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MelodicRose(OD)

Calc 2 Struggles...

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Hello folks! So I have a rather extensive Calc 2 project that's due on Friday that I am struggling massively with. Mainly the first problem (I've gotten a lot of the other problems worked on and can easily go to the math lab tomorrow for clarification and checking for correctness). I've got a picture of the entire first problem, I'm not looking for answers but mainly pointers on what to do/work on. I struggled with problems like this on my first exam (I failed with a 24%..) so I'm really really looking for some kind of help so I can understand what on Earth I am doing..

Problem : https://gyazo.com/8521c9aa24280fe91fe3ddd3c4407300

I know this thread hasn't been updated in a while, but I figured I'd give it a shot x-x Thank you in advance

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So which parts of the problem do you need help with? Everything?

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I have never taken a Calculus course.. and all of the College level math I've done stopped at the first question. I can help you with the focus for the Parabola but that's probably about it. Sorry... lol. I can start you there if you need it though.

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Any help would be <3 I don't even know what the focus is. The really crappy part is that the stuff on the project are stuff we haven't learned in class.. Like hyperbolic functions (what even)

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Ok they would give you a Geometry/Algebra II Trig problem that I haven't seen in roughly 15 years. My gut tells me he's given you an Ellipse problem. An Ellipse is a circle whose radius is not constant as you rotate around. Namely it has a Major and Minor axis. I'd have to work this problem out myself and verify if I am right(I could be wrong though). Here's some info regarding Focus. To convert square inches into square meters requires dimensional analysis and knowing that 1 inch = 2.54 centimeters or .0254 meters.

PS: X-Axis is 48 inches long, Y-Axis looks to be 18 inches or major and minor axis respectively after you rotate around the Y-Axis. To find Focus points, use the equation in that link.

Edited by DarkHelmet

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The focus by itself is hard to conceptualize. It properly define the focus you need its counterpart: the directrix. The focus is a point while the directrix is a line. If you measure the distance from the directrix to the parabola (measuring perpendicularly from the directrix) that distance will be the same as the distance from that point on the parabola to the focus.

Here's an image to visualize it

focus-directrix-locus.jpg

As for the integral step, it's not terribly bad. You need to integrate with respect to y which means you need to rewrite your problem so that y is the independent variable (f(y) instead of f(x)). From there you need to rewrite your integral. Instead of summing the distance from the y axis to the parabola you want to treat that distance as a radius to a circle. Using that radius, you want to sum up the circumferences of those circles to get the surface area. (Remember that integrals are just another way to perform sums.)

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Don't have the time to explain this... lol, disregard and talk to Altros. REALLY busy atm lol

Edited by DarkHelmet

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So I managed to get the focus and the equation of the function..

y = 1/128 x^2.. or x = (128y)^1/2 <-- If I did my math correctly.

The focus is (0,32) <-- which seems strange but was taken as correct.

I was told by other classmates that I needed to do the Surface Area formula which is : https://gyazo.com/a97a6b8240db5d3073dc379cecb2694a

So.. I'm assuming since I need to go around the Y-axis, not the x-axis, I do need to use the "x = " function. f(y) = (128y)^1/2.. f'(y) = 4sqrt(2)/sqrt(y)? <-- I hate square roots.. with a passion! So you square f'(y)^2 = 32/y

SO SA = https://gyazo.com/c21135353220de5879dfa6ea20c84c18 <-- The 32/x in there should be 32/y, I mis-typed. They should all be in terms of "y" tbh, I just typed in x cuz I'm stupid sometimes. and the first function should be the (128y) ^1/2

I don't know what the limits would be in terms of Y.. Could I do u-sub to finish that out? What units would it be in as well?

Edited by MelodicRose(OD)

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SA = Integral(2*pi*r*dS)

r = f(y) in this case

dS = sqrt(1 + f'(y)**2) dy in this case

In the formula you linked (first one) the addition of 1 was missing.

Deriving the relationship between dS and dx or dy is rather straightforward.

Imagine you have a curve and you want to approximate the length of the curve. You draw a bunch of lines from point to point and sum the length of those lines. As you draw more and more lines your approximation becomes better and better.

So now we'll look at one of these lines. We shall saw that it ranges from x1 to x2 and y1 to y2. We shall say that it is of length P. By Euclidean distance: P = sqrt( (x2 - x1)**2 + (y2 - y1)**2 ).

Next we look at Mean Value Theorem, which states that for a range of values, there exists some point (zm) such that: f(z2) - f(z1) = f'(zm) * (z2 - z1)

(z2 - z1) = dz

So now it's just a bit of substitution.

For dx:

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

for dy:

P = dS = sqrt( (f'(y)dy)**2 + dy**2 ) = sqrt ( (f'(y)**2 + 1) * dy**2 ) = sqrt (1 + f'(y)**2) * dy

For the ranges, you look at the range of values, along your axis of integration, that you want to integrate across. (The whole parabola in this case.)

I would use u-substitution to integrate. If you need help choosing u I can do my best to point you in the right direction without spoiling it.

Edited by Altros(OD)

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So it is not SA = integral 2*pi*f(x) * sqrt((1+ f'y)^2)?

I thought you needed the original function ahead of the square root too, was what I did somewhat right? x-x

(please don't get frustrated, I just really don't understand)

I currently have SA = Integral 2*pi*(128y)^1/2 * sqrt((1+32/y))

The 32/y is f'(y)^2 if I did the math correctly. Idk how a u sub would make that work :/

Edited by MelodicRose(OD)

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So it is not SA = integral 2*pi*f(x) * sqrt((1+ f'y)^2)?

I thought you needed the original function ahead of the square root too, was what I did somewhat right? x-x

(please don't get frustrated, I just really don't understand)

You have it almost right. I need to go for just over an hour but when I get back I'll type up a full breakdown of how the integral is constructed. I can, however, quickly fix that function.

SA = integral( 2 * pi * f(y) * sqrt(1 + f'(y)^2) dy)

That function should yield the correct surface area when integrated.

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So it is not SA = integral 2*pi*f(x) * sqrt((1+ f'y)^2)?

I thought you needed the original function ahead of the square root too, was what I did somewhat right? x-x

(please don't get frustrated, I just really don't understand)

You have it almost right. I need to go for just over an hour but when I get back I'll type up a full breakdown of how the integral is constructed. I can, however, quickly fix that function.

SA = integral( 2 * pi * f(y) * sqrt(1 + f'(y)^2) dy)

That function should yield the correct surface area when integrated.

Without limits on the integral? :S

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The integral should be definite. Your axis of integration is the y-axis and you are integrating across the whole parabola as defined by the problem (ends at (-48, 18);(48,18) with vertex at (0,0)). So your limits are from y=0 to y=18.

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Ok now that I have some time(and the fact I had to pull out an old Pre-Calc book), I read up on Parabolas in Analytic Geometry chapter. As Altros demonstrated, you got a parabola whose end points end at -48,18 and 48,18 respectively. Its vertex is at the origin so you know the equation is in standard form or x^2 = 4py with p being the value of the focus. You are give two points on the parabola so use them in the equation and solve for P. Doing so yields P = 32, hence 0,32 is the Focus :)

Now to take the integral for Surface Area, we need to get its equation, hence substitute 32 for P to get x^2 = 128y or y = [(x^2)/128)]. Take the derivative to get y' = x/64

Area of a Surface of Revolution

What you get from that link above is that to get the surface area of the parabola as you rotate it around the y-axis is the following equation(equation#8 in link)

SA = Integral of 2pi * x * ds dx. 2pi=constant(2 times pi), x being just that x and ds = SQRT of (1 + (dy/dx)^2) and dx = integral with respect to x. You will be taking the integral from 0 to 48(value of x at vertex out to its end-point). Now plug the derivative into the equation and use U-substitution. U = 1 + (x^2 / 4096 <- 64 squared), DU = x / 2048 and limits of 1 and 1.5625. Solve for SA. To convert square inches into square meters, use dimensional analysis.

Altros, nothing is straightforward in mathematics. Got to translate the mathematical jargon into stuff people will understand ;). Also if I misread the problem and its around x-axis(which wouldn't make sense frankly) you'd create a funky conic that I can't really visualize well.

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You guys are legit doing two different things.. altros is doing it across the y axis, so I had to turn everything in terms of y.. with y limits of integration. Dark is doing it with respect to x with x limits of integration. Why..

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I had to draw a picture to help with visualizing my explanation. I probably won't have enough time to edit it into this post so I'll be posting it in a few minutes.

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HViQGbD.png

Man that image is ginormous.

Okay so now I can explain how this works.

We have a parabola with the y-axis as its axis of symmetry. We want the surface area of the object created by rotating the parabola about the y-axis.

We have f(x) = y(x) = y = x**2 / 128. We can rewrite this to h(y) = x(y) = x = 8 * sqrt(2y).

Now, we know the circumference of the circle in the picture above. It's 2 * pi * h(y). Now, unfortunately we cannot multiply this by dy as ds does not equal dy (see zoomed in picture). So we get 2 * pi * h(y) * ds.

However, ds, dy and dx make a right triangle. ds**2 = dx**2 + dy**2. So ds = sqrt(1 + (dx/dy) ** 2) * dy. (My previous post has the proof for this.

So now we have this to represent the infinitely small circular cross-section: 2 * pi * h(y) * sqrt(1 + (dx/dy) ** 2) * dy.

dx/dy = h'(y) = 8 / sqrt(2y)

(dx/dy) ** 2 = 64 / 2y = 32 / y

Thus: 2 * pi * 8 * sqrt(2y) * sqrt(1 + 32 / y) * dy = 16 * pi * sqrt(2y * (1 + 32 / y) ) * dy
= 16 * pi * sqrt(2y + 64) * dy

Now we can integrate this to obtain the surface area

SA = integral(16 * pi * sqrt(2y + 64) * dy)

But we have a limited range. We want to integrate across the whole parabola and the parabola's y-values range from 0 to 18 ( (0, 0) to (+/- 48, 18) ). Thus:

SA = integral[0:18](16 * pi * sqrt(2y + 64) dy) = 16pi * integral[0:18](sqrt(2y + 64) dy)

You'll want to use u-substitution to complete that integral. If something still doesn't make sense, feel free to keep asking questions.

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I don't understand how you got that when you changed it to x = .. I got something completely different.

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I don't understand how you got that when you changed it to x = .. I got something completely different.

y = x**2 / 128

x**2 = 128y

x = +/- sqrt(128y)

x = sqrt(128y)

x = sqrt(64 * 2 * y)

x = sqrt(64) * sqrt(2y)

x = 8 * sqrt(2y) = x(y) = h(y)

dx/dy = 8 * (1/2) * (2y) ** (-1/2) * 2 <--- from chain rule

dx/dy = x'(y) = h'(y) = 8 / sqrt(2y)

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I wish I could algebra that well. the square roots trip me up something fierce. Let me work on this part and I will poke you with what I come up with. I appreciate the step by step, it helps a lot x-x

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Think of roots as exponents.

sqrt(x) = x**(1/2)

cbrt(x) = x**(1/3)

nrt(x) = x**(1/n)

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"Thus: 2 * pi * 8 * sqrt(2y) * sqrt(1 + 32 / y) * dy = 16 * pi * sqrt(2y * (1 + 32 / y) ) * dy
= 16 * pi * sqrt(2y + 64) * dy"

I'm sorry, can you explain this part? Did you combine the square roots?

Also would you set u = 2y +64

Edited by MelodicRose(OD)

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Yes. Rewriting as exponents:

(2y) ** (1/2) * (1 + 32 / y) ** (1/2) = (2y * (1 + 32 / y)) ** (1/2)

Edited by Altros(OD)
  • Upvote 1

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