The length of the curve \(x = t^3, y = \frac {3t^2}2,\;0 \le t \le 1\

1.	The length of the curve \(x = t^3, y = \frac {3t^2}2,\;0 \le t \le 1\) will be:1. 23 / 2 - 1 units2. 33 / 2 - 2 units3. 53 / 2 + 1 units4. 33 / 2 + 1 units
Answer» Correct Answer - Option 1 : 2^{3 / 2} - 1 units Concept: We have the formula of arc length in parametric form as, \(\hbox{Arc length }=\int_a^b\; \sqrt{\left({dx\over dt}\right)^2+\left({dy\over dt}\right)^2}\;dt\) Calculation: Given: \(x = t^3, y = \frac {3t^2}2,\;0 \le t \le 1\) \(\frac{dx}{dt}\) = 3t2 and \(\frac{dy}{dt}\) = 3t Placing, these values in the above equation, we get \({ arc\;length }=\int_0^1\; \sqrt{\left({3t^2}\right)^2+\left({3t}\right)^2}\;dt\) \({ arc\;length }=\mathop \smallint \limits_0^1 3t\sqrt {{t^2} + 1} dt\) taking u = t2 + 1, \(\frac{du}{dt}=2t\) and \(dt=\frac{1}{2t}du\) \(arc\;length = \frac{3}{2}\mathop \smallint \limits_1^2 \sqrt u du\) \(arc\;length = \|{u^{{3}/{2}}}\|_1^2\) \(arc\;length = {2^{{3}/{2}} }- 1\) Hence the required arc length will be (23/2 - 1) unit.

The length of the curve \(x = t^3, y = \frac {3t^2}2,\;0 \le t \le 1\) will be:1. 23 / 2 - 1 units2. 33 / 2 - 2 units3. 53 / 2 + 1 units4. 33 / 2 + 1 units

Answer» Correct Answer - Option 1 : 2^{3 / 2} - 1 units

Concept:

We have the formula of arc length in parametric form as,

\(\hbox{Arc length }=\int_a^b\; \sqrt{\left({dx\over dt}\right)^2+\left({dy\over dt}\right)^2}\;dt\)

Calculation:

Given:

\(x = t^3, y = \frac {3t^2}2,\;0 \le t \le 1\)

\(\frac{dx}{dt}\) = 3t2 and \(\frac{dy}{dt}\) = 3t

Placing, these values in the above equation, we get

\({ arc\;length }=\int_0^1\; \sqrt{\left({3t^2}\right)^2+\left({3t}\right)^2}\;dt\)

\({ arc\;length }=\mathop \smallint \limits_0^1 3t\sqrt {{t^2} + 1} dt\)

taking u = t2 + 1,

\(\frac{du}{dt}=2t\) and \(dt=\frac{1}{2t}du\)

\(arc\;length = \frac{3}{2}\mathop \smallint \limits_1^2 \sqrt u du\)

\(arc\;length = |{u^{{3}/{2}}}|_1^2\)

\(arc\;length = {2^{{3}/{2}} }- 1\)

Hence the required arc length will be (23/2 - 1) unit.