Question

How do you find the lengths of the curve `y=(4/5)x^(5/4)` for `0<=x<=1`?

Answer 1

`8/15(1+sqrt(2))`

Explanation:

We Need the formula `s=int_a^bsqrt(1+(f'(x))^2)dx`.

given `f(x)=4/5*x^(5/4)`
using that `(x^n)'=nx^(n-1)` we get

`f'(x)=4/5*5/4*x^(5/4-1)`

`f'(x)=x^(1/4)` so we have to integrate

`int_0^1sqrt(1+sqrt(x))dx`
to calculate the indefinite integtral we Substitute

`t=sqrt(1+sqrt(x))`

so `x=(t^2-1)^2`

`dx=4t(t^2-1)dt`

and we have `4*int t^2(t^2-1)dt=4int(t^4-t^2dt`

this is `4(t^5/5-t^3/3)+C=4/15(1+sqrt(x))^(3/2)(3sqrt(x)-2)+C`
and we have

`[4/5sqrt(1+sqrt(x))^5-4/3sqrt(1+sqrt(x))^3]_0^1=8/15(1+sqrt(2))`

Answer 2

This is where the formula that is in Sonnhard's solution comes from.

Explanation:

Set the total length of the curve `C` over the interval `a and b` as `L_c`

Let us consider a minuscule portion of the curve that is so short and greatly magnified that it looks as though it is straight.

Set its length as `deltas`

Using Pythagoras we have `(deltas)^2=(deltax)^2+(deltay)^2`

`lim_(deltas->0)(deltas)^2->(ds)^2= (dx)^2+(dy)^2`

Take the square root of both sides

`ds=sqrt((dx)^2+(sy)^2)`

Multiply by 1 and you do not change the value. However 1 comes in many forms.

`ds=1xxsqrt((dx)^2+(sy)^2) color(white)("dd")->color(white)("dd")ds=dx/dxxxsqrt((dx)^2+(sy)^2)`

`color(white)("ddddddddddddddddddddd")->color(white)("d")ds=dxxxsqrt((dx)^2/(dx)^2+(dy)^2/(dx)^2)`

`color(white)("ddddddddddddddddddddd")->color(white)("d")ds=dxxxsqrt(1+(dy)^2/(dx)^2)`

But `(dy)^2/(dx)^2` may be written as `(f'(x))^2`

`color(white)("ddddddddddddddddddddd")->color(white)("d")ds=dxxxsqrt(1+(f'(x))^2 )`

`color(white)("ddddddddddddddddddddd")->color(white)("d")ds=color(white)("dd")sqrt(1+(f'(x))^2 ) xxdx`

Multiply `ds` by 1 and use the dot instead of `xx`

`color(white)("ddddddddddddddddddddd")->color(white)("d")1.ds=color(white)("dd")sqrt(1+(f'(x))^2 ) .dx`

Sum up all the `ds's` using integration

`color(white)("ddddddddddd")->color(white)("d")int_a^b color(white)(.)1.ds=L_c=color(white)("dd")int_a^bsqrt(1+(f'(x))^2 ) .dx`

As in Sonnhard's solution