Find The Exact Length Of The Curve X=3t^2 And Y=2t^3 From 0

Find the exact length of the curve: x=e^t + e^-t, y=5 - 2t, between 0 and 3?

int over [0,3] of {sqrt[x'^2 + y'^2} dt.....{....} = e^t + e^(-t)
int over [0,pi/2] of {sin {2t} dt} - int over [ pi/2,pi] of { sin {2t} dt} or int over [0,pi] of |sin 2t| dt...here {...} = sqrt [sin 2t]^2...you can do the integration....note cos t cos 3t - sin t sin 3t = cos 4t=
1 - [sin 2t]^2

Find the exact length of the curve x=3cos(T)-cos(3T) y=3sin(T)-sin(3T) O

√[(dx/dt)^2 + (dy/dt)^2]
= √[(-3 sin t - 3 sin(3t))^2 + (3 cos t - 3 cos(3t))^2]
= 3 √[(sin t + sin(3t))^2 + (cos t - cos(3t))^2]
= 3 √[(sin^2(t) + 2 sin t sin(3t) + sin^2(3t)) + (cos^2(t) - 2 cos t cos(3t) + cos^2(3t))]
= 3 √[2 + 2 sin t sin(3t) - 2 cos t cos(3t)]
= 3 √[2 - 2 (cos t cos(3t) - sin t sin(3t))]
= 3 √[2 - 2 cos(t + 3t)], via sum of angles identity
= 3 √[2(1 - cos(4t))]
= 3 √[2 * 2 sin^2(2t)], via half angle identity
= 6 |sin(2t)|.

Hence, the arc length equals
∫(t = 0 to π) 6 |sin(2t)| dt
= 2 * ∫(t = 0 to π/2) 6 sin(2t) dt, via symmetry (see graph)
= -6 cos(2t) {for t = 0 to π/2}
= 12.

I hope this helps!

Find the exact length of the curve. x = 4 + 3t2, y = 3 + 2t3, 0 ≤ t ≤ 2?

ds = √(dx² + dy²)
dx = 6tdt
dy = 6t²dt
ds = 6t√(1+t²)dt
let u = t²
du = 2tdt

ds = 3√(1+u)du
s = 2(1+u)^(3/2) + C
= 2(1+t²)^(3/2) + C

for 0 ≤ t ≤ 2
s = 2[5^(3/2) - 1]
s = 10√5 - 2

Find the length of the loop of the given curve. x = 9t - 3t^3 y = 9t^2?

Since x = 3t(3 - t^2) and y = 9t^2.

Note that the beginning and the end of the loop occur when
x = 0 (and t nonzero) ==> t = ± √3.

Link of plot:
http://www.wolframalpha.com/input/?i=plo...

So, the arc length is given by
∫ √[(dx/dt)^2 + (dy/dt)^2] dt
= ∫(t = -√3 to √3) √[(9 - 9t^2)^2 + (18t)^2] dt
= 2 ∫(t = 0 to √3) √[(9 - 9t^2)^2 + (18t)^2] dt, by evenness of the integrand
= 2 ∫(t = 0 to √3) 9 √[(1 - t^2)^2 + (2t)^2] dt
= 18 ∫(t = 0 to √3) √(t^4 + 2t^2 + 1) dt
= 18 ∫(t = 0 to √3) √(t^2 + 1)^2 dt
= 18 ∫(t = 0 to √3) (t^2 + 1) dt
= 18 (t^3/3 + t) {for t = 0 to √3}
= 6(t^3 + 3t) {for t = 0 to √3}
= 36√3.

I hope this helps!

Find the arc length of the curve, x = 3 cos t - cos 3t; y = 3 sin t -sin 3t?

x = 3cos(t) - cos(3t)
dx = (-3sin(t) + 3sin(3t)) dt

y = 3sin(t) - sin(3t)
dy = (3cos(t) - 3cos(3t)) dt

L = ∫ √( [dy/dt]² + [dx/dt]² )
L = ∫ √( [(3cos(t) - 3cos(3t)) dt]² + [(-3sin(t) + 3sin(3t)) dt]² )
L = ∫ √( [3cos(t) - 3cos(3t)]² + [-3sin(t) + 3sin(3t)]² ) dt, over [0, 2π]

If you have a picture of the curve, you'll notice that the interval 0 ≤ t ≤ π is reflected over the x-axis, and you can take advantage of symmetry here:
L = 2 ∫ √( [3cos(t) - 3cos(3t)]² + [-3sin(t) + 3sin(3t)]² ) dt, over [0, π]
L = 2 ∫ √(9cos²(t) - 18cos(t)cos(3t) + 9cos²(3t) + 9sin²(t) - 18sin(t)sin(3t) + 9sin²(3t)) dt, over [0, π]
L = 2 ∫ √(9(cos²(t) + sin²(t)) + 9(cos²(3t) + sin²(3t)) - 18(cos(t)cos(3t) + sin(t)sin(3t))) dt, over [0, π]
L = 2 ∫ √(9(1) + 9(1) - 18(cos(t)cos(3t) + sin(t)sin(3t))) dt, over [0, π]

cos(t)cos(3t) + sin(t)sin(3t) = cos(t - 3t), if you recall the angle sum-difference identities.
cos(-2t) = cos(2t) = cos²(t) - sin²(t)

L = 2 ∫ √(18 - 18(cos²(t) - sin²(t))) dt, over [0, π]
L = 2 √18 ∫ √(1 - cos²(t) + sin²(t)) dt, over [0, π]
L = 2 √18 ∫ √(sin²(t) + sin²(t)) dt, over [0, π]
L = 2 √18 ∫ √(2sin²(t)) dt, over [0, π]
L = 2 √18 √2 ∫ √(sin²(t)) dt, over [0, π]
L = 2 √36 ∫ sin(t) dt, over [0, π]
L = 2(6) [-cos(t)] over [0, π]
L = 12 (-cos(π) + cos(0))
L = 12 (1 + 1)
L = 12(2)
L = 24

More convenient if you have a calculator handy.

Find the exact length of the curve. Please show all steps to help me figure this stuff out.?

(1) With x = 4 + 3t^2 and y = 5 + 2t^3, we see that:
dx/dt = 6t and dy/dt = 6t^2.
(Note that this curve is transversed exactly once as dx/dt and dy/dt are positive)

Then, plugging dx/dt, dy/dt, and the limits into the arc-length formula yields:
L = ∫ √[(dx/dt)^2 + (dy/dt)^2] dt (from t=a to b)
= ∫ √[(6t)^2 + (6t^2)^2] dt (from t=0 to 1)
= ∫ √(36t^4 + 36t^2) dt (from t=0 to 1)
= 6 ∫ t√(t^2 + 1) dt (from t=0 to 1).

Now, substitute u = t^2 + 1 ==> du = 2t dt. Changing the bounds:
(a) Lower bound: t = 0 ==> u = 0^2 + 1 = 1
(b) Upper bound: t = 1 ==> u = 1^2 + 1 = 2.

This yields:
L = 6 ∫ t√(t^2 + 1) dt (from t=0 to 1)
= 3 ∫ √(t^2 + 1) (2t dt) (from t=0 to 1)
= 3 ∫ √u du (from u=1 to 2)
= 2u^(3/2) (evaluated from u=1 to 2)
= 4√2 - 1.

(2) and (3) can be done using:
L = ∫ √[1 + (dy/dx)^2] dx (from x=a to b).

I hope this helps!

Find arc-length of x=1+3t^2, y=4+2t^3, 0<=t<=1?

Note that, if a relation given by x = f(t) and y = g(t) is transversed exactly once on [a, b], the arc-length of the relation on the interval [a, b] is:
L = ∫ √[(dx/dt)^2 + (dy/dt)^2] dt (from t=a to b).

With x = 1 + 3t^2 and y = 4 + 2t^3, we see that:
dx/dt = 6t and dy/dt = 6t^2.

Since dx/dt ≥ 0 and dy/dt ≥ 0 when t is in [0, 1], this is transversed once.

Hence, the arc-length of x = 1 + 3t^2 and y = 4 + 2t^3 on 0 <= t <= 1 is:
L = ∫ √[(dx/dt)^2 + (dy/dt)^2] dt (from t=a to b)
= ∫ √[(6t)^2 + (6t^2)^2] dt (from t=0 to 1)
= ∫ √(36t^2 + 36t^4) dt (from t=0 to 1)
= ∫ √[36t^2 * (1 + t^2)] dt (from t=0 to 1)
= ∫ 6t√(1 + t^2) dt (from t=0 to 1) (note that |6t| = 6t on 0 <= t <+ 1)
= 3 ∫ √(1 + t^2) d(1 + t^2) (from t=0 to 1)
= 3(1 + t^2)^(1/2 + 1)/(1/2 + 1) (evaluated from t=0 to 1)
= 2(1 + t^2)^(3/2) (evaluated from t=0 to 1)
= 2(2)^(3/2) - 2
= 4√2 - 2.

I hope this helps!

Calculus problem, parametric equations - find the length of the curve?

You are using the correct formula.

With x = 3t^2 and y = 2t^3, differentiating each equation with respect to t yields:
dx/dt = 6t and dy/dt = 6t^2.

Since we want the arc length of the curve from t = 0 to 2, the arc length is given by:
L = ∫ √[(6t)^2 + (6t^2)^2] dt (from t=0 to 2)
= ∫ √(36t^4 + 36t^2) dt (from t=0 to 2)
= ∫ √[36t^2 * (t^2 + 1)] dt (from t=0 to 2), by factoring
= 6 ∫ t√(t^2 + 1) dt (from t=0 to 2), since √(36t^2) = 6t for t > 0.

At this point, apply the following u-substitution
u = t^2 + 1 ==> du = 2t dt.

Changing the bounds:
(a) Lower bound: t = 0 ==> u = 0^2 + 1 = 1
(b) Upper bound: t = 2 ==> u = 2^2 + 1 = 5.

Hence:
L = 6 ∫ t√(t^2 + 1) dt (from t=0 to 2)
= 3 ∫ √(t^2 + 1) (2t dt) (from t=0 to 2)
= 3 ∫ √u du (from u=1 to 5), by applying substitutions
= 2u^(3/2) (evaluated from u=1 to 5), by integrating
= 2(5)^(3/2) - 2
= 10√5 - 2, since 5^(3/2) = 5 * 5^(1/2) = 5√5
= 2(5√5 - 1).

I hope this helps!