Advertisements
Advertisements
प्रश्न
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
Advertisements
उत्तर
Here x = a sec θ ...(1)
y = b tan θ ....(2)
Differentiating (1) and (2) w.r.t. θ, we get
`dx/(d θ)` = a sec θ tan θ and `dy/(d θ)` = b sec2 θ
`therefore dy/dx = (dy/(d θ))/(dx/(d θ))`
= `(b sec^2 θ)/(a sec θ tan θ)`
= `(b sec θ)/(a tan θ)`
= `b/a` sec θ cot θ
= `b/a xx 1/(cos θ) xx (cos θ)/(sin θ)`
= `b/a xx 1/(sin θ)`
= `b/a` cosec θ
APPEARS IN
संबंधित प्रश्न
If `log_10((x^3-y^3)/(x^3+y^3))=2 "then show that" dy/dx = [-99x^2]/[101y^2]`
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 y =1 − cos θ, x = 1 − sin θ, then `dy/dx "at" θ =pi/4` is ______
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 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 = (sin^3t)/sqrt(cos 2t), y = (cos^3t)/sqrt(cos 2t)`
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 = a (2θ – sin 2θ) and y = a (1 – cos 2θ), find `dy/dx` when `theta = pi/3`
If X = f(t) and Y = g(t) Are Differentiable Functions of t , then prove that y is a differentiable function of x and
`"dy"/"dx" =("dy"/"dt")/("dx"/"dt" ) , "where" "dx"/"dt" ≠ 0`
Hence find `"dy"/"dx"` if x = a cos2 t and y = a sin2 t.
IF `y = e^(sin-1x) and z =e^(-cos-1x),` prove that `dy/dz = e^x//2`
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'?
If y = sin -1 `((8x)/(1 + 16x^2))`, find `(dy)/(dx)`
x = `"t" + 1/"t"`, y = `"t" - 1/"t"`
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 = asin2t (1 + cos2t) and y = b cos2t (1–cos2t), show that `("dy"/"dx")_("at t" = pi/4) = "b"/"a"`
Differentiate `x/sinx` w.r.t. sin x
If x = t2, y = t3, then `("d"^2"y")/("dx"^2)` is ______.
Derivative of x2 w.r.t. x3 is ______.
If x = `a[cosθ + logtan θ/2]`, y = asinθ then `(dy)/(dx)` = ______.
Under what condition is the formula \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\) directly applicable?
Which formula is used after finding \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)?
If required, in terms of which variables may the final answer be expressed instead of the parameter?
For \(x=a\cos^3\theta\) and \(y=a\sin^3\theta\), what are \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\)?
For \(x=a\cos^3\theta\) and \(y=a\sin^3\theta\), which expression gives \(\frac{dy}{dx}\) in terms of \(x\) and \(y\)?
If \(x=a\cos t\) and \(y=a\sin t\), what is \(\frac{dy}{dx}\)?
What should always be checked before using the main formula for the derivative of parametric functions?
