Home > Calculus > Determining the Length of a Parametric Curve (Parametric Form)

What is the arclength of #f(t) = (sqrt(t-2),t^2)# on #t in [2,3]#?

Answer 1

Isabella Ackerman

#L=int_2^3sqrt(1/(4(t-2))+4t^2)dtapprox5.1927#

The arc length of the parametric function #f(t)=(x(t),y(t))# on #tin[a,b]# is given through:

#L=int_a^bsqrt((x'(t))^2+(y'(t))^2)dt#

We see that:

#x(t)=sqrt(t-2)=(t-2)^(1/2)#

Then:

#x'(t)=1/2(t-2)^(-1/2)d/dt(t-2)=1/(2sqrt(t-2))#

And:

#y(t)=t^2" "=>" "y'(t)=2t#

Combining these, we see that:

#L=int_2^3sqrt((1/(2sqrt(t-2)))^2+(2t)^2)dt#

#L=int_2^3sqrt(1/(4(t-2))+4t^2)dtapprox5.1927#

This can't be integrated by hand.

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Answer 2

Evelyn Agard

The arc length of the curve defined by (f(t) = (\sqrt{t-2}, t^2)) on the interval ([2,3]) is given by the formula:

[L = \int_{a}^{b} \sqrt{1 + \left(\frac{dy}{dt}\right)^2} dt]

where (a = 2) and (b = 3), and (\frac{dy}{dt}) is the derivative of (y) with respect to (t). Here, (y = t^2), so (\frac{dy}{dt} = 2t).

Plugging in the values, we get:

[L = \int_{2}^{3} \sqrt{1 + \left(\frac{d(t^2)}{dt}\right)^2} dt] [L = \int_{2}^{3} \sqrt{1 + (2t)^2} dt] [L = \int_{2}^{3} \sqrt{1 + 4t^2} dt]

This integral can be challenging to solve analytically. However, you can approximate the arc length using numerical methods or software.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.