If each side of an equilateral triangle increases at the rate of 2 cm / sec , find the rate of increase of its area when its side of length 3 cm .

If x cm is the side of the equilateral triangle and A is its area, then A = 3 4 x 2 . Differentiating w.r.t t , we get d A d t = 3 4 × 2 x . d x d t d A d t = 3 2 × x . d x d t - - - - - ( 1 ) Now, we know that, d x d t = 2 cm / sec and x = 3 cm ∴ Equation ( 1 ) gives. d A d t = 3 2 × 3 × 2 d A d t = 3 6 2 cm 2 / sec Hence, rate of increasing of the area of equilateral triangle is 3 6 2 cm 2 / sec .

A man of height 1 . 5 meters walks toward a lamp post of height 4 . 5 meters, at the rate of 3 4 meter / sec . Find the rate at which (i) his shadow is shortening. (ii) the tip of the shadow is moving.

Given, Let O A be the lamp post, M N the man, M B = x , his shadow and O M = y the distance of the man from lamp post at time t . Then d y d x = 3 4 is the rate at which the man is moving towards from the lamp post. d y d x is the rate at which his shadow is shortening. B is the tip of the shadow and it is at a distance of x + y from the post. d d t x + y = d x d t + d y d t is the rate at which the tip of the shadow is moving. From the figure, x 1 . 5 = x + y 4 . 5 ∴ 4 . 5 x = 1 . 5 x + 1 . 5 y ∴ 45 x = 15 x + 15 y [ By multiplying 10 on both sides ] ∴ 30 x = 15 y ∴ x = 1 2 y Differentiating both sides w.r.t. t , we get ∴ d x d t = 1 2 d y d t = 1 2 3 4 m / s = 3 8 m / sec d x d t + d y d t = 3 8 + 3 4 = 9 8 m / sec Hence, the rate at which his shadow is shortening is 3 8 m / sec and the rate at which the tip of the shadow is moving is 9 8 m / sec .

Show that y = log 1 + x - 2 x 2 + x , x > - 1 , is an increasing function of x throughout its domain.

For the given function y = log 1 + x - 2 x 2 + x On differentiating both sides w.r.t. x we get d y d x = d d x log ( 1 + x ) - 2 x 2 + x d y d x = d d x log 1 + x - d d x 2 x 2 + x d y d x = 1 1 + x - 2 + x d d x 2 x - 2 x d d x 2 + x 2 + x 2 ∵ d d x a b = b d d x a - a d d x b b 2 d y d x = 1 1 + x - 2 + x 2 - 2 x 2 + x 2 d y d x = 1 1 + x - 4 + 2 x - 2 x 2 + x 2 d y d x = 1 1 + x - 2 2 + x 2 d y d x = 2 + x 2 - 4 1 + x 1 + x 2 + x 2 d y d x = x 2 + 4 x + 4 - 4 - 4 x 1 + x 2 + x 2 d y d x = x 2 1 + x 2 + x 2 d y d x = x 2 2 + x 2 1 1 + x For y to be increasing d y d x > 0 , Then x 2 2 + x 2 1 1 + x > 0 If 1 + x > 0 ⇒ x > - 1 ∵ x 2 + x 2 is always + v e . Then, f ' x > 0 Hence, The function is increasing function of x throughout its domain.