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### Course: Calculus 2>Unit 5

Lesson 3: Arc length: parametric curves

# Parametric curve arc length

Conceptual introduction to the formula for arc length of a parametric curve.

## Want to join the conversation?

• Why doesn't the integral give us the area under the curve?
• Normally, you're integrating a quantity like f(x) dx, which is a rectangle of height f(x) and infinitesimal width dx, so the sum of infinitely many rectangles gives you area. However, here you're integrating a quantity ds = sqrt(dx^2 + dy^2), which is a line segment of infinitesimal length; the sum of infinitely many line segments gives you length. Let me know if this helps.
• Hi, About 2 minutes in to the video Sal explains that dx=(dx/dt)*dt and that dy=(dx/dt)*dt. I have a hard time understanding how that works and why that is the case. Can someone help me understand this better?
• In the video, Dx is the rate of change our function X. Our function X is written in terms of t, so the derivative of X(t) will be dx/dt, the derivative of our function X with respect to t, multiplied by dt, the derivative or rate of change of the variable t, which will always be equal to 1 here. It's basically the same thing as taking the derivative of any other function with the variable x in it, but in this case its replaced with the variable t. For example, the derivative of x^2 is equal to 2x(dx) , where d/dx=2x and dx=1. So in the video, dx/dt is like d/dx and dt=dx.
• It doesn't make sense to me conceptually that dx=(dx/dt)*dt. Why are we multiplying times dt, besides the fact that it cancels out with dx/dt to give dx? Is there another reason besides this, or is that the reason? Please explain.
• You know how Distance = Velocity * Time?
Think of dx as the distance along x axis, dx/dt as the velocity along x axis, and dt as the time.
• Why can't you just find dy/dx (by doing (dy/dt)/(dx/dt) then use the arc length equation: int(sqrt(1+(dy/dx)^2)). You can also find your limits by plugging the starting and end point of the interval in your paramteric equations and solving.
• Why didn't we solve it by turning the parametric equation into y(x)= expression with x? Would it still work if we did like that or is it problematic because we lose the information of time related to position?
(1 vote)
• This isn't always possible because y may not be a function of x. Consider the curve given by
<x, y>=<tcos(t), tsin(t)>. This is a spiral centered on the origin, so it fails both the vertical line test and the horizontal line test infinitely many times.

We use parametric equations because there are lots of curves that just can't be described by y as a function of x. This gives us a more powerful language, but it also means we can't always convert back to the 'familiar' y=f(x) setting.
• What is the formula of area under a parametric curve ?
• If x=f(t) and y=g(t) and the parametric curve is traced out exactly once as t increases from a to b, then the area is the integral from a to b of g(t)*f'(t) dt (not sure how to do integral formatting).
(1 vote)
• can we use the arc length formula to get the arc length here? if no, why not?
(1 vote)
• By arc length formula, do you mean sqrt(1+(dy/dx)^(2))? If so, see that to use that, you need dy/dx. For that, you need to use x(t) and y(t) to find y(x). That works too though, You're still bound to get the same answer.

However, this'll only work if you're able to express y as a function of x. There could be cases where you can't. In such cases, using this formula would be the better choice.