मराठी

If x = ecos2t and y = esin2t, prove that dydxdydx=-ylogxxlogy - Mathematics

Advertisements
Advertisements

प्रश्न

If x = ecos2t and y = esin2t, prove that `"dy"/"dx" = (-y log x)/(xlogy)`

बेरीज
Advertisements

उत्तर

Given that: ecos2t and y = esin2t

⇒ cos 2t = log x and sin 2t = log y.

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

`"dx"/"dt" = "e"^(cos2"t") * "d"/"dt" (cos 2"t")`

= `"e"^(cos 2"t") (- sin 2"t") * "d"/"dt" (2"t")`

= `- "e"^(cos2"t") * sin 2"t" * 2`

= `2"e"^(cos2"t") * sin 2"t"`

Now y = esin2t

`"dy"/"dt" = "e"^(sin2"t") * "d"/"dt"(sin 2"t")`

= `"e"^(sin2"t") * cos 2"t" * "d"/"dt"(2"t")`

= `"e"^(sin2"t") * cos 2"t" * 2`

= `2"e"^(sin2"t") * cos 2"t"`

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

= `(2"e"^(sin2"t") * cos2"t")/(-2"e"^(cos2"t") * sin 2"t")`

= `("e"^(sin2"t") * cos2"t")/(-"e"^(cos2"t") * sin2"t")`

= `(y cos 2"t")/(-x sin 2"t")`

= `(y log x)/(-x log y)`   ......`[(because cos 2"t" = log x),(sin 2"t" = log y)]`

Hence, `"dy"/"dx" = - (y log x)/(x log y)`.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 5: Continuity And Differentiability - Exercise [पृष्ठ ११०]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
पाठ 5 Continuity And Differentiability
Exercise | Q 49 | पृष्ठ ११०

संबंधित प्रश्‍न

If  `log_10((x^3-y^3)/(x^3+y^3))=2 "then show that"  dy/dx = [-99x^2]/[101y^2]`


find dy/dx if x=e2t , y=`e^sqrtt`


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 x=at2, y= 2at , then find dy/dx.


 

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

 

If x=α sin 2t (1 + cos 2t) and y=β cos 2t (1cos 2t), show that `dy/dx=β/αtan t`


If x = a sin 2t (1 + cos2t) and y = b cos 2t (1 – cos 2t), find the values of  `dy/dx `at t = `pi/4`


If x = cos t (3 – 2 cos2 t) and y = sin t (3 – 2 sin2 t), find the value of dx/dy at t =4/π.


Derivatives of  tan3θ with respect to sec3θ at θ=π/3 is

(A)` 3/2`

(B) `sqrt3/2`

(C) `1/2`

(D) `-sqrt3/2`


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

x = 4t, y = `4/y`


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

x = `a(cos t + log tan  t/2)`, y = a sin t


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

x = a sec θ, y = b tan 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 `y = e^(sin-1x)   and  z =e^(-cos-1x),` prove that `dy/dz = e^x//2`


If y = sin -1 `((8x)/(1 + 16x^2))`, find `(dy)/(dx)`


Evaluate : `int  (sec^2 x)/(tan^2 x + 4)` dx


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


sin x = `(2"t")/(1 + "t"^2)`, tan y = `(2"t")/(1 - "t"^2)`


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


If x = 3sint – sin 3t, y = 3cost – cos 3t, find `"dy"/"dx"` at t = `pi/3`


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


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


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 ____________.


Form the point of intersection (P) of lines given by x2 – y2 – 2x + 2y = 0, points A, B, C, Dare taken on the lines at a distance of `2sqrt(2)` units to form a quadrilateral whose area is A1 and the area of the quadrilateral formed by joining the circumcentres of ΔPAB, ΔPBC, ΔPCD, ΔPDA is A2, then `A_1/A_2` equals


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


Let a function y = f(x) is defined by x = eθsinθ and y = θesinθ, where θ is a real parameter, then value of `lim_(θ→0)`f'(x) is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×