Find the arclength of y= 4x^(3/2) on [2,3] x-axis
what is 15 x 15
wrong answer by the bot
y = 4x^(3/2)
y' = 6x^(1/2) , so (y')^2 = 36x
arclength = ∫ √(1 + 36x) dx from 2 to 3
= [ 1/54 (√(1+36x)^(3/2) ] from 2 to 3
= 1/54( 109^(3/2) - 72^(3/2) )
= appr 9.5
To find the arclength of a curve, we need to use a calculus method called integration. The formula for arclength is given by:
L = ∫ sqrt(1 + (dy/dx)^2) dx
In this case, we need to find dy/dx for the given curve y = 4x^(3/2). Let's calculate that derivative first:
dy/dx = d(4x^(3/2))/dx
To find the derivative, we can apply the power rule of differentiation:
dy/dx = (3/2)(4)x^(3/2 - 1)
= 6x^(1/2)
Now that we have dy/dx, we can substitute it into the formula for arclength:
L = ∫ sqrt(1 + (dy/dx)^2) dx
= ∫ sqrt(1 + (6x^(1/2))^2) dx
= ∫ sqrt(1 + 36x) dx
Now, let's calculate this integral over the interval [2,3] with respect to x:
L = ∫[2,3] sqrt(1 + 36x) dx
To compute this integral, we can use integration techniques such as substitution or integration by parts.
After evaluating the definite integral, we will obtain the arclength of the curve y = 4x^(3/2) on the x-axis over the interval [2,3].