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

Application of derivatives

6.3. Tangents and Normals.

Q.1. If x = a sin 2t (1 + cos 2t) and y = b cos 2t (1 – cos 2t), show that [dy/dx]at t=π/4 = b/a.

Solution :

We have, x = a sin 2t (1 + cos 2t)
Therefore, dx/dt = a. [2 cos 2t (1 + cos 2t) + sin 2t (– 2sin 2t)]
= 2a [cos 2t + cos2 2t – sin2 2t]
= 2a [cos 2t + cos 4t]
and y = b cos 2t (1 – cos 2t)
then dy/dt = b [– 2 sin 2t (1 – cos 2t) + cos 2t. 2 sin 2t]
= 2b [– sin 2t + 2 sin 2t cos 2t]
= 2b [– sin 2t + sin 4t]
Therefore, dy/dx = (dy/dt)/(dx/dt)
= [2b {– sin 2t + sin 4t}]/[2a{cos 2t + cos 4t}]
Therefore, [dy/dx]at t = π/4 = b/a [{– sinπ/2 + sin π}/{cos π/2 + cos π}]
= b/a[{– 1 + 0}{0 – 1}]
= b/a. [Proved.]

Q.2. If x = a(cos θ + log tan θ/2) and y = a sin θ, find the value of dy/dx at θ = π/4.

Solution :

We have, x = a( cos θ + log tan θ/2),
dx/dθ = a [– sin θ + 1/(tan θ/).sec2 θ/2.(1/2)]
= a [– sin θ + {(cos θ/2)/(sin θ/2)}.1/(cos2 θ/2).1/2]
= a [– sin θ + 1/(2 sin θ/2.cos θ/2)]
= a [– sin θ + 1/sin θ ]
= a [{– sin² θ + 1)/sin θ]
= a(cos² θ/sin θ).
and y = a sin θ
dy/dθ = a cos θ.
Therefore, dy/dx = (dy/dθ)/(dx/dθ) = [ a cos θ]/[a cos² θ/sin θ] = sin θ/cos θ = tan θ.
Therefore, [dy/dx]atθ = π/4 = tan π/4 = 1. [Ans.]

Q.3. Find the slope of the tangent to the curve y = 3x⁴ – 4x at x = 1.

Solution :

We have, y = 3x⁴ – 4x
dy/dx = 12x³– 4.
Slope = [dy/dx]_{x =1} = 12(1)³ – 4 = 12 – 4 = 8. [Ans.]

Q.4. For the curve y = 3x² + 4x, find the slope of the tangent to the curve at the point whose x-coordinate is – 2.

Solution:

We have y = 3x² + 4x
Therefore, dy/dx = 6x + 4
Slope of tangent to the curve at x = – 2 is
m = [dy/dx]at x = –2 = 6(– 2) + 4
= –12 + 4 = – 8. [Ans.]

Paper By Mr. M.P.Keshari
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