NEET Physics Solutions For Class 11 Chapter 10 Mathematical Tools – Double Differentiation

NEET Physics Class 11 Chapter 10 Mathematical Tools – Double Differentiation

If f is a differentiable function, then its derivative f’ is also a function, so f’ may have a derivative of its own, denoted by (f’)’ = f’’.

This new function f’’ is called the second derivative of f because it is the derivative of the derivative of f. Using Leibniz notation, we write the second derivative of y = f (x) as

⇒ \(\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d^2 y}{d x^2}\)

Another notation is f ’’ (x) = D₂ f (x) = D²f(x)

Interpretation Of Double Derivative

We can interpret f ’’ (x) as the slope of the curve y = f ’(x) at the point (x, f ’(x)). In other words, it is the rate of change of the slope of the original curve y = f (x).

• In general, we can interpret a second derivative as a rate of change of a rate of change. The most familiar example of this is acceleration, which we define as follows.
• If s = s(t) is the position function of an object that moves in a straight line, we know that its first derivative represents the velocity v(t) of the object as a function of time:

v(t) = s’(t) = \(\frac{\mathrm{ds}}{\mathrm{dt}}\)

• The instantaneous rate of change of velocity concerning time is called the acceleration a(t) of the object.
• Thus, the acceleration function is the derivative of the velocity function and is therefore the second derivative of the position function:

a (t) = v’(t) = s’’(t)

or in Leibniz notation, a = \(\frac{d v}{d t}=\frac{d^2 s}{d t^2}\)

Question 1. If f (x) = x cos x, find f ’’ (x).

Answer:

Using the Product Rule, we have

f’ (x) = \(x \frac{d}{d x}(\cos x)+\cos x \frac{d}{d x}(x)\)

= – x sin x + cos x

To find f’’ (x) we differentiate f’(x):

f’’(x) = \(\frac{d}{d x}\) (–x sin x + cos x)

⇒ \(-x \frac{d}{d x}(\sin x)+\sin x \frac{d}{d x}(-x)+\frac{d}{d x}(\cos x)\)

= – x cos x – sin x – sin x

= – x cos x – 2 sin x

Question 2. The position of a particle is given by the equation s = f (t) = t³ – 6t² + 9t. where t is measured in seconds and s in meters. Find the acceleration at time t. What is the acceleration after 4s?

Answer:

The velocity function is the derivative of the position function :

s = f (t) = t³ – 6t² + 9t

⇒ v(t) = \(\frac{d s}{d t}=3 t^2-12 t+9\)

The acceleration is the derivative of the velocity function :

⇒ \(a(t)=\frac{d^2 s}{d t^2}=\frac{d v}{d t}=6 t-12\)

⇒ \(a(4)=6(4)-12=12 \mathrm{~m} / \mathrm{s}^2\)

Application Of Derivatives

Differentiation As A Rate Of Change

⇒ \(\frac{d y}{d x}\) is rate of change of ‘y’ with respect to ‘x’ :

For examples:

1. v =\(\frac{d x}{d t}\) This means velocity ‘v’ is the rate of change of displacement ‘x’ concerning time ‘t’
2. a = \(\frac{d v}{d t}\) This means acceleration ‘a’ is the rate of change of velocity ‘v’ concerning time ‘t’.
3. F = \(\frac{d p}{d t}\) This means force ‘F’ is the rate of change of momentum ‘p’ concerning time ‘t’.
4. τ = \(\frac{d L}{d t}\) This means torque ‘τ ’ is the rate of change of angular momentum ‘L’ concerning time ‘t’
5. Power = \(\frac{d W}{d t}\) This means power ‘P’ is the rate of change of work ‘W’ concerning time ‘t’
6. Ι = \(\frac{d q}{d t}\) This means current ‘Ι’ is the rate of flow of charge ‘q’ concerning time ‘t’

Question 1. The area A of a circle is related to its diameter by the equation A = \(\frac{\pi}{4} D^2\). How fast is the area changing concerning the diameter when the diameter is 10 m?

Answer:

The (instantaneous) rate of change of the area concerning the diameter is

⇒ \(\frac{\mathrm{dA}}{\mathrm{dD}}=\frac{\pi}{4} 2 \mathrm{D}=\frac{\pi \mathrm{D}}{2}\)

When D = 10 m, the area is changing at a rate (π/2) 10 = 5π m²/m. This means that a small change

ΔD m in the diameter would result in a change of about 5π ΔD m² in the area of the circle.

Question 2. Experimental and theoretical investigations revealed that the distance a body released from rest falls in time t is proportional to the square of the amount of time it has fallen. We express this by saying that

s = \(\frac{1}{2} \mathrm{gt}^2\)

where s is distance and g is the acceleration due to Earth’s gravity. This equation holds in a vacuum, where there is no air resistance, but it closely models the fall of dense, heavy objects in air. The figure shows the free fall of a heavy ball bearing released from rest at time t = 0 sec.

1. How many meters does the ball fall in the first 2 sec?
2. What is its velocity, speed, and acceleration then?

Answer:

1. The free–fall equation is s = 4.9 t².

During the first 2 sec. the ball falls

s= 4.9(2)² = 19.6 m,

2. At any time t, velocity is derivative of displacement:

v(t) = s’(t) = \(\frac{d}{d t}\)(4.9t²) = 9.8 t.

At t = 2, the velocity is v= 19.6 m/sec in the downward (increasing s) direction. The speed at t = 2 is

speed = |v(2)| = 19.6 m/sec.

a = \(\frac{\mathrm{d}^2 \mathrm{~s}}{\mathrm{dt}^2}\)

= 9.8m/s²

Maxima And Minima:

Suppose a quantity y depends on another quantity x in a manner shown in the figure. It becomes maximum at x₁ and minimum at x₂. At these points, the tangent to the curve is parallel to the x−axis, and hence its slope is tan θ = 0. Thus, at a maximum or a minimum,

slope = \(\frac{d y}{d x}=0\)

Application Of Derivatives Maxima:

Just before the maximum, the slope is positive, at the maximum, it is zero and just after the maximum, it is negative. Thus, \(\frac{d y}{d x}\) decreases at a maximum and hence the rate of change of \(\frac{d y}{d x}\) is negative at a maximum i.e. \(\frac{d}{d x}\left(\frac{d y}{d x}\right)\) < 0 at maximum.

The quantity \(\frac{d}{d x}\left(\frac{d y}{d x}\right)\) is the rate of change of the slope. It is written as \(\frac{d^2 y}{d x^2}\)

Conditions for maxima are: \(\frac{d y}{d x}=0\)

⇒ \(\frac{d^2 y}{d x^2}<0\)

Application Of Derivatives Minima:

Similarly, at a minimum the slope changes from negative to positive. Hence with the increases of x., the slope is increasing which means the rate of change of slope concerning x is positive

hence \(\frac{d}{d x}\left(\frac{d y}{d x}\right)>0\)

Conditions for minima are: \(\frac{d y}{d x}=0\)

⇒ \(\frac{d^2 y}{d x^2}>0\)

Quite often it is known from the physical situation whether the quantity is a maximum or a minimum.

The test on \(\frac{d^2 y}{d x^2}\) may then be omitted.

Question 3. Find minimum value of y = 1 + x² – 2x

⇒ \(\frac{d y}{d x}=2 x-2\) for minima \(\frac{d y}{d x}=0\)

2x-2 = 0

x = 1

⇒ \(\frac{d^2 y}{d x^2}=2\)

⇒ \(\frac{d^2 y}{d x^2}>0\)

at x =1, there are minima

for a minimum value of y

y_minima = 1 + 1 – 2 = 0