मराठी

If x = sint and y = sin pt, prove that dydxdydxp(1-x2)d2ydx2-xdydx+p2y = 0 - Mathematics

Advertisements
Advertisements

प्रश्न

If x = sint and y = sin pt, prove that `(1 - x^2) ("d"^2"y")/("dx"^2) - x "dy"/"dx" + "p"^2y` = 0

बेरीज
Advertisements

उत्तर

Given that: x = sin t and y = sin pt

Differentiating both the parametric functions w.r.t. t

`"dx"/"dt"` = cos t and `"dy"/"dt"` = cos pt. p = p cos pt

`"dy"/"dx" = ("dy"/"dt")/("dx"/"dt")`

= `("p" cos "pt")/cos "t"`

∴ `"dy"/"dx" = ("p" cos "pt")/cos"t"`

Again differentiating w.r.t. x,

`"d"/"dx"("dy"/"dx") = "p"*"d"/"dx"(cos"pt"/cos"t")`

⇒ `("d"^2y)/("dx"^2) = "p"[(cos "t" * "d"/"dx" (cos "pt") - cos "pt" * "d"/"dx" (cos "t"))/(cos^2"t")]`

= `"p"[(cos"t"(- sin "pt") * "p" "dt"/"dx" - cos "pt"(- sin "t") * "dt"/"dx")/(cos^2"t")]`

= `"p"[(-"p" cos "t" sin "pt" + cos "pt" sin "t")/(cos^2"t")]"dt"/"dx"`

= `"p"[(-"p" cos "t" sin "pt" + cos "pt" sin "t")/(cos^2"t")]* 1/cos"t"`

= `"p"[(-"p" cos "t" sin "pt" + cos "pt" sin "t")/cos^3"t"]`

Now we have to prove that

`(1 - x^2) * ("d"^2y)/("dx"^2) - x * "dy"/"dx" + "p"^2y` = 0

L.H.S. = `(1 - x^2) ["p"((-"p" cos "t" sin "pt" + cos "pt" sin "t")/cos^3"t")] - x."p" (cos "pt")/cos"t" + "p"^2y`

⇒ `(1 - sin^2"t") ["p"((-"p" cos "t" sin "pt" + cos "pt" sin "t")/cos^3"t")] - ("p" sin "t" * cos "pt")/cos"t" + "p"^2 * sin "pt"`

⇒ `cos^2"t" [(-"p"^2 cos "t" sin "pt" + cos "pt" sin "t")/(cos^3"t")] - ("p" sin "t" * cos "pt")/cos"t" + "p"^2 * sin "pt"`

⇒ `(-"p"^2 cos "t" sin "pt" + "p" cos "pt" sin "t")/cos "t" - ("p" sin "t" cos "pt")/cos"t" + "p"^2 sin "pt"`

⇒ `(-"p"^2 cos "t" sin "pt" + "p" cos "pt" sin "t" - "p" sin "t" cos "pt" + "p"^2 sin "pt" cos "t")/cos "t"`

⇒ `0/cos"t"` = 0 = R.H.S.

Hence proved.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 5: Continuity And Differentiability - Exercise [पृष्ठ ११३]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
पाठ 5 Continuity And Differentiability
Exercise | Q 81 | पृष्ठ ११३

संबंधित प्रश्‍न

If x = f(t), y = g(t) are differentiable functions of parammeter ‘ t ’ then prove that y is a differentiable function of 'x' and  hence, find dy/dx if x=a cost, y=a sint


If `ax^2+2hxy+by^2=0` , show that `(d^2y)/(dx^2)=0`


 

 If x=a sin 2t(1+cos 2t) and y=b cos 2t(1cos 2t), find `dy/dx `

 

If x = a sin 2t (1 + cos 2t) and y = b cos 2t (1 – cos 2t) then find `dy/dx `

 


Find the value of `dy/dx " at " theta =pi/4 if x=ae^theta (sintheta-costheta) and y=ae^theta(sintheta+cos theta)`


If x and y are connected parametrically by the equations, without eliminating the parameter, find `bb(dy/dx)`.

x = a cos θ, y = b cos θ


If x and y are connected parametrically by the equations, without eliminating the parameter, find `bb(dy/dx)`.

x = sin t, y = cos 2t


If x and y are connected parametrically by the equations, without eliminating the parameter, find `bb(dy/dx)`.

x = cos θ – cos 2θ, y = sin θ – sin 2θ


If x and y are connected parametrically by the equations, without eliminating the parameter, find `bb(dy/dx)`.

x = a (θ – sin θ), y = a (1 + cos θ)


If x and y are connected parametrically by the equations, without eliminating the parameter, find `bb(dy/dx)`.

x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ)


If `x = acos^3t`, `y = asin^3 t`,

Show that `(dy)/(dx) =- (y/x)^(1/3)`


If x = a (2θ – sin 2θ) and y = a (1 – cos 2θ), find `dy/dx` when `theta = pi/3`


The cost C of producing x articles is given as C = x3-16x2 + 47x.  For what values of x, with the average cost is decreasing'?  


x = `"t" + 1/"t"`, y = `"t" - 1/"t"`


x = 3cosθ – 2cos3θ, y = 3sinθ – 2sin3θ


x = `(1 + log "t")/"t"^2`, y = `(3 + 2 log "t")/"t"`


Differentiate `tan^-1 ((sqrt(1 + x^2) - 1)/x)` w.r.t. tan–1x, when x ≠ 0


Derivative of x2 w.r.t. x3 is ______.


If `"x = a sin"  theta  "and  y = b cos"  theta, "then"  ("d"^2 "y")/"dx"^2` is equal to ____________.


If y `= "Ae"^(5"x") + "Be"^(-5"x") "x"  "then"  ("d"^2 "y")/"dx"^2` is equal to ____________.


If x = `a[cosθ + logtan  θ/2]`, y = asinθ then `(dy)/(dx)` = ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×