y = 0.03 * ( -16666 + 0)

y = 0.03 * -16666

y = -499.98

So...uh...I guess the length from y(i) to y(i + 1280) is 1280 meters, regardless of i.

I'm pretty certain your professor typo'd that problem but I have no idea what his intended hyperbola was and that is what his current equation is...so...

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I, uh, don't know how to say this...but that function is a straight horizontal line...

y = 0.03 * ( -16666 + 250 * ( x/(e**500) + (-x)/(e**500) ) )

y = 0.03 * ( -16666 + 250 * (x - x) / (e**500) )

y = 0.03 * ( -16666 + 250 * (0) / (e**500) )

y = 0.03 * ( -16666 + 0)

y = 0.03 * -16666

y = -499.98

So...uh...I guess the length from y(i) to y(i + 1280) is 1280 meters, regardless of i.

I'm pretty certain your professor typo'd that problem but I have no idea what his intended hyperbola was and that is what his current equation is...so...

https://gyazo.com/e6e2c856d51bbb79bbca2b722dd2ccc7 No it isn't :/

https://gyazo.com/73fe0ce760a4a9665ccb009a387066cb

These are the definitions of the hyperbolic functions on the beginning part of this section. The end of the y= of my project problem looks alot like sinhx.. but idk what to do xD

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Oh, woops. Completely misread that. I thought it was x/e**(500) not e**(x/500). That makes much more sense. Alright:

y = 0.03 * ( -16666 + 250 * ( e**(x/500) + e**(-x/500) ) )

y' = 7.5 * ( e**(x/500)/500 - e**(-x/500)/500 )

y' = 7.5/500 * (e**(x/500) - e(-x/500))

Now, as for what the bounds of the integral are...I would have to guess from y-intercept to y-intercept.

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I need to square that y' correct? and thanks youuuuuuuu

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Yeah I legitimately don't know how to do this problem cuz when I go looking for x - intercepts my online thingy says there are none, there is only one y-intercept and it's below (0,0)

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Oh woops, from x-intercept to x-intercept. There should be 2 and yes, you'll need to square y'.

Unfortunately, I'm not that great at solving hyperbolic functions for x-intercepts.

y = 0.03 * (-16666 + 250 * (e**(x/500) + e**(-x/500)))

y = 0.03 * (-16666 + 500 * cosh(x/500))

0 = 0.03 * (-16666 + 500 * cosh(x/500))

0 = -499.98 + 15 * cosh(x/500)

499.98 / 15 = cosh(x/500)

+/-arccosh(499.98 / 15) = x / 500

x = 500 * +/-arccosh(499.98 / 15)

x = 500 * arccosh(499.98 / 15) and 500 * -arccosh(499.98 / 15)

arccosh(z) = ln(z + sqrt(z - 1) * sqrt(z + 1))

cosh(z) = (e**z +e**-z) / 2

I don't know if you've covered hyperbolic trig functions or not. I tried to solve this without using them but I ran into a logarithm that I did not know how to reduce. That ultimately forced me to use hyperbolic trig functions.

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Oh woops, from x-intercept to x-intercept. There should be 2 and yes, you'll need to square y'.

Unfortunately, I'm not that great at solving hyperbolic functions for x-intercepts.

y = 0.03 * (-16666 + 250 * (e**(x/500) + e**(-x/500)))

y = 0.03 * (-16666 + 500 * cosh(x/500))

0 = 0.03 * (-16666 + 500 * cosh(x/500))

0 = -499.98 + 15 * cosh(x/500)

499.98 / 15 = cosh(x/500)

+/-arccosh(499.98 / 15) = x / 500

x = 500 * +/-arccosh(499.98 / 15)

x = 500 * arccosh(499.98 / 15) and 500 * -arccosh(499.98 / 15)

arccosh(z) = ln(z + sqrt(z - 1) * sqrt(z + 1))

cosh(z) = (e**z +e**-z) / 2

I don't know if you've covered hyperbolic trig functions or not. I tried to solve this without using them but I ran into a logarithm that I did not know how to reduce. That ultimately forced me to use hyperbolic trig functions.

We didn't cover them, that's why I have no idea what I am doing, I don't know how to derive them, integrate them, anything

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The hyperbolic trig functions were just to find the ranges for the definite integral. That x = ... is the x values for the x-intercepts. You can convert arccosh to ln using the formula I gave you. I also showed the formula I used to convert the original hyperbola function to cosh.

If your professor hasn't covered them then that's pretty typical. Most of the time hyperbolic functions would be covered in Calc I for maybe 1 lecture. They really don't put much emphasis on them but they're really use for conics.

Hyperbolic Trig Functions

Inverse Hyperbolic Trig Functions

If you want to try and memorize the formulas for them. I don't actually know the proofs for them; I believe they're pretty involved with logarithms and complex-space.

Anyway, my previous post gives you the 2 x-intercepts to use as bounds. I believe one of my previous posts showed how to take the derivative of y to get y'.

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Okay, I'm going to attempt it :3 will update if I need more help (bound to happen..)

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Hello again beautiful people, I'm back with only ONE problem today! I've done the rest of the project on my own and the problems were correct according to my professor (woot!). But I have one more.. I have no idea, so I am looking for more insight! This one is a differential equation of hell

https://gyazo.com/96b9214613af241eb16c757a039b7364

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Small picture is small. Any chance we can get a higher resolution picture?

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Yes, that's better though I can't see part C.

For part A it looks like you should plug in 0 for dv/dt and solve for v.

For part B you need to separate the variables. So we have m * (dv/dt) = m * g - k * v**2. Start by separating the differential (multiply both sides by dt). From there, you want all "v"s on the same side as dv (specifically you want them being multiplied/dividing dv) and everything else with dt. Normally you would want all "t"s with dt but there are no "t"s in this case so that simplifies it. Once you have it like this: h(v)dv = f(t)dt then you simply integrate both sides and solve for v.

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Part C says : suppose m=70kg , g=9.8m/sec2, find the terminal velocity and a general formula for v

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I also have 0 idea how to separate the variables in this, we were told it might have a natural log in it and I'm sitting here like i have no idea..

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Okay, I think I misunderstood the problem. You aren't supposed to integrate until part 3. Part 3 also looks like it's impossible without being given a value for k. Regardless.

Part 1: assume dv/dt = 0. Solve for v. This is your vf aka your terminal velocity.

Part 2: leave dv/dt in. Solve for v. Reduce your solution to include as little "m"s, "g"s and "k"s as possible by using your equation for vf to substitute them as vf. This is what they mean by "write in terms of vf," I think. I couldn't actually get the entire thing to be just "vf"s and "dv/dt"s.

Part 3: use the given values in your equation from part 1 to calculate vf. Now take your equation in part 2 and rewrite it so that it looks like this: h(v)dv = f(t)dt. (Note that f(t) and h(v) do not technically need to contain t or v.) Integrate both sides and solve for v. That will get you the general equation for v. (There is an easier way to go about this, but that involves differential equations, something that is typically taught after calculus.)

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This problem is technically a differential equation problem.. we learned basics for some oddball reason

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They taught you differential equations in Calc II? DiffEqs are sliced right out of linear algebra, they don't really have a place being taught in a Calc II lecture. Which method did he teach you for solving DiffEqs?

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I literally dunno how to do part b still lol

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Okay. So I was initially right about part b then. Separation of variables works like this:

If you have a equation of this form: dy/dx = f(x) * h(y) then you can separate the two sides.

divide both sides by h(y): (1 / h(y)) * dy/dx = f(x)

multiply both sides by dx: (1 / h(y)) * dy = f(x) * dx

Once in this form, you then integrate both sides, left with respect to y and right with respect to x.

For this problem you have things in terms of v and t. There are also no "t"s in the differential so f(t) = 1. This means that we have this:

m * (dv/dt) = mg - kv**2

dv/dt = g - kv**2 / m

h(v) = g - kv**2 / m

f(t) = 1

What I don't know is if you should then integrate in part 2 or in part 3. He says to solve for v and write in terms of vf so I'd assume you integrate in part 2.

Edited by Altros(OD)

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