Q1: A circular disc of radius 3 cm is being heated. Due to expansion, its radius increases at a rate of 0.05 cm per second. Find the rate at which its area is increasing if the radius is 3.2 cm.
Ans: Let us assume that “r” be the radius of the given disc and “A” be the area, then the area is given as:
A = πr^{2}
By using the chain rule,
Then dA/dt = 2πr(dr/dt)
Thus, the approximate rate of increase of radius = dr = (dr/dt) ∆t = 0.05 cm per second
Hence, the approximate rate of increase in area is:
dA = (dA/dt)(∆t) = 2πr[(dr/dt) ∆t ]
= 2π (3.2) (0.05)
= 0.320π cm
per second.
Therefore, when r= 3.2 cm, then the area is increasing at a rate of 0.320π cm^{2}/second.
Q2: What is the equation of the normal to the curve y = sin x at (0, 0)?
Ans: Given that, y = sin x
Hence, dy/dx = cos x
Thus, the slope of the normal = (1/cos x)_{x =0} = 1
Therefore, the equation of the normal is y0 = 1(x0) or x + y = 0
Q3: Show that the function f(x) = tan x – 4x is strictly decreasing on [π/3, π/3]
Ans: Given that, f(x) = tan x – 4x
Then, the differentiation of the function is given by:
f’(x)= sec^{2}x – 4
When π/3 <x π/3, 1<sec x <2
Then, 1<sec^{2}x <4
Hence, it becomes 3 < (sec^{2}x4)<0
Hence, for π/3 <x π/3, f’(x)<0
Therefore, the function “f” is strictly decreasing on [π/3, π/3]
Q4: Determine all the points of local maxima and local minima of the following function: f(x) = (¾)x^{4} – 8x^{3} – (45/2)x^{2} + 105
Ans: Given function: f(x) = (¾)x^{4} – 8x^{3} – (45/2)x^{2} + 105
Thus, differentiate the function with respect to x, we get
f ′ (x) = –3x^{3 }– 24x^{2} – 45x
Now take, 3x as common:
= – 3x (x^{2} + 8x + 15)
Factorise the expression inside the bracket, then we have:
= – 3x (x +5)(x+3)
f ′ (x) = 0
⇒ x = –5, x = –3, x = 0
Now, again differentiate the function:
f ″(x) = –9x^{2} – 48x – 45
Take 3 outside,
= –3 (3x^{2} + 16x + 15)
Now, substitue the value of x in the second derivative function.
f ″(0) = – 45 < 0. Hence, x = 0 is point of local maxima
f ″(–3) = 18 > 0. Hence, x = –3 is point of local minima
f ″(–5) = –30 < 0. Hence, x = –5 is point of local maxima.
Q5: A stone is dropped into a quiet lake and waves move in the form of circles at a speed of 4 cm/sec. At the instant, when the radius of the circular wave is 10 cm, how fast is the enclosed area increasing?
Ans: We know that the area of a circle with radius “r” is given by A = πr^{2}.
Hence, the rate of change of area “A’ with respect to the time “t” is given by:
dA/dt = (d/dt) πr^{2}
By using the chain rule, we get:
(d/dr)(πr^{2}). (dr/dt) = 2πr.(dr/dt)
It is given that, dr/dt = 4 cm/sec
Therefore, when r = 10 cm,
dA/dt = 2π. (10). (4)
dA.dt = 80 π
Hence, when r = 10 cm, the enclosing area is increasing at a rate of 80π cm^{2}/sec.
Q6: For the given curve: y = 5x – 2x^{3}, when x increases at the rate of 2 units/sec, then how fast is the slope of the curve change when x = 3?
Ans: Given that, y = 5x – 2x^{3}
Then, the slope of the curve, dy/dx = 56x^{2}
⇒ d/dt [dy/dx]= 12x. dx/dt
= 12(3)(2)
= 72 units per second
Hence, the slope of the curve is decreasing at the rate of 72 units per second when x is increasing at the rate of 2 units per second.
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