Question

Integral for calculating length of arc of parabola y^2=4x, cut of by the line

Answer 1

Answer:

Step-by-step explanation:

Putting y=2a⋅tan θ  

=4a⋅∫π402asecθ⋅sec2θdθ  

Let J =  ∫secθ⋅sec2θdθ  

Applying integration by parts will give

J=12⋅[secθtanθ+ln|secθ+tanθ|]  

Then,

I=4a⋅12⋅[secθtanθ+ln|secθ+tanθ|]π40  

=2a⋅(2–√+ln(1+2–√)−0)  

=a⋅(22–√+ln(3+22–√)) ,

The required length.

Answer 2

Step-by-step explanation:

y^2 = 4ax => 2y(dy/dx)=4a

=> dy/dx= 4a/2y=2a/y.

√{1+(dy/dx)^2} = √{1 + (2a/y)^2}

=√{1+(4a^2/y^2) = √{1 + 4a^2/4ax}

= √{(x+a)/x}=√(x+a)/√x. But the length of the curve

= integ.(a to b)√{1+(dy/dx)^2} dx

So the required length

=2.integ.(0 to a)√(x+a)/√x dx

Let √x =t => 1/2√x dx = dt=> dx/√x = 2dt

x = t^2 > x+a = t^2+a. When x=0, t=0 and when x=a, t=√a. The required length

=2.integ.(0 to √a)√(t^2+a)2dt

=4[(t/2)√(t^2+a) + (a/2)log{t+√(t^2+a)}] (0 to √a)

=4[(√a/2)√(2a)+(a/2)log(√a+√(2a)] -4[(a/2)log√a

=2√2.a+2alog{√a(1+√2)}-2alog√a

=2√2.a +2alog(1+√2).

If there is any correction please correct it.