# find the length of the arc of the curve from point P to point Q. y=1/2x^2, P(-7,49/2), Q(7,49/2) please explain the "why" as you solve The graph is a parabola:

The parabola is symmetric, so we may instead find the length of the curve from (0,0) to (7, 49/2) and double it:

A formula for arc length is:

int_a^b sqrt(1+(dy/dx)^2) dx

Thus we want to find:

2 int _0 ^7 sqrt(1+(dy/dx)^2) dx

y=(1/2)x^2

dy/dx...

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The graph is a parabola:

The parabola is symmetric, so we may instead find the length of the curve from (0,0) to (7, 49/2) and double it:

A formula for arc length is:

int_a^b sqrt(1+(dy/dx)^2) dx

Thus we want to find:

2 int _0 ^7 sqrt(1+(dy/dx)^2) dx

y=(1/2)x^2

dy/dx = x

1+(dy/dx)^2 = 1+x^2

So we want to find:

2 int_0^7 sqrt(1+x^2) dx

We ignore the constant, and the definite integral for now:

int sqrt(1+x^2)dx

We can make the trig substitution:

x="tan" theta , dx = "sec"^2 theta d theta

The integral becomes:

int sqrt(1+"tan"^2 theta) "sec"^2 theta d theta

=int "sec"^3 theta d theta

This is an integration by parts problem:

u= "sec" theta               dv="sec"^2 theta d theta

du = "sec" theta "tan" theta   v= "tan" theta

int "sec"^3 theta d theta = "sec" theta "tan" theta - int "sec" theta "tan"^2 theta d theta

="sec" theta "tan" theta - int "sec" theta ("sec"^2 theta - 1) d theta

="sec" theta "tan" theta - int "sec"^3 theta d theta + int "sec" theta d theta

="sec" theta "tan" theta - int "sec" ^3 theta d theta + "ln" |"sec" theta + "tan" theta|

Now, we have a int "sec"^3 theta d theta  on both sides of the equation, so:

2 int "sec"^3 theta d theta = "sec" theta "tan" theta + "ln" |"sec" theta + "tan" theta|

int "sec"^3 theta = (1/2) ( "sec" theta "tan" theta + "ln" |"sec" theta + "tan" theta| ) +C

We have our answer in theta but we want it in x.  Our original substitution was:

x="tan" theta

Another way to say this is "tan" theta = x/1

That is, if we had a right triangle, we could make x the opposite side of the angle, and 1 the adjacent side, and our triangle would represent this substitution.

Then the hypotenuse would be sqrt(x^2+1)

So "sec" theta would be (sqrt(x^2+1))/1

So:

int sqrt(1+x^2) dx = (1/2)( (sqrt(x^2+1))(x) + "ln" |(sqrt(1+x^2))+(x)| )

=(1/2)(xsqrt(1+x^2) + "ln"|sqrt(1+x^2)+x|)+C

Now we go back to our original problem:

int_0^7 sqrt(1+x^2) = 

(1/2)(7 sqrt(50)+"ln"|sqrt(50)+7|)-(1/2)(0sqrt(1) + "ln"(1+0))

=(1/2)(7sqrt(50) + "ln" |sqrt(50)+7|)`

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