Class 12 Maths NCERT Solutions for Chapter 5 Continuity and Differentiability Exercise 5.6
Continuity and Differentiability Exercise 5.6 Solutions
1. If x and y are connected parametrically by the equation, without eliminating the parameter, find dy/dx .
x = 2at2 , y = at4
Solution
The given equations are x = 2at2 , y = at4
2. If x and y are connected parametrically by the equations x = a cos θ, y = b cos θ, without eliminating the parameter, find dy/dx .
Solution
The given equations are x = a cos θ, y = b cos θ
3. If x and y are connected parametrically by the equation, without eliminating the parameter, find dy/dx .
x = sin t, y = cos 2t
Solution
The given equations are x = sin t, y = cos 2t
Solution
x = 4t, y = 4/y
x = cos θ- cos 2θ, y = sin θ - sin 2θ
6. If x and y are connected parametrically by the equation, without eliminating the parameter, find dy/dx.
x = a (θ – sin θ), y = a (1 + cos θ)
Solution
The given equations are x = a (θ – sin θ), y = a (1 + cos θ)
7. If x and y are connected parametrically by the equation, without eliminating the parameter, find dy/dx.
x = sin3t/√(cos 2t) , y = cos3t/√(cos 2t)
Solution
The given equations are x = sin3t/√(cos 2t) and y = cos3t/√(cos 2t)
8. If x and y are connected parametrically by the equation, without eliminating the parameter, find dy/dx.
x = a[cos t + log tan (t/2) ], y = a sin t
Solution
The given equation are x = a[cos t + log tan (t/2) ], y = a sin t
9. If x and y are connected parametrically by the equation, without eliminating the parameter, find dy/dx.
x = a sec θ, y = b tan θ
Solution
The given equations are x = a sec θ, y = b tan θ
10. If x and y are connected parametrically by the equation, without eliminating the parameter, find dy/dx.
x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ)
Solution
The given equations are x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ)
11. If x = √(asin – 1), y = √(acos – 1) show that dy/dx = -y/x
Solution
The given equations are x = √(asin – 1) and y = √(acos – 1)
Hence, proved.