Blending of Doppler's Effect and Calculus

Well this is easy.

First minimum frequency occurs when the source has maximum velocity of seperation that is when tangent to the parabola at the point where the source is present passes through the point where the observer is present.

It is clear that $(-1,0)$ lies on the intersection of the directrix and axis of the parabola, hence the point of contact of the tangent is the end point of latus rectum(Simple properties of parabola alternatively you can solve it by other methods)

Co-ordinate of end point of latus rectum is $(1,2)$

Hence our first task is find the time source will take to reach the end point of latus rectum.

Here formula for finding arc length of a curve will help us here.

$L = \displaystyle \int _{ 0 }^{ { x }_{ 0 } }{ \sqrt { 1+{ (\frac { dy }{ dx } ) }^{ 2 } } dx }$

In this question we have $y=2\sqrt{x} , {x}_{0}=1$ .

Our integral becomes :

$L=\displaystyle \int _{ 0 }^{ 1 }{ \sqrt { 1+\frac { 1 }{ x } } dx }$

Put $x=\tan^{2}(\theta)$

$L= \displaystyle 2\int _{ 0 }^{ \pi /4 }{ { sec }^{ 3 }\theta d\theta }$

Applying integration by parts :

$u = sec\theta , v' = sec^{2}\theta$

$L =\displaystyle 2( sec\theta tan\theta \overset { \pi /4 }{ \underset { 0 }{ | } } -\int _{ 0 }^{ \pi /4 }{ sec\theta { tan }^{ 2 }\theta d\theta } )$

$L = \displaystyle 2(sec\theta tan\theta \overset { \pi /4 }{ \underset { 0 }{ | } } -\int _{ 0 }^{ \pi /4 }{ { sec }^{ 3 }\theta d\theta } +\int _{ 0 }^{ \pi /4 }{ sec\theta d\theta })$

$2L = \displaystyle 2(sec\theta tan\theta +ln(sec\theta +tan\theta )\overset { \pi /4 }{ \underset { 0 }{ | } } )$

$\Rightarrow L= \sqrt { 2 } +ln(\sqrt { 2 } +1)$

Time taken = $\dfrac{L}{\dfrac{v}{2}}$ = $\dfrac{2\sqrt{2}+2ln(\sqrt{2}+1)}{v}$

Now the net time after which the observer hears the minimum frequency is the taken taken by source to reach the point and the time in which the sound reaches the observer.

Time taken by sound to reach the observer.

=Distance between $(-1,0)$ and $(1,2)$ / Speed of sound

$=\frac{2\sqrt{2}}{v}$

Now net time ${T}_{net} = \dfrac{ 4\sqrt{2} + 2ln( \sqrt{2} +1 ) }{v}$

The physics of this problem is as follows:

The sound source is moving along the parabola and is continuously emitting sound waves which travel to the observer.

The important thing to note hear is that the sound pulse with the minimum $apparent$ (or Doppler shifted) frequency is emitted by the source at the instant when it is moving away from the source(i.e. when it's velocity is along the line joining it to the observer).

As is obvious, the point on the parabola where this happens corresponds to the point where the tangent drawn from $(-1,0)$ touches the parabola.

Therefore what we have to find, is the (time taken by the sound source to travel along the parabola to this point) + (the time taken for the sound pulse emitted at this point to reach the observer).

That's the physics done. For the math, the first step is to find the point where the tangent from $(-1,0)$ touches the parabola.

$y^2=4x$ can be parametrised as $(t^2,2t)$ . Let the point to be found be $P(t)$ .

The tangent at P is: $ty=x+t^2$ This passes through $(-1,0)$ . We get $t=\pm 1$ .

Either value of t is equivalent. So let's work with t=1.

Thus, $P\equiv(1,2)$ .

To find the time taken by the sound source to go from the origin to P, we need to find the arc length of the parabola between x=0 and x=1 (as this is the distance travelled by the source).

The arc length s, is given by: $s=\int_{0}^{1}\sqrt{1+(\frac{dy}{dx})^2}\quad dx$

Here, $\frac{dy}{dx}=\frac{1}{\sqrt x}$ $\quad (as \quad y=2\sqrt x)$ .

Thus, $s=\int_{0}^{1}\sqrt{1+\frac{1}{x}}\quad dx= \sqrt{2}+ log(\sqrt{2}+1)$

Note that the corresponding time is $\frac{s}{\frac{v}{2}}=\frac{2s}{v}$ .

Now, the final part is very easy: We find the distance between P(1,2) and (-1,0) which is equal to $2\sqrt 2$ and the corresponding time is $\frac{2\sqrt 2}{v}$

Add the two, and you're done.