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

What is the arclength of #f(t) = (t+sqrt(lnt),t-sqrtlnt))# on #t in [1,e]#?

Answer 1

The arc length between #t=a# and #t=b# on the parametric curve #y=f(t),x=g(t)# is given by #int_a^bsqrt((dy/dt)^2+(dx/dt)^2)\ dt#

So, from the question, we have #t# from #1# to #e# and #y=t-sqrt(ln(t)),x=t+sqrt(ln(t))#. Then, #dy/dt=1-1/(2tsqrt(ln(t)))# and #dx/dt=1+1/(2tsqrt(ln(t)))#.

Thus, we need to find #int_1^esqrt((1-1/(2tsqrt(ln(t))))^2+(1+1/(2tsqrt(ln(t))))^2)\ dx# #=int_1^esqrt(2+1/(2t^2ln(t)))\ dx#

Using numerical approximation methods, we find that the integral is approximately equal to #3#.

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Answer 2

Emilia Aguilar

To find the arc length of the curve (f(t) = (t + \sqrt{\ln t}, t - \sqrt{\ln t})) on the interval (t \in [1, e]), use the formula for arc length:

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

Substitute the expressions for (x(t)) and (y(t)) into this formula, then integrate from (1) to (e).

[ L = \int_{1}^{e} \sqrt{\left(1 + \frac{1}{{2\sqrt{\ln t}}}\right)^2 + \left(1 - \frac{1}{{2\sqrt{\ln t}}}\right)^2} , dt ]

Evaluate this integral to find the arc length.

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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.