Calculus in Project Arrhythmia... Integral Cube

Рет қаралды 3,198

Жыл бұрын

It's like that one level Algebra but way worse.
PLAY THE LEVEL: steamcommunity.com/sharedfile...
==CREDITS==
Level by @MOTIONIII
Song "Integral Cube" by Function Phantom
Play Project Arrhythmia: vitamin.games
Since you read to the bottom of this...join DXL's store, my discord server? And subscribe, of course :D / discord

Пікірлер: 26

@Official_Dre Жыл бұрын

we need more videos of dxl playing levels i love your reactions lmao. as someone who is still only learning basic geometry, this level scares me

@nottelling8695 Жыл бұрын

as someone who just learned what numbers are I'm terrified

@robloxhood101 Жыл бұрын

Me who decided to type out the answers in the comments

@Minemaker0430 Жыл бұрын

Funniest level I've ever seen. I was laughing so hard during the stream lol

@sijam2m59 Жыл бұрын

Same

@metalpipelol69 Жыл бұрын

Algebra from pa memories 1.0 but harder

@nabeel01gcg43 Жыл бұрын

Yeah totally from Memories 1.0

@MochaSlushes Жыл бұрын

thank you for uploading the best part of the stream lmao

@IcefangGD Жыл бұрын

Evaluate the line integral where C is the given curve. We're integrating over the curve C, y to the third ds, and C is the curve with parametric equations x = t cubed, y = t. We're going from t = 0 to t = 2. So we're going to integrate over that curve C of y to the third ds. We're going to convert everything into our parameter t in terms of our parameter t. So I'm going to be integrating from t = 0 to t = 2. Those will be my limits of integration. Now y is equal to t, so I'm going to replace y with what it's equal to in terms of t. So I'm going to be integrating the function t to the third. Now ds we're going to write as a square root of dx dt squared + dy dt squared, squared of all that as we said dt. So we're integrating now everything with respect to t. So this is going to be equal to the integral from 0 to 2 of t to the third times the square root of -- see the derivative of x with respect to t is 3 t squared. So we have 3 t squared squared + dy dt; well, that's just 1 squared dt. So we have the integral from 0 to 2 of t to the third times the square root of 9 t to the fourth + 1 dt. So this is a pretty straightforward integration here. We're going to let u be equal to 9 t to the fourth + 1 then du is equal to 36 t to the third dt and so that tells me I can replace a t to the third dt with a du over 36. And so we're going to have the integral then from -- well, new limits of integration. I'm just going to put some squiggly marks there to remind myself that we switched variables. So I'm not going from t = 0 to t = 2. I'm doing things in terms of you right now. But I have a 1 over 36. I'll put that out front, and we're going to have the square root of u. So u to the 1/2, t to the third dt was replaced by du over 36. We got the 36 out front. And so now this is a pretty easy antiderivative in terms of u. It's u to the 3/2 times 2/3. And again, different limits of integration. We could figure out what they are in terms of u, but I'm going to convert back into t. So we're going to have 1 over 36 times 2/3 times u to the 3/2. Now, u is 9 t to the fourth + 1, that to the 3/2 power. And now we can go ahead and go from original limits of integration 0 to 2. So let's see, when I put a 2 in here, we're going to have -- 1 over 36 times 2/3. That's going to be 1 over 54, isn't it? So we'll have 1 over 54 times -- putting a 2 in, we have 9 times 2 to the fourth. That's 9 times 16, which is 144 + 1, is 145. So we put the 2 in there, we get 145 to the 3/2 minus, putting the 0 in, we get 9 x 0 to the fourth. That's 0. 0 + 1 is 1. So we just get 1 to the 3/2 or 1. So let's see, what's the best way to write this. How about 1 over 54 -- I guess we could leave it like that. We could also write 145 to the 3/2 as 145 times the square root of 145 and then minus 1. And that is that line integral of y to the third ds over the given curve C.

@noskillman6507 Жыл бұрын

whar

@nabeel01gcg43 Жыл бұрын

The answer is 21

@robloxhood101 Жыл бұрын

Me kinda knowing what some parts mean and other bits are either just restating facts or I have no clue wtf is going on

@robloxhood101 Жыл бұрын

ANSWERS WITH EXPLANATIONS, DO NOT READ IF YOU HAVE NOT COMPLETED THE LEVEL (Unless you don’t want to do calculations to finish this level) First part: 1. d/dx (6x^2 + 3) = 12x (Derivative of a constant(Aka any number) = 0, d/dx (x^n) = nx^(n-1), d/dx k(f(x)) = kf’(x), where f(x) is any function, and f’(x) = d/dx (f(x)), d/dx (f(x) + g(x)) = f’(x) + g’(x),where f(x), g(x) are functions, and f’(x) = d/dx (f(x)), g’(x) = d/dx (g(x))) 2. ∫4x dx = 2x^2 + c (∫ x^n dx= (x^(n+1))/(n+1) + c, ∫kf(x) dx= k*∫f(x) dx + c, REMEMBER TO PUT IN + c FOR ALL INDEFINITE INTEGRALS. (Aka Antiderivatives, c is an arbitrary constant of integration as the derivative of a constant is 0.) ) 3. d/dx cos(x) = -sin(x) (Standard derivative, watch out for the negative sign, do not mix up with d/dx sin(x) = cos(x).) 4. ∫sec^2(x) dx = tan(x)+c (Standard Integral, this comes from the fact that d/dx tan(x) = sec^2(x) and the fact that integration is sort of the reverse operation of differentiation, which can be proven using the quotient rule for derivatives, but it takes way too long to be calculated in the short amount of time that is given in each question. REMEMBER TO PUT IN + c FOR ALL INDEFINITE INTEGRALS. (Aka Antiderivatives, c is an arbitrary constant of integration as the derivative of a constant is 0.) ) 5. lim x-> 5 (3x+2) = 17 (Can be understood as asking what is the value of 3x+2 as x gets closer to 5) (Formal definition of limits takes too long to prove this in the short amount of time of the question. This is easier done by substituting x=5 into 3x+2 as the function y=3x+2 is defined at x=5.) 6. d^2/dx^2 (10x^2) = 20 (Just like functions, d^2/dx^2 means take the derivative twice. (Or you could call it the second derivative.) So, d^2/dx^2 (10x^2) = d/dx (d/dx (10x^2)) = d/dx (2(10x)) = d/dx (20x) = 20.) 7. lim x-> ∞ 1/x = 0 (Can be understood as basically what does 1/x approach as x becomes a very large POSITIVE number.) (Formal definition of limits takes too long to prove this in the short amount of time of the question. Although you cannot plug in infinity for x, you can plug in bigger and bigger numbers for the value x and see where the answer trails off to.) 8. d/dx (e^x) = e^x (Standard derivative.)

@robloxhood101 Жыл бұрын

Second part: 1. d/dx (5(2x+3)^3) = 30(2x+3)^2 (First instance of what’s called the chain rule for derivatives. Basically: d/dx f(g(x)) = f’(g(x))*(g’(x)), where f(x) and g(x) are functions, and f’(x) = d/dx f’(x). If we let y=f(g(x)), u=g(x), then y= f(u). We have du/dx = g’(x) (Basically means differentiate u in terms of x, i.e. du/dx = d/dx (g(x)) and from earlier, d/dx g(x) = g’(x)). We also have dy/du = f’(u) (Basically means differentiate y in terms of u, i.e. dy/du = d/du (f(u)). From earlier, d/dx f(x) = f’(x), so we can just change the variable to say that d/du f(u) = f’(u)). The chain rule thus states that: dy/dx = dy/du * du/dx (Basically they act like fractions) = f’(u)*g’(x) = f’(g(x))*g’(x) (Since u=g(x).) To solve this: Let u= 2x+3. Then y = 5u^3. du/dx = 2, dy/du = 15u^2 (Refer back to the 1st question of the first part) Then, dy/dx = dy/du * du/dx = 15u^2 * 2 = 30u^2 = 30(2x+3)^2 (Since u= 2x+3)) 2. ∫ 5 ∫ 7 dx = 21 ∫ 2 (This is the first definite integral. Normally, ∫ 7 dx = 7x + c (Standard result, look at first part). However, definite integrals basically say, what is the area bounded by the graph between x=2 and x=5. This can be figured out by substituting x=2 and x=5 into the integrated result and taking the difference. We also ignore the constant c because it is the same for both x values. Alternatively, this is also the same as the area of a rectangle with height 7 and a length 3 (5-2) if the graph could be viewed. In general, if F(x)= ∫f(x) dx, AND F(a), F(b) are defined, then: ∫ b ∫ f(x) dx = F(b) - F(a). ∫ a Where f(x) is any function, a,b are 2 constants.) 3. d/dx [(x^3)(e^x)] = (3x^2)(e^x) + (x^3)(e^x) (First instance of the product rule for derivatives. For any two functions f(x), g(x), where f’(x) = d/dx f(x), g’(x) = d/dx g’(x) d/dx (f(x)g(x)) = f’(x)g(x) + f(x)g’(x). Alternatively, letting u= f(x), v = g(x), u’ = f’(x), v’ = g’(x), d/dx uv = u’v + uv’. Letting u = x^3, v = e^x, then u’ = 3x^2, v’ = e^x. Then: d/dx (x^3)(e^x) = (3x^2)(e^x) + (x^3)(e^x) 4. lim x->2 ((x^2-4)/(x-2)) = 4 (The original function is undefined when x=2 since the denominator of the fraction will equal 0, so we are not able to substitute x=2 immediately into the function to get the limit. However, algebra still functions as normal in limits. With (a^2-b^2) = (a-b)(a+b) and noticing that 4=2^2, (x^2-4) = (x^2-2^2) = (x-2)(x+2). The fraction becomes: ((x-2)(x+2))/(x-2), which reduces to (x+2). Then, lim x->2 ((x^2-4)/(x-2)) = lim x->2 (x+2) = 4 (Since the new function in the limit is now well defined at x=2, we can just substitute x=2 into the function.) In general, the limit of a function at a certain x value is equivalent to what the function approaches, regardless of what value or whether the original function is defined at that specific x value.)

@robloxhood101 Жыл бұрын

Maze portion: 1. d/dx sin(5x) = 5cos(5x) (Application of chain rule, with u = 5x, y = sin(u), du/dx = 5, dy/du = cos(u).) 2. d/dy (9y^3-7y^2+69) = 27y^2-14y (Refer to the first question of the first part) 3. Calculating the shaded area bounded by the line y=3 and y=3x^2. Meeting points have been clearly stated at x=-1, x=1. Answer: 4 (units^2) (Basically shaded area = (area under line y=3) - (area under curve y=3x^2) The area under the horizontal line y=3 is basically a rectangle, with height 3 and length 2 (1-(-1)=2). It’s area = 3*2 = 6. Meanwhile area under the curve is actually the definite integral of 3x^2 from -1 to 1. ∫3x^2 dx = x^3 + c, in which case the c can be ignored for definite integrals. So area under the curve = 1^3 - (-1)^3 = 1 -(-1) = 2. So shaded area = 6 - 2 = 4 units^2.)

@robloxhood101 Жыл бұрын

BONUS: The Trick question (Which is solvable, but requires the usage of a graphing software like GeoGebra/Desmos) ‘Part b’ (Aka the last jk question that shows up here) Two trains, Train A and Train B, depart simultaneously from two stations in opposing directions. Train A’s velocity can be modelled by the equation dy/dt = 0.2t(e^0.2t), while Train B’s velocity is modelled by dy/dt = -0.5tln(0.38t). If the two stations are 100 units apart and the trains depart at time t=0, at what time will they meet? (Leave in 5s.f.) Answer: t=11.434 (If this was an actual real question, everyone would be so pissed as this is the only question to include INTEGRATION BY PARTS AND needs graphing software to do. This formula is derived from integrating the product rule of derivatives. The product rule of derivatives state that: d/dx (f(x)g(x)) = f’(x)g(x) + f(x)g’(x). Letting d/dx (f(x)g(x)) = (f(x)g(x))’ gives (f(x)g(x))’ = f’(x)g(x) + f(x)g’(x). Integrating both sides gives: ∫(f(x)g(x))’ dx = ∫(f’(x)g(x) + f(x)g’(x)) dx f(x)g(x) = ∫(f’(x)g(x))dx + ∫(f(x)g’(x))dx, which can be rewritten to become ∫(f(x)g’(x))dx = f(x)g(x) - ∫(f’(x)g(x))dx. Letting u=f(x), v= g(x), du/dx = f’(x) and dv/dx = g’(x) -> du = f’(x) dx, dv = g’(x) dx yields the most common version of the integration by parts formula. In general, integration by parts says that: ∫u dv = uv - ∫v du, where u and v are ANY choices of functions of x. The solve: dy/dt = 0.2t(e^0.2t) y = ∫0.2t(e^0.2t) dt Let u = t, dv = 0.2(e^0.2t). (Pick choices of u and dv that make finding du, v easy.) Then du = 1, v = e^0.2t (Remember that this choice of v requires you to remember that the chain rule exists!) y = ∫0.2t(e^0.2t) dt = t(e^0.2t) - ∫(e^0.2t) dt = t(e^0.2t) - 5(e^0.2t) + c = (t-5)(e^0.2t) + c When t=0, y=0 (Train A) (t-5)(e^0.2t) + c = 0 (0-5)(e^0.2(0)) + c = 0 (-5)(e^0) + c = 0 (-5)(1) + c = 0 (-5) + c = 0 c = 5 Train A’s displacement: y = (t-5)(e^0.2t)+5 dy/dt = -0.5tln(0.38t) = -0.5t (ln(0.38) + ln(t)) = (-0.5ln(0.38))t -0.5tln(t) y = ∫ (-0.5ln(0.38))t -0.5tln(t) dt = 1/4ln(50/19)t^2 -0.5 ∫tln(t) dt Let u = ln(t), dv = t (We can’t integrate ln very easily, it requires a integration by parts on its own) du = 1/t, v = t^2/2, y = 1/4ln(50/19)t^2 -0.5 ∫tln(t) dt = 1/4ln(50/19)t^2 -1/2((t^2/2)(ln(t)) - ∫t/2 dt) = t^2ln(50/19)/4 - (t^2/4)(ln(t)) + t^2/8 + c When t=0, y=100, t^2ln(50/19)/4 - (t^2/4)(ln(t)) + t^2/8 + c = 100 0 + c = 100 c = 100 Train B: y = t^2ln(50/19)/4 - (t^2/4)(ln(t)) + t^2/8 + 100 Plotting both graphs gives that t = 11.434

@MotionIII Жыл бұрын

Holy shit dude you're actually insane. AND you solved the train question?! I tip my hat to you.

@robloxhood101 Жыл бұрын

@@MotionIIIThanks!