BRIDGE COURSE

UNDERSTANDING DIFFERENTIATION

𝒅𝒚
Geometrical significance of
𝒅𝒙
RULES OF DIFFERENTIATION
𝑑𝑦
1. Constant rule: If y=a (where a is a constant) , 𝑑𝑥 =0
𝑑𝑦
2. Power rule: If y=axn ,𝑑𝑥 =n×axn−1
𝑑𝑦 𝑑𝑢 𝑑𝑣
3. Sum or difference rule: If y = u ± v, then𝑑𝑥 = 𝑑𝑥 ±𝑑𝑥

4. Chain Rule: If y is a function of u and u is a function of x, then

𝑑𝑦 𝑑𝑦 𝑑𝑢
= 𝑑𝑢 .𝑑𝑥
𝑑𝑥

5. Product rule: If y=uv , then dy/dx = u (dv/dx) + v (du/dx)

6.Quotient rule: If y=u/v, then
𝑑𝑢 𝑑𝑣
𝑑𝑦 𝑣 𝑑𝑥 −𝑢 𝑑𝑥
=
𝑑𝑥 𝑣.𝑣

INTEGRATION

RULES OF INTEGRATION

∫(𝑢 ± 𝑣)𝑑𝑥 = ∫ 𝑢𝑑𝑥±∫ 𝑣𝑑𝑥

CONDITIONS OF MAXIMA AND MINIMA OF A FUNCTION:

𝑑𝑦
For maxima = 0 andd2y/dx2< 0
𝑑𝑥
𝑑𝑦
For maxima 𝑑𝑥 = 0 andd2y/dx2> 0
1 MARK QUESTIONS
1. Given that the position of a particle varies as cube of time. What type of motion does the particle
have?
2. Is instantaneous speed equal to magnitude of instantaneous velocity? Give reason for your answer.
3. Distinguish between average velocity and instantaneous velocity.
4. Velocity of an object does not change from instant to instant. What can you infer about the average
velocity of the object?
1
5. Given s = ut + 2 at2. Obtain the expression for velocity as a function of time.

2 AND 3 MARK QUESTIONS

1. Given x = t3 + 4t2-5t + 15. Find the average acceleration between time t = 1s and t = 3s.
2. A particle moves along a straight line such that its displacement at any time t is given by:
S = t3 – 6t2 +3t +4. Find the velocity of the particle when its acceleration is zero.
3. A body moving along a straight line according to the relation v = 4t + 2. If x = 3 m at t = 2s, find
its position and acceleration at t = 3s.
 4. A travelling wave in a string is described by the equation y = A sin (ωt – kx). Find the maximum
particle velocity.
5. The position of a particle varies as y = at3 – bt2. At what time will the acceleration of the particle
be zero?
6. Power of an engine is 250 W. What is the work done by the engine in the time interval of t =2s to t =
5 s.
7. The growth of bacterial population is represented by P (t) = 800t + 100 t2. Find the instantaneous rate
of growth.
8. Find the dimensions of a rectangle with perimeter 1000 metres so that the area of the rectangle is a
maximum.
5 MARK QUESTIONS
1. A car is moving on a straight road along x- axis in such a way that its coordinate ‘x’ varies with time
‘t’ according to the expression
x (t) = 3 + 5t2+ t
(a) Find the initial velocity
(b) Find the initial acceleration
(c) Is the motion of the car uniform? If yes, why and if no, why not?
(d) Is the motion of the car uniformly accelerated? If yes, find the acceleration of the car.
CASE STUDY BASED QUESTIONS

1. The equation of a straight line passing through a given point (x0,y0) having finite slope m is given
Δ𝑦
by y-y0 = m (x-x0). For a straight line graph, the slope is given by Δ𝑥 whereas for a non-linear graph,
the slope of the tangent to the curve y =f(x) at the point(x0,y0) is given by dy/dx=f(x0). Also the area
𝑥2
under the curve from x = x1 to x = x2 is given by∫𝑥1 𝑦𝑑𝑥 .Based on the above information answer the
following questions:

(i) Slope of the tangent to the curve y =3x4- 4x at x = 4 is:

(a) 761 (b) 764 (c) 167 (d) 716

(ii) Slope of the tangent to the curve y = x3-x+1 at the point whose x coordinate is 2

(a) 11 (b) 9 (c) 3 (d) 12

(iii) Area enclosed by the curve given by y = 3x2 -1 and the x axis from x = 0 to x = 4 is
(a) 64 (b) 65 (c) 63 (d) 24
(iv) The position of a body moving along x-axis at time t is given by x = (t2 - 4t + 6) m. The distance
travelled by the body in time interval t= 0 to t= 3s
(a) 5m
(b) 7m
(c) 4m
(d) 3m
(v) If v is the velocity of a body moving along x-axis then acceleration of the body is
𝑑𝑣
(a)
𝑑𝑥
𝑑𝑣
(b) 𝑣 𝑑𝑥
 𝑑𝑢
(c) x
𝑑𝑥
𝑑𝑥
(d) 𝑣
𝑑𝑣

2. Differential calculus is the study of the definition, properties, and applications of the derivative of a
function. The process of finding the derivative is called differentiation. Given a function and a point in the
domain, the derivative at that point is a way of encoding the small-scale behaviour of the function near that
point. By finding the derivative of a function at every point in its domain, it is possible to produce a new
function, called the derivative function or just the derivative of the original function. Based on the above
answer the following Questions,

(i) The position of a particle moving on a straight track is given by: x=6 + 9t + 3t2 metre. Its acceleration at
t = 2s is

(a) 6 m/s2 (b) 9 m/s2 (c) 54 m/s2 (d) 27 m/s2

(ii) The velocity of a particle is v = 5+ (a1 + a2t ) where a1 and a2 are constants and t is time. The
acceleration of the particle is:

(a) a1 (b) a2 (c) zero (d) a1a2

(iii) The displacement-time graph of a moving particle is shown ahead. The instantaneous velocity of
the particle is negative at the point

(a) E (b) F (c) C (d) D

(iv) The position x of a particle varies with time t as x = at2 – bt3. The acceleration of the particle will
be zero at time t equal to:

(a) a/b (b) 2a/3b (c) a/3b (d) Zero

(v) Displacement time graph for motion of a particle is as shown in the figure . What can you say about the
instantaneous velocities at points A, B and C ?

(a) positive, negative, positive

(b) positive , positive , zero
(c) negative, zero, positive
(d) positive, zero , negative
HOTS
When a particle is projected at an angle near the earth surface, it moves simultaneously in horizontal and
vertical directions. Motion of such a particle is called a projectile motion. Since there is no force acting in the
horizontal direction, the velocity remains constant in the horizontal direction. In the vertical direction the
gravitational pull of earth produces deceleration. Such motion is helpful in determining various parameters
such as maximum height, range etc in firing of shells from large guns.The effect of air resistance is neglected
while deriving the equation of motion for a projectile and acceleration due to gravity is taken as constant. The
path (trajectory) of the projectile is parabolic in shape and is given by the equation:
y= x tanθ -gx2/ (2u2 cos2θ)

θ
where x and y are the coordinates in X and Y planes, θ is the angle of projection measured with respect to the
horizontal direction, u is initial velocity with which the particle is projected and g is the acceleration due to
gravity.
(i) Using the above equation, the value of dy/dx will be:
(a) dy/dx= tanθ -gx/( u2 cos2 θ)
(b) dy/dx= x sec2θ -gx2/( u2 2cosθ sinθ )
(c) dy/dx= tanθ -gx/( 2u cos2 θ)
(d) dy/dx= 2x sec2θ -gx/( u2 2cosθ sinθ )
(ii) The value of slope at x=0 and t=0 i.e. at the time of initial position will be:
(a) dy/dx= cos θ
(b) dy/dx= sin θ
(c) dy/dx= tan θ
(d) dy/dx=0
(iii) Using the product of (dy/dx)t=0 (dy/dx)t=t = -1 & taking x=(u cos θ) t, time t in terms of u, θ and g
can be given by:
(a) t= (u2/g) sin2θ
(b) t= (g sinθ) / u
(c) t= (ucos θ) / g
(d) t= (u/g) sinθ
(iv) The value of dy/dx at the highest point when the tangent drawn to the curve is parallel to the x axis
is 0. Using the value of dy/dx at the highest point of the projectile, the value of x at which maximum
height is reached can be find out. This value of x is :
(a) x= u2sinθ / 2g
(b) x= u2 sin 2θ / 2g
(c) x= u2 sinθ /g – sin2θ
(d) x= u t +1/2 gt2
 ASSERTION-REASON TYPE QUESTION
In the following questions 1 and 2, a statement of assertion followed by a statement of reason is
given, choose the correct answer out of the following choices

(A) Both the Assertion and the Reason are correct and the Reason is the correct explanation of the
Assertion.
(B) The Assertion and the Reason are correct but the Reason is not the correct explanation of the
Assertion.

(C) Assertion is correct but the Reason is wrong statement.

(D) Both assertion and reason are wrong statements.

1. Statement 1. The linear momentum of body increases by 50%. The corresponding increase in
kinetic energy is 100%.
Statement 2. Kinetic energy , K = p2/2m, then it may be written as ∆𝐾/K = 2∆P/P.
2. Statement 1. The average and instantaneous velocities have the same value in a uniform motion.
Statement 2. In uniform motion, the velocity of an object increases uniformly.