Question

1) if x = Sint , y = Sinpt , prove that ( 1- x 2 ) y2 -xy , + p2y = 0 AND 2) if ( x - a )2 + ( y - b )2 = c2 , prove that ( 1+(dy/ dx )2 )3/2 divided by d2y/ dx2 is a constant independent of a & b . PLEASE REPLY SOON.

Answer 1

Please check the attachments. Thank you.

Answer 2

Step-by-step explanation:

First part:

Given:

x = sint

y = sin pt

Solution:

$\frac{d x}{d t}=\cos t$

$\frac{d y}{d t}=p \cos p t$

Now,

$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{p \cos p t}{\cos t}=y_{1}$

Now,

$\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{d y}{d x}\right)$

$\frac{d^{2} y}{d x^{2}}=\frac{d}{d t}\left(\frac{p \cos p t}{\cos t}\right) \frac{d t}{d x}$

$\frac{d^{2} y}{d x^{2}}=\left(\frac{\left((\cos t) \times\left(p^{2}(-\sin p t)\right)\right)-((p \cos p t) \times(-\sin t))}{\cos ^{2} t}\right)\left(\frac{1}{\cos t}\right)=y_{2}$

$\left(1-x^{2}\right)+y_{2}-x y_{1}+p^{2} y=\left(1-\sin ^{2} t\right) \cdot\left(\frac{(\cos t) \times\left(p^{2}(-\sin p t)\right)-((p \cos p t) \times(-\sin t)}{\cos ^{2} t}\right) )\left(\frac{1}{\cos t}\right)-\left((\sin t) \times\left(\frac{p \cos p t}{\cos t}\right)\right)+p^{2} \sin p$

$=\left(\cos ^{2} t\right) \cdot\left(\frac{-p^{2}(\cos t \cdot \sin p t)+p(\cos p t \cdot \sin t)}{\cos ^{3} t}\right)-\left(\frac{p(\sin t \cos p t)}{\cos t}\right)+p^{2} \sin p t$

$=\left(\frac{-p^{2}(\cos t \cdot \sin p t)+p(\cos p t \cdot \sin t)}{\cos t}\right)-\left(\frac{p(\sin t \cos p t)}{\cos t}\right)+p^{2} \sin p t$

$=\left(\left(\frac{-p^{2}(\cos t \cdot \sin p t)}{\cos t}\right)+\left(\frac{p(\cos p t \cdot \sin t)}{\cos t}\right)\right)-\left(\frac{p(\sin t \cos p t)}{\cos t}\right)+p^{2} \sin p t$

$=\left(-p^{2} \sin p t\right)+\left(\frac{p(\cos p t \cdot \sin t)}{\cos t}\right)-\left(\frac{p(\sin t \cos p t)}{\cos t}\right)+p^{2} \sin p t$

$\left(1-x^{2}\right)+y_{2}-x y_{1}+p^{2} y=0$

Hence proved.

Second Part: $(x-a)^{2}+(y-b)^{2}=c^{2}$

The above is the equation of a circle with centre (a, b) and radius c.

(To eliminate the arbitrary constants we differentiate the equation twice.)

Differentiating with respect to x on both sides.

$2(x-a)+2(y-b)\left(\frac{d y}{d x}\right)=0$

$x-a+(y-b)\left(\frac{d y}{d x}\right)=0$

$\frac{d y}{d x}=-\frac{x-a}{y-b}=\frac{a-x}{y-b}$

Differentiating the above equation again with respect to x

$\left(\frac{d^{2} y}{d x^{2}}\right)=\frac{(y-b)\left(\frac{d y}{d x}\right)(a-x)+(a-x)\left(\frac{d y}{d x}\right)(y-b)}{(y-b)^{2}}$

$\frac{d^{2} y}{d x^{2}}=\frac{y-b(-1)+(a-x) \frac{d y}{d x}}{(y-b)^{2}}$

$=\frac{(a-x) \frac{d y}{d x}-(y-b)}{(y-b)^{2}}$

$=\frac{\left((a-x) \times \frac{a-x}{y-b}\right)-(y-b)}{(y-b)^{2}}$

$\Rightarrow\left(\frac{d^{2} y}{d x^{2}}\right)=\frac{(a-x)^{2}-(y-b)^{2}}{(y-b)^{3}}$

We know that, $(x-a)^{2}+(y-b)^{2}=c^{2}$

$=\frac{-(x-a)^{2}-(y-b)^{2}}{(y-b)^{3}}$

$\therefore \frac{d^{2} y}{d x^{2}}=-\frac{c^{2}}{(y-b)^{3}}$

$L H S=\frac{\left(1+\left(\frac{d y}{d x}\right)^{2}\right)^{\frac{3}{2}}}{\frac{d^{2} y}{d x^{2}}}$

$=\frac{\left(1+\left(\frac{(x-a)}{(y-b)}\right)^{2}\right)^{3 / 2}}{-\frac{c^{2}}{(y-b)^{3}}}$

$=\frac{\left(1+\left(\frac{(x-a)^{2}}{(y-b)^{2}}\right)\right)^{3 / 2}}{-\frac{c^{2}}{(y-b)^{3}}}$

$=-\frac{\left(\frac{(y-b)^{2}+(x-a)^{2}}{(y-b)^{2}}\right)^{\frac{3}{2}} \times(y-b)^{3}}{c^{2}}$

$=-\frac{\left(\frac{(y-b)^{3}+(x-a)^{3}}{(y-b)^{3}}\right) \times(y-b)^{3}}{c^{2}}$

$=-\frac{(y-b)^{3}+(x-a)^{3}}{c^{2}}$

$=-\frac{c^{3}}{c^{2}}$

$\frac{\left(1+\left(\frac{d y}{d x}\right)^{2}\right)^{\frac{3}{2}}}{\frac{d^{2} y}{d x^{2}}}=-c$

Where c is a constant independent of a and b.

Hence, proved.