# Differentiation and Its Applications Questions and Answers

This course contains many exam standard worked out questions on differentiation and its applications.

All the questions have well elaborated video answers with explanations.

Most of the questions have been derived from the past KCSE exams.

More questions are being added continuously.

Lessons (**20**) * SHARE*

- 1.
The equation of a curve is given by #y= x^3 – 4x^2 – 3x#
(a) Find the value of y when x = -1. (1 mark)
(b) Determine the stationary points of the curve. (5 marks) (c) Find the equation of the normal to the curve at x = 1

12m 44s - 2.
The gradient of the curve #y = 2x^3 – 9x^2 + px -1# at x =4 is 36.
(a) Find: (i) The value of p; (3 marks) (ii) The equation of the tangent to the curve at x = 0.5.
(b) Find the co-ordinates of the turning points of the curve.

11m 8s - 3.
The gradient of the tangent to the curve # y = ax^3 + bx # at the point (1,1) is -5. Calculate the values of a and b.

3m 45s - 4.
The equation of a curve is # y = 2x^3 + 3x^2 #. (a) Find: (i) The x-intercept of the curve: (ii) The y-intercept of the curve. (1 mark) (b) (i) Determine the stationary points of the curve. (ii) For each point in (b) (i) above, determine whether it is a maximum or a minimum. (c) Sketch the curve.

8m 36s - 5.
The equation of a curve is given as #y = 2x^3 - (9)/(2)x^2 - 15x + 3# (a) Find: (i) The value of y when x = 2; (2 marks) (ii) The equation of the tangent to the curve at x = 2. (b) Determine the turning points of the curve.

12m 12s - 6.
The equation of a curve is given as #y = (1)/(3)x^3 – 4x + 5#. Determine: (a) The value of y when x = 3; (b) The gradient of the curve at x = 3; (c) The turning points of the curve and their nature.

6m 58s - 7.
Find the coordinates of the turning points of the curve whose #y=6+2x-4x^2#

2m 52s - 8.
a).i).Find the coordinates of the stationary points on the curve #y =x^3 – 3x +2#. ii). For each stationary point determine whether it is minimum or maximum. b).Sketch the graph of the function #y = x^3 -3x +2#.

7m 37s - 9.
The velocity V m/s, of a moving body at time t seconds is given by #V = 5t^2– 12t +7#.Calculate the acceleration when t = 2 seconds.

1m 26s - 10.
A stone is thrown vertically upwards from a point O. After t seconds, the stone is S metres from O. Given that #S= 29.4t-4.9t^2#,find the maximum height reached by the stone.

2m 42s - 11.
A curve is represented by the function #y =(1)/(3)x^3 + x^2 – 3x +2.# a). Find #(dy)/(dx)# b).Determine the values of y at the turning points of the curve. c).Sketch the curve #y = (1)/(3)x^3 + x^2 – 3x + 2#

12m 16s - 12.
A particle moves along a straight line such that its displacement s metres from a given point is #S=t^3 -5t^2 +3t +4# where t is time in seconds. Find: a).The displacement of the particle at t =5 seconds. b).The velocity of the particle when t =5 seconds. c) The values of t when the particle is momentarily at rest. d).The acceleration of the particle when t=2 seconds.

7m 26s - 13.
The sum of two numbers x and y is 40.Write down the expression, in terms of x for the sum of the squares of the two numbers. Hence determine the minimum value of #x^2 +y^2#

4m 49s - 14.
The distance s metres from a fixed point O, covered by a particle after t seconds is given by the equation #s= t^3 -6t^2 + 9t +5#. a).Calculate the gradient of the curve at t=0.5 seconds. b).Determine the values of s at the maximum and the minimum turning points of curves. c).Sketch the curve of #s= t^3 -6t^2 +9t+5#.

12m 37s - 15.
A rectangular box open at the top has a square base. The internal side of the base is x cm long and the total internal surface area of the box is #432 cm^2# a) Express in terms of x i).The internal height, h of the box. ii).The internal volume, V of the box. b).Find: i).The value of x for which the internal volume V box is maximum. ii).The maximum internal volume of the box.

9m 31s - 16.
The displacement,s metres,of a moving particle after t secs is given by #S =2t^3 -5t^2 + 4t +2# Determine : a)The velocity of the particle when t=3 seconds b)The value of t when the particle is momentarily at rest. c).The displacement when the particle is momentarily at rest. d).The acceleration of the particle when t=3 seconds.

10m 4s - 17.
The acceleration of a body moving along a straight line is #(4-t) ms^2# and its velocity is v m/s after t seconds. a).i) If the initial velocity of the body is 3m/s, express the velocity v in terms of t. ii).Find the velocity of the body after 2 seconds. b) Calculate: i).The time taken to attain maximum velocity. ii).The distance covered by the body to attain the maximum velocity.

9m 46s - 18.
The displacement of, s metres, of a moving particle from the point O, after t seconds is given by, #s = t^3 – 5t^2 + 3t + 10# a) Find s when t= 2 b) Determine: i)The velocity of the particle when t=5. ii)The value of t when the particle is momentarily at rest. c).Find the time, when the velocity of the particle is maximum.

7m 26s - 19.
Given the curve #y= 2x^3 + (1)/(2) x^2 – 4x + 1#. , find the: i).Gradient of the curve at #(1,-frac(1)(2))# ii).Equation of the tangent to the curve at #(1,-frac(1)(2) )#

3m 53s - 20.
The displacement s metres of a particle moving along a straight line after t seconds is given by #s = 3t + (3)/(2)t^2 – 2t^3# a).Find the initial acceleration. b).Calculate: i).The time when the particle was momentarily at rest. ii). Its displacement by the time it comes to rest momentarily. c).Calculate the maximum speed attained.

8m 57s

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