Calculus!! Find the length of the curve defined by?

Question

Find the length of the curve defined by: y=3x^(3/2)+3 from x=4 and x=7. Can someone explain how to do this problem???

Thanks a ton for the help.

Answer 1

Suppose ds is a differntial element of the length of the curve.

let ds = (( 1 + (ds/dy)^2 )^(1/2))dy
The length of the curve, s, is the integral of ds between points a and b.
For the problem above,
ds/dy = (9/2)*x^(1/2)
so ds = ((1 + ((9/2)*x^(1/2))dy
s will be the integral of ds between 4 and 7

Answer 2

length curve=integral from 4 to 7 of square root(1+square of derivative)

y'=3/2 *3 x^(1/2)
y'=9/2*square root(x)

y'^2= 81/4*x

1+y'^2= 81/4*x +1

curve length=integral from x=4 to x=7 of
square root(81/4*x +1) dx

Try to calculate the value of this integral

Answer 3

The length of a curve is defined by the following fundamental: L = ??[a million + (dy/dx)^2] dx. I received't tutor a thanks to derive it, yet do only not ignore that it really is with regard to the pythagorean theorem. so commence by differentiating y with appreciate to x, providing you with dy/dx = 2(3/2 x^(a million/2)) = 3x^a million/2 Now positioned dy/dx into the formula, providing you with L = ??[a million + (3x^a million/2)^2] dx carry out the mathematics operations, providing you with L = ??[a million + 9x] dx Now carry out a u-substitution (opposite of the chain rule), take u = a million + 9x, du = 9dx, dx = du/9 that promises L = ??[u] * a million/9 du carry out the blending and replace u interior the equation. i imagine you may take it from there.

Answer 4

sqrt(1+F'(x)^2)

sqrt(1+(9/2X^(1/2))^2) then just subtract what you get from 7 from what you get in 4.