Missing dt in the integrand.Jd0g33 said:Homework Statement
Calculate the arc length of [itex]<2t,t^2,lnt>[/itex] from [itex]1=<t=<e[/itex]
Homework Equations
Arc length=[itex]∫√{(x')^2 + (y')^2 + (z')^2}[/itex]
Write the quantity in the radical as 4t2 + 4 + 1/t2, and then factor it.Jd0g33 said:The Attempt at a Solution
So I have gotten to this point:
[itex]∫√{4 + 4t^2 + \frac{1}{t^2}}[/itex]
Am I on the right track, and if so, how do I integrate that?
You can't do that, but you can multiply by t2 over itself. OTOH, the expression can be factored directly, to (2t + 1/t)2.Jd0g33 said:Nevermind, got it. multiply t^2
Arc length is the distance along a curve or arc, measured as the length of the curve from one point to another.
A vector function is a mathematical function that takes in one or more inputs and produces a vector output.
The arc length of a vector function can be calculated using the formula:
S = ∫√(1 + f'(x)^2)dx
where f'(x) is the derivative of the vector function and the integral is taken over the given interval.
No, the arc length of a vector function cannot be negative as it represents a physical distance and must be positive.
Calculating the arc length of a vector function is important in many fields of science and engineering, as it helps in understanding the shape and properties of curves, calculating work done by a force, and finding the shortest distance between two points in space.