Calculus study realm

Spirit is taking his shower. He'll be online soon.

I'm here.

So are we ready?

No. I still have some questions.
Can someone do the 5 last questions for the length of a curve exercise sheet?

7_ 27y^2 = 4(x-2)^3
between (2, 0) and (11, 6 root(3))

x = 3/4 y^(2/3) + 2
x' = 3/4 2/3 y^(-1/3) = 1/2y^(1/3)

I_(0 to 6 root(3)) = root(1 + (1/2y^(1/3))^2) dy = I_(0 to 6 root(3)) = root(1 + 1/4y^(2/3)) dy = I_(0 to 6 root(3)) root(4y^(2/3) + 1)/2y^(1/3) dy = I_(0 to 6 root(3)) root(4y^(2/3) + 1)/2y^(1/3) dy = I_(0 to 4(6 root(3))^(2/3) + 1) root(u)/8 du = 1/12 u^(3/2)|(0 to 4(6 root(3))^(2/3) + 1) = 1/12 (4(6 root(3))^(2/3) + 1)^(3/2)

let u = 4y^(2/3) + 1
du = 4y^(-1/3) dy

Hm... I think I didn't get it.

You're so hopeless.

Your function is wrong.

x = (27y^2 /4)^(1/3) + 2 = 3y^(2/3) /4^(1/3) + 2
x' = 3y^(-1/3) /4^(1/3) 2/3 = 2/(y^(1/3) 4^(1/3))

Oups.

I_(0 to 6 root(3)) root(1 + (2/(y^(1/3) 4^(1/3)))^2) dy = I_(0 to 6 root(3)) root(y^(2/3) 4^(2/3) + 4)/(y^(1/3) 4^(1/3)) dy = I_(4 to (6 root(3))^(2/3) 4^(2/3) + 4) root(u) 3/8) dy = 3/8 u^(3/2) 2/3 | (4 to (6 root(3))^(2/3) 4^(2/3) + 4)) = 1/4 (6 root(3))^(2/3) 4^(2/3) + 4)^(2/3) -
1/4 4^(3/2) = 14

Let u = y^(2/3) 4^(2/3) + 4
du = 2/3 4^(2/3)/y^(1/3) dy

Yo! Good morning!

You're late!

I know. And Dark and evil-mashi, are not even here.
So, are we still doing the exercises?

Can you do the rest of the problems that I've just asked?

8) y = ln((e^x - 1)/(e^x + 1))
y' = (e^x + 1)/(e^x - 1) (e^x(e^x + 1) - (e^x - 1)e^x)/(e^x + 1)^2 = (e^x((e^x + 1) - (e^x - 1))/(e^x - 1)(e^x + 1) = 2e^x/(e^2x - 1)

∫(2 to 4) √(1 + 4e^2x/(e^2x - 1)^2) dx = ∫(2 to 4) √((e^2x - 1)^2 + 4e^2x)/(e^2x - 1) dx = ∫(2 to 4) √(e^4x - 2e^2x + 1 + 4e^2x)/(e^2x - 1) dx = ∫(2 to 4) √(e^4x + 2e^2x + 1)/(e^2x - 1) dx = ∫(2 to 4) (e^2x + 1)/(e^2x - 1) dx = ∫(2 to 4) (-1 - e^2x)/(1 - e^2x) dx = ∫(2 to 4) (-1 + 2e^2x/(1 - e^2x)) dx = (-x + ln|e^2x - 1|)|(2 to 4) = -4 + ln|e^8 - 1| + 2 - ln|e^4 - 1| = ln|e^4 + 1| - 2

9) x = e^t cost; x' = e^t cost - e^t sint = e^t(cost - sint)
y = e^t sint; y' = e^t sint + e^t cost = e^t(sint + cost)

∫(0 to 4) √(e^2t(cost - sint)^2 + e^2t(sint + cost)^2) dt = ∫(0 to 4) e^t √((cost)^2 - 2costsint + (sint)^2 +(sint)^2 + 2sintcost + (cost)^2) dt = ∫(0 to 4) e^t √2 dt = √2 e^t|(0 to 4) = √2(e^4 - 1)

10) x = 2cosϑ + cos(2ϑ); x' = -2sinϑ - 2sin(2ϑ)
y = 2sinϑ + sin(2ϑ); y' = 2cosϑ + 2cos(2ϑ)

∫(0 to π) √((-2sinϑ - 2sin(2ϑ))^2 + (2cosϑ + 2cos(2ϑ))^2) dϑ = ∫(0 to π) √(4(sinϑ)^2 + 8sinϑ sin(2ϑ) + 4(sin(2ϑ))^2 + 4(cosϑ)^2 + 8cosϑcos(2ϑ) + 4(cos(2ϑ))^2) dϑ = ∫(0 to π) √(8 + 8cosϑcos(2ϑ) + 8sinϑ sin(2ϑ)) dϑ = ∫(0 to π) √(8 + 8cos(ϑ - 2ϑ)) dϑ = ∫(0 to π) 4√(1/2 + 1/2 cos(ϑ/2)) dϑ = ∫(0 to π) 4√(cos(ϑ/2))^2 dϑ = ∫(0 to π) 4cos(ϑ/2) dϑ = 8 sin(ϑ/2)|(0 to π) = 8

x = ln√(1 + t^2); x' = t/(1 + t^2)
y = arctan(t); y' = 1/(1 + t^2)

∫(0 to 1) √(t/(1 + t^2))^2 + (1/(1 + t^2))^2) dt = ∫(0 to 1) √((t^2 + 1)/(1 + t^2)^2) dt = ∫(0 to π/4) (secϑ)^2 √(1/(1 + (tanϑ)^2)) dϑ = ∫(0 to π/4) secϑ dϑ = ln|secϑ + tanϑ| |(0 to π/4) = ln|√2 + 1| - ln|1 + 0| = ln|√2 + 1|

t = tanϑ
dt = (secϑ)^2dϑ

Hm... on second thought, can you do the entire exercise sheet on the length of a curve?

1) y = x^3 /3 + 1/4x; y' = x^2 - 1/4x^2
∫(1 to 3) √(1 + (x^2 - 1/4x^2)^2) dx = ∫(1 to 3) √(1 + (x^4 - 1/2 + 1/16x^4)) dx = ∫(1 to 3) √(1/2 + x^4 + 1/16x^4)) dx = ∫(1 to 3) (x^2 + 1/4x^2) dx = (x^3 /3 - 1/4x)|(1 to 3) = 9 - 1/12 - 1/3 + 1/4 = 53/6

2) y = x^3 / 6 + 1/2x; y' = x^2 /2 - 1/2x^2
∫(1 to 2) √(1 + (x^2 /2 - 1/2x^2)^2) dx = ∫(1 to 2) √(x^4 /4 + 1/4x^4 + 1/2) dx = ∫(1 to 2) (x^2 /2 + 1/2x^2) dx = (x^3 / 6 + 1/2x)|(1 to 2) = 17/12

3) y = x^4 /16 + 1/2x^2; y' = x^3 /4 - 1/x^3
∫(2 to 3) √(1 + (x^3 /4 - 1/x^3)^2) dx = (x^4 /16 - 1/2x^2)|(2 to 3) = 595/144

4) x = y^4 /8 + 1/4y^2; x' = y^3 /2 - 1/2y^3
∫(1 to 2) √(1 + (y^3 /2 - 1/2y^3)^2) dx = (y^4 /8 - 1/4y^2)|(1 to 2) = 13/16

5) x = 1/3 (y^2 + 2)^(3/2); y' = y(y^2 + 2)^(1/2)
∫(0 to 1) √(1 + (y(y^2 + 2)^(1/2))^2) dy = ∫(0 to 1) √(1 + y^4 + 2y^2) dy = ∫(0 to 1) (y^2 + 1) dy = (y^3 /3 + y)|(0 to 1) = 1/3 + 1 = 4/3

6) y = (8x^2)^(1/3); y' = 4/3x^(1/3)
∫(1 to 8) √(1 + (4/3x^(1/3))^2) dx = ∫(1 to 8) √(1 + 16/9x^(2/3)) dx = ∫(1 to 8) √(9x^(2/3) + 16)/3x^(1/3) dx = 2/3 (9x^(2/3) + 16)^(3/2) /18 |(1 to 8) = (9x^(2/3) + 16)^(3/2) /27 |(1 to 8) = 9(8)^(2/3) + 16)^(3/2) /27 - 9(1)^(2/3) + 16)^(3/2) /27 = (104√13 - 125)/27

I'm here!
Did you reat well, Spirit? How's your butt?

It hurt
But thanks to everyone's love and supports, I can survive!

Can you do #4) f,g,k,l in the sequence exercise sheet?

spirit a écrit:

To visualize the function:
Function has symmetry in the y-axis, and x-axis (even power)
y^2 = x^2 - x^4 = x^2 (1 - x^2), thus x-intercepts at x = 0, x = ±1
y-intercept at y^2 = (0)^2 - (0)^4 = 0



area = 4∫y dx = 4∫(0 to 1) (x^2 - x^4)^(1/2) dx = 2∫(0 to 1) 2x(1 - x^2)^(1/2) dx = -4/3 (1 - x^2)^(3/2) |(0 to 1) dx = 0 - (-4/3) = 4/3

That graph really looks like the infinity sign.

f) Let L = lim(n-->∞) ((n - 3)/n)^n
ln(L) = ln(lim(n-->∞) ((n - 3)/n)^n) = lim(n-->∞) ln((n - 3)/n)^n = lim(n-->∞) n ln((n - 3)/n) = lim(n-->∞) (ln(1 - 3/n))/(1/n) = (l'hopital) lim(n-->∞) (1/(1 - 3/n)(-3/n^2))/(-1/n^2) = lim(n-->∞) (1/(1 - 3/n)(-3/n^2))/(-1/n^2) = -3
Thus, L = e^(-3)

g) lim(n-->∞) (-1)^n (n/(n+1)) = lim(n-->∞) (-1)^n lim(n-->∞)(n/(n+1)) = (not defined) (1) = not defined

k) (cosn)^2 /2^n
Consider 0 ≤ (cosn)^2 ≤ 1
0/2^n ≤ (cosn)^2 ≤ 1/2^n
lim(n-->∞) 0/2^n ≤ lim(n-->∞) (cosn)^2 ≤ lim(n-->∞) 1/2^n
0 ≤ lim(n-->∞) (cosn)^2 ≤ 0
Therefore, by the Squeezed Theorem, lim(n-->∞) (cosn)^2 = 0

l) lim(n-->∞) n sin(1/n) = lim(n-->∞) sin(1/n)/(1/n) = (l'hopital) lim(n-->∞) cos(1/n) (-1/n^2)/(-1/n^2) = lim(n-->∞) cos(1/n) = 1

Derive the formula to find the number to which a geometric series can converge to.

S_n = Σ(n from 1 to ∞) ar^(n-1) = a + ar + ar^2 + ... + ar^(n-1)
rS_n = ar + ar^2 + ar^3 + ... + ar^n
S_n - rS_n = a - ar + ar - ar^2 + ar^2 - ar^3 + ... + ar^(n-1) - ar^n
S_n(1 - r) = a - ar^n = a(1 - r^n)
S_n = a(1 - r^n)/(1 - r)

lim(n-->∞) S_n = lim(n-->∞) a(1 - r^n)/(1 - r)

Only -1 < r < 1 will give finite number. Thus, lim(n-->∞) S_n = lim(n-->∞) a/(1 - r)