find the lenght of the parametric curve x = 3t^2+2, y=2t^3-1/2, for 1<=t<=4?

Question

I'm thinking is this formula: (dy/dx) = (dy/dt)/(dx/dt), hence:
x(t) = 3t^2 + 2 = (dx/dt) = 6t
y(t) = 2t^3 - 1/2 = (dy/dt) = 6t^2, then plug in the respective values to the formula above: (6t^2/6t), am I on the right track? any hint is appreciated, thanks!

Actually, I'm kinda confused to which formula I should use, I was thinking: L= ∫ sqrt (dx/dt)^2 + (dy/dx)^2) dt, from alpha to theta as my upper and lower limits, which would be 1 and 4 respectively. However, aren't alpha and theta radians?

Answer 1

there are a few ways to do this...but i believe the method you're trying to use may require implicit differentiation...

yes... (dy/dx) = (dy/dt)/(dx/dt)...this is just the chain rule...

but notice something:

You're trying to use the arclength formula which integrates
Sqrt[1+(dy/dx)^2],
which is integrated with respect to x (or if you replace (dy/dx) with (dx/dy) is integrated with respect to y), and not t.

There is another method (2 representations, identical method) which will be more convenient, at least in the setup:

Integrate Sqrt[(dy/dt)^2 + (dx/dt)^2] over 1<=t<=4 which is exactly the same as the following:

define r(t) = (x(t), y(t)). Then immediately, r'(t) = (x'(t), y'(t)). The arclength of r(t) over the interval (1,4) is then defined as:
the integral of Sqrt[] over 1<=t<=4
where = |r'(t)|^2 is just the inner product, and you always take the principal branch (positive) of the square root. I'm not going to explicitly solve the problem, but that'll get you going in the right direction...

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In response to your latest addendum, you are correct in that you want to integrate Sqrt[(dx/dt)^2 + (dy/dt)^2] from 1 to 4, or from alpha to beta, or alpha to theta, or t0 to t. These symbols do not imply units, nor do they restrict the curves for which this formula is valid, unless the interval specified traverses a point over which the parametrized curve is not differentiable. But this curve is C^infinity (meaning it is continuous and all derivatives of the function are continuous out to infinite degree), though not regular (in the sense that there are points at which the first derivative will vanish, one being t=0). What you added in your addendum is just the arclength formula I gave above in this post...there are no units implied in alpha, beta, or whatever you choose to call the bounds. They simply represent an interval over which you wish to calculate the arclength of a curve...no more, no less...

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And a hint on the calculation of that integral...

Int[Sqrt[(dx/dt)^2 + (dy/dt)^2] = Int[Sqrt[36t^2 + 36t^4]]
= Int[6*t*Sqrt[1+t^2]]

You can try t = tan(phi), dt = (sec(phi))^2 to make the thing a bit more friendly looking...it's still gonna look unpleasant, but it's a bit easier to do outright at that point...in fact, from there, it's just a u-substitution with
u = sec(phi), du = sec(phi)*tan(phi)...

It sucks and is rather tedious, but all that's left at that point is going back to the original coordinate system...good luck...

Answer 2

Perimeter = sum of three sides 9x + 1/4 = x + 1/2 + 2x – 1 + third side = 3x – 1/2 + third side third side = 9x + 1/4 – 3x + 1/2 = 6 x + 3/4

Answer 3

Nick gave you a god answer. So, let's go straight to the point.
We want to integrate sqrt(1 + ((dy/dt)/(dx/dt))^2) from 1 to 4.

dy/dt = 6t^2
dx/dt = 6t

So, by the chain rule, dy/dx = (6t^2)/(6t) = t And

sqrt(1 + ((dy/dt)/(dx/dt))^2) = sqrt(1 + t^2)
And we have to compute Integral (1 to 4) sqrt(1 + t^2) dt

Put t = sinh u, so that 1 + t^2 = cosh^2(u) and dt = cosh(u) du
And our integral becomes Int (arcsinh(1) to arcsinh(4) cosh(u) cosh(u) du = Int (arcsinh(1) to arcsinh(4) cosh(u)^2 du = (1/2) Int (arcsinh(1) to arcsinh(4) (1 + cosh(2u)) du = (1/2) ( [u + 1/2sinh(2u)] (arcsinh(1) to arcsinh(4)

Now, you plug in the limits, subtract and you have the length. Observe that sinh(2arcsin(u))= 2 sinh(arcsinh(u)) cosh(arcsinh(u)) = 2 u sqrt(1 + u^2).