Advertisements
Advertisements
प्रश्न
Find y2 for the following function:
x = a cosθ, y = a sinθ
Advertisements
उत्तर
x = a cos θ, y = a sin θ
`"dx"/("d"θ)`= a(-sinθ) = -a sinθ …….. (i)
`"dy"/("d"θ)` = a(cosθ)
`therefore y_1 = "dy"/"dx" = ("dy"/("d"θ))/("dx"/("d"θ)) = ("a" cos theta)/(- "a" sin theta)`
`y_1 = "dy"/"dx"` = - cot θ
`y_2 = ("d"^2y)/"dx"^2 = - (- "cosec"^2 theta) ("d"theta)/"dx"`
`= "cosec"^2theta ("d"theta)/"dx"`
`= "cosec"^2theta 1/("dx"/("d"theta))`
`=> "cosec"^2 theta xx ("cosec" theta)/(- "a")`
`= (- 1)/"a" "cosec"^3 theta`
APPEARS IN
संबंधित प्रश्न
Differentiate the following with respect to x.
`(3 + 2x - x^2)/x`
Differentiate the following with respect to x.
`e^x/(1 + x)`
Differentiate the following with respect to x.
`e^x/(1 + e^x)`
Differentiate the following with respect to x.
x3 ex
Differentiate the following with respect to x.
sin(x2)
Differentiate the following with respect to x.
`1/sqrt(1 + x^2)`
If y = 500e7x + 600e-7x, then show that y2 – 49y = 0.
If y = `(x + sqrt(1 + x^2))^m`, then show that (1 + x2) y2 + xy1 – m2y = 0
If y = sin(log x), then show that x2y2 + xy1 + y = 0.
If y = 2 sin x + 3 cos x, then show that y2 + y = 0.
