1try
y-x=√cos2t
x = sin2t / √cos 2t
y= cos2t / √cos 2t find the value of the product of the cube of y and the second derivative of y wrt x
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7 Answers
11y3 X d2y/dx2
=cos6t/(cos2t)3/2 X P
to find P
find dy/dt and then dx/dt then divide them to get dy/dx then take the derivative of dy /dx to get P
u will find that dy/dt=-cos2tsin2t+sin2tcos2t/cos2t√cos2t
dx/dt=cos2tsin2t-sin2tcos2t/cos2t√cos2t
dy/dx=-1
so d2y/dx2=0
so answer is 0
62\\x = \frac{sin^2t}{\sqrt{cos 2t}} \\y= \frac{cos^2t}{\sqrt{cos 2t}} \\y-x=\frac{cos^2t-sin^2t}{\sqrt{cos 2t}} \\y-x=\frac{cos 2t}{\sqrt{cos 2t}} \\y-x=\sqrt{cos 2t}
62\\x = \frac{sin^2t}{\sqrt{cos 2t}} \\y= \frac{cos^2t}{\sqrt{cos 2t}} \\y+x=\frac{cos^2t+sin^2t}{\sqrt{cos 2t}} \\y+x=\frac{1}{\sqrt{cos 2t}}
62Thus, (x+y)(x-y)=-1
x2-y2=-1
2x-2y(dy/dx)=0
dy/dx=x/y
again,
differentiating 2x-2y(dy/dx)=0
1-(dy/dx)2-y(d2y/dx2)=0
thus, 1-(x/y)2=y(d2y/dx2)
thus, y3(d2y/dx2) = y2-x2 = 1
