x=a sin t - b cot t , y=a cot t + b sin y.Prove d 2 y ÷ dx 2 =(-x 2 -y 2 )÷y 3

Question

Aditya Gupta · Answer

note that ques is wrong and it should be cosx instead of cotx.

x=asint-bcost
y=acost+bsint
squaring and then adding both eqs., we get
x2+y2=a2+b2
Now diff. w. r. to x
2x+2y.dy/dx=0
dy/dx=-x/y
diff .w.r.to x again
d2y/dx2= -{(y.1-x.dy/dx)/y2}
= -{(y+x2/y)/y2}=-(x2+y2)/y3

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x=a sin t - b cot t , y=a cot t + b sin y.Prove d 2 y ÷ dx 2 =(-x 2 -y 2 )÷y 3

x=a sin t - b cot t , y=a cot t + b sin y.Prove d²y ÷ dx²=(-x²-y²)÷y³

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x=a sin t - b cot t , y=a cot t + b sin y.Prove d 2 y ÷ dx 2 =(-x 2 -y 2 )÷y 3

x=a sin t - b cot t , y=a cot t + b sin y.Prove d2y ÷ dx2=(-x2-y2)÷y3

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x=a sin t - b cot t , y=a cot t + b sin y.Prove d²y ÷ dx²=(-x²-y²)÷y³