If the radius of a sphere is measured as 7 cm with an error of 0 . 01 cm, then find the approximate error in calculating its volume.

Radius of a sphere = 7 cm Error = 0.01 cm Let r be the radius of the sphere and Δ r be the error in measuring the radius. Then, r = 7 cm and Δ r = 0.01 cm Now, the volume V of the sphere is given by, V = 4 3 π r 3 ∴ d V d r = 4 π r 2 ∴ d V = d V d r Δ r = 4 π r 2 Δ r = 4 π 7 2 0.02 cm 3 = 3.92 π cm 3 Hence, the approximate error in calculating the volume is 3.92 π cm 3 .

Find the equation of tangent of a curve y = x 2 - 2 x + 3 which is parallel to the line 2 x - y + 9 = 0 .

Given equation of the curve: y = x 2 - 2 x + 3 and equation of the line: 2 x - y + 9 = 0 . Slope of the given line is 2 Since tangent is parallel to the line, slope of tangent will be also 2 Now, slope of tangent using given curve is m = d y d x = 2 x - 2 ∴ d y d x = 2 ⇒ 2 x - 2 = 2 ⇒ x = 2 Putting x = 2 in equation of curve will give y = 3 Therefore, the line is passing through ( 2 , 3 ) , So, the equation of the tangent is given by, y - 3 = m ( x - 2 ) ⇒ y - 3 = 2 ( x - 2 ) ⇒ 2 x - y = 1

Prove that the value of function x 1 + x tan x is maximum at x = cos x .

Let f x = x 1 + x tan x For the value to be maximum d f x d x = 0 So d d x x 1 + x tan x = 0 Applying quotient rule of differentiation, we get 1 + x tan x d x d x - x d d x 1 + x tan x 1 + x tan x 2 = 0 ⇒ 1 + x tan x × 1 - x d 1 d x + d x tan x d x 1 + x tan x 2 = 0 ⇒ 1 + x tan x - x 0 + x d tan x d x + tan x d x d x = 0 [Applying product rule of differentiation] ⇒ 1 + x tan x - x x sec 2 x + tan x 1 = 0 ⇒ 1 + x tan x - x 2 sec 2 x - x tan x = 0 ⇒ 1 - x 2 sec 2 x = 0 ⇒ 1 = x 2 sec 2 x ⇒ 1 sec 2 x = x 2 ⇒ cos 2 x = x 2 ⇒ cos x = ± x Therefore, the value of function x 1 + x tan x is maximum at x = cos x .

Find the intervals in which the function f given by f x = sin x + cos x ; 0 ≤ x ≤ 2 π is strictly decreasing.

For the given function f x = sin x + cos x On differentiating both sides w.r.t. x , we get f ' x = d d x sin x + cos x ⇒ f ' x = d d x sin x + d d x cos x ⇒ f ' x = cos x - sin x ∵ d d x sin x = cos x , d d x cos x = - sin x For the critical points of the function over the interval 0 ≤ x ≤ 2 π is given by f ' x = 0 ⇒ cos x - sin x = 0 ⇒ cos x = sin x ⇒ x = π 4 , 5 π 4 as 0 ≤ x ≤ 2 π The points x = π 4 and 5 π 4 divides the interval 0 , 2 π into three disjoint intervals. i.e., 0 , π 4 , π 4 , 5 π 4 and 5 π 4 , 2 π If 0 < x < π 4 , f ' x = cos x - sin x > 0 ∵ cos x > sin x ⇒ f ' x > 0 ⇒ f x is strictly increasing. If π 4 < x < 5 π 4 , f ' x = cos x - sin x < 0 ∵ cos x < sin x ⇒ f x is strictly decreasing. If 5 π 4 < x < 2 π , f ' x = cos x - sin x > 0 ∵ cos x > sin x ⇒ f x is again strictly increasing. Hence, the given function f x = sin x + cos x ; 0 ≤ x ≤ 2 π is strictly decreasing for x ∈ π 4 , 5 π 4 .

Find the intervals in which the function f given by f x = sin x + cos x ; 0 ≤ x ≤ 2 π is strictly increasing.

For the given function f x = sin x + cos x On differentiating both sides w.r.t. x , we get f ' x = d d x sin x + cos x ⇒ f ' x = d d x sin x + d d x cos x ⇒ f ' x = cos x - sin x ∵ d d x sin x = cos x , d d x cos x = - sin x For the critical points of the function over the interval 0 ≤ x ≤ 2 π is given by f ' x = 0 ⇒ cos x - sin x = 0 ⇒ cos x = sin x ⇒ x = π 4 , 5 π 4 as 0 ≤ x ≤ 2 π The points x = π 4 and 5 π 4 divides the interval 0 , 2 π into three disjoint intervals. i.e., 0 , π 4 , π 4 , 5 π 4 and 5 π 4 , 2 π If 0 < x < π 4 , f ' x = cos x - sin x > 0 ∵ cos x > sin x ⇒ f ' x > 0 ⇒ f x is strictly increasing. If π 4 < x < 5 π 4 , f ' x = cos x - sin x < 0 ∵ cos x < sin x ⇒ f x is strictly decreasing. If 5 π 4 < x < 2 π , f ' x = cos x - sin x > 0 ∵ cos x > sin x ⇒ f x is again strictly increasing. Hence, the given function f x = sin x + cos x ; 0 ≤ x ≤ 2 π is strictly increasing for x ∈ 0 , π 4 and 5 π 4 , 2 π .