CBSE Guess > Papers > Important Questions > Class XII > 2010 > Maths > Mathematics By Mr. M.P.Keshari

Application of derivatives

Q.5. Find the equation of the tangent and the normal to the curve y = x³ at the point P(1, 1).

Solution :

We have the curve y = x³ ----------------------------- (1)
=> dy/dx = 3x²
[dy/dx](1, 1) = 3 = Slope of tangent. Slope of normal = – 1/3.
Equation of tangent to (1) at (1, 1) is given by
(y – 1) = 3(x – 1) => y = 3x – 2. [Ans.]
equation of normal to (1) at (1, 1) is given by
(y – 1) = – 1/3(x – 1) => 3y – 3 = – x + 1 => x + 3y = 4. [Ans.]

Q.6. Find the equation of the tangent to the curve: x = θ + sin θ, y = 1 + cos θ at θ = π/4.

Solution :

We have, x = θ + sin θ, y = 1 + cos θ
=> dx/dθ = 1 + cos θ, dy/dθ = – sin θ
=> dy/dx = (dy/dθ)/(dx/dθ) = – sin θ/(1 + cos θ).
[dy/dx]atθ=π/4 = (– 1/√2)/(1 + 1/√2) = – 1/(√2 + 1) = 1 – √2.
Also at π/4 x = π/4 + sin π/4 = π/4 + 1/√2.
y = 1 + cos π/4 = (√2 + 1)/√2 .
Therefore, equation of tangent at θ = π/4 is :
y – [(√2 + 1)/√2] = (1 – √2)(x – π/4 – 1/√2)
=> y – 1 – 1/√2 = (1 – √2)x + (√2 – 1)π/4 – 1/√2 + 1
=> y = (1 – √2)x + (√2 – 1)π/4 + 2. [Ans.]

Q.7. Find the equation of the tangent to the curve x = sin 3t, y = cos 2t, at t = π/4.

Solution :

x = sin 3t, dx/dt = 3 cos 3t;
y = cos 2t, dy/dt = – 2 sin 2t
dy/dx = (dy/dt)/(dx/dt)
= – 2sin 2t/3cos 3t
Therefore, slope = m = [dy/dx]atπ/4
= [– 2sin2(π/4)]/[3cos 3(π/4)]
= [– 2sin(π/2)]/[3cos(3π/4)]
= {– 2(1)}/{3(–1/√2)}
= (2√2)/3.
Now, x1 = sin 3π/4 = 1/√2;
y1 = cos 2π/4 = cos π/2 = 0
Therefore, required point is (1/√2, 0) and the equation of tangent at (1/√2, 0) is
y – 0 = (2√2/3)(x – 1/√2) [Using : y – y1 = m (x – x1)]
Or, 3y = 2√2 x – 2
Or, 2√2 x – 3y – 2 = 0. [Ans.]

Paper By Mr. M.P.Keshari
Email Id : [email protected]
Ph No. : 09434150289