How do you find the arc length of #x=2/3(y-1)^(3/2)# between #1

Question

How  do you find the arc length of #x=2/3(y-1)^(3/2)# between #1<=y<=4#?

Cameron Alexander · Accepted Answer

It can be found by #L=int_0^4sqrt{1+(frac{dx}{dy})^2}dy#. Let us evaluate the above definite integral. By differentiating with respect to y,
#frac{dx}{dy}=(y-1)^{1/2}# So, the integrand can be simplified as
#sqrt{1+(frac{dx}{dy})^2}=sqrt{1+[(y-1)^{1/2}]^2}=sqrt{y}=y^{1/2}# Finally, we have
#L=\int_0^4y^{1/2}dy=[frac{2}{3}y^{3/2}]_0^4=frac{2}{3}(4)^{3/2}-2/3(0)^{3/2}=16/3# Hence, the arc length is #16/3#. I hope that this helps.

Cameron Alexander · Answer

undefined To find the arc length of the curve defined by $x = \frac{2}{3}(y - 1)^{3/2}$ between $y = 1$ and $y = 4$, we use the formula for arc length:

$$ L = \int_{y_1}^{y_2} \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy $$

First, we find $\frac{dx}{dy}$ by differentiating $x$ with respect to $y$:

$$ \frac{dx}{dy} = \frac{d}{dy}\left(\frac{2}{3}(y - 1)^{3/2}\right) $$

$$ = \frac{2}{3} \times \frac{3}{2}(y - 1)^{1/2} \times \frac{d}{dy}(y - 1) $$

$$ = (y - 1)^{1/2} $$

Now, we substitute $\frac{dx}{dy} = (y - 1)^{1/2}$ into the formula for arc length:

$$ L = \int_{1}^{4} \sqrt{1 + (y - 1)} \, dy $$

$$ = \int_{1}^{4} \sqrt{y} \, dy $$

$$ = \frac{2}{3}y^{3/2} \bigg|_1^4 $$

$$ = \frac{2}{3}(4^{3/2} - 1) $$

$$ = \frac{2}{3}(8 - 1) $$

$$ = \frac{2}{3} \times 7 $$

$$ = \frac{14}{3} $$

Therefore, the arc length of the curve between $y = 1$ and $y = 4$ is $\frac{14}{3}$.

How do you find the arc length of #x=2/3(y-1)^(3/2)# between #1&lt;=y&lt;=4#?

How do you find the arc length of #x=2/3(y-1)^(3/2)# between #1<=y<=4#?