Download presentation

Presentation is loading. Please wait.

Published byJavier Hollern Modified over 2 years ago

1
Chapter : Differentiation

2
Questions 1. Find the value of Answers (1) 2. If f(x) = 2[2x + 5]⁴, find f ‘ (-2) Answers(2) 3. If y = 2s⁶ and x = 2s – 1, find dy/dx in terms of s Answers(3) 4. Given that y = 15x [3-x], calculate a) the value of x when y is maximum b) the maximum value of y Answers(4)

3
Questions 5. Find the coordinate of the point on the curve y = 32x² +1 at which the tangent is parallel with the x-axis Answers(5) 6. Find the coordinates of the points on the curve y = [3x – 1]³ where the gradient of the tangent to the curve is 81 Answers (6) 7. Given that y = x + 3x, use differentiation to find the small change in y when x increases 2 to 2.01. Answers (7)

4
Questions 8. Find the equation of the normal to the curve y = 2x² - 3x+2 at the point where its x coordinate is 3. Answers(8) 9. Find the coordinates of the turning point of y = x³ + 15 x² - 24, 2 and state whether it is a maximum or minimum point. Answers (9)

5
Questions 10. The radius of a ball is increasing at a rate of 0.8cm¯¹. Find the rate of change of its surface area when its radius is 8cm. Answer (10)

6
Question 1 Back To Questions

7
Question 2 If f(x) = 2[2x + 5]⁴, find f ’ (-2) If f(x) =2[2x + 5]⁴ f ’ (x) = 8[2x + 5]³. 2 = 16 [2x + 5]³ f ’ (-2) = 16[2(-2) + 5]³ = 16[1]³ = 16 answer (2) Back To Questions

8
Question 3 If y = 2s⁶ and x = 2s -1, find dy/dx in terms of s. From x = 2s – 1 2s = x + 1 S =x+1 2 Therefore y = 2s⁶ y = 2 dy/dx = 12 1 X 2 so we replace with S, = 6 = 6s⁵ Back to Questions

9
Question 4 Given that y=15x [3 – x] calculate a) The value of x when y is a maximum From y = 15x [3-x] = 45x – 15x² dy/dx = 45-30x = 0 45 = 30x x = 3/2 answer (a) b) The maximum value of y Substitute x = 3/2 into the equation y = 15x [3 –x] = 15(3/2) (3-3/2) = 45/2 [3/2] = 33.75 answer (b) Back to Questions

10
Question 5 Find the coordinates of the point on the curve y = 32x² + 1/x at which the tangent is parallel to the x- axis. y = 32x²+ 1/x = 32x²+ x¯¹ dy/dx = 64x –x¯² = 64x - 1/x² = 0 64x = 1/x² 64x³=1 x³=1/64 x = ∛(1/64) = 1/4 when x = ¼ y = 32x²+ 1/x substitute x = ¼ into x in the equation y = 32( 1/4 )²+ 1/(1/4) y = 32 ( 1/16) + 4 y = 32/16 + 4 = 96/16 = 6 therefore, the coordinates of the point is (x, y ) = (1/4, 6) answer (5) Back to Questions

11
Question 6 Find the coordinates of the points on the curve y = [3x-1]³ where the gradient of the tangent to the curve is 81 y = [3x-1]³ dy/dx = 3[3x-1] ². 3 = 9[3x-1]² given dy/dx = 81 9[3x-1]²=81 [3x-1]=9 3x-1=√9 3x-1=3 3x=4 x=4/3 with x = 4/3 y = [3(4/3)-1]³ = 3³ = 27 so, the coordinates of the point is [4/3, 27] answer (6) Back to Questions

12
Question 7 Given that y = x+3x, use differentiation to find the small change in y when x increases from 2 to 2.01. δy/δx = dy/dx δy = dy/dx × δx y = x² + 3x dy/dx = 2x + 3 δx = 2.01 – 2 = 0.01 substitute the information into the formula δy=[2x+3]×0.01 = 2[2]+3×0.01 (original x is 2) = 7×0.01 = 0.07 answer (7) Back to Questions

13
Question 8 Find the equation of the normal to the curve y = 2x² - 3x + 2 at the point where its x coordinate is 3. y = 2x² - 3x + 2 dy/dx = 4x + 3 substitute x = 3 into y y = 2[3]² + 3[3] + 2 = 11 coordinate of the point is (3, 11) substitute x = 3 into dy/dx dy/dx = 4[3]- 3 = 9 So, the gradient of the curve at the point is 9 Thus, the gradient of the normal to that is -1/9 General form of equation y - y1 = m[x-x1] y – 11 = -1/9 [x-3] gradient with negative regression y= -1/9 x + 34/3 y- intersect answer (8) Back to Questions

14
Question 9 Find the coordinates of the turning point of y = x³ +15/2 x² - 24, and state whether it is a maximum or minimum point. From y = x³ +15/2 x² - 24 dy/dx = 3x²+15x At turning point, dy/dx = 0 3x² + 15x = 0 x[3x+15] = 0 so, x=0 or 3x+15 = 0 3x = -15 x = -5 Coordinates of turning point when y = [0]³ + 15/2 [0]² - 24 = -24 Coordinates of the point (0, -24) Coordinates of turning point when x = -5 y = [-5]³ + 15/2 [-5]² - 24 = -24 y = -125 + 375/2 – 24 = 77/2 Coordinates of the point (-5, 77/2) Substitute the coordinates to determine the min and max points At point (0,-24) dy/dx = 3x² + 15x dy²/dx² = 6x + 15 = 6[0] + 15 = 15>0 (minimum point) At point (-5, 77/2) dy/dx = 3x² + 15x dy²/dx² = 6x + 15 = 6[-5] + 15 = -15 < 0 (maximum point) Back to Questions

15
Question 10 The radius of a ball is increasing at a rate of 0.8cms¯¹. Find the rate of change of its surface area when its radius is 8cm. 0.8cms¯¹ As we know the area of sphere = 4πr² Rate of change its surface area (dA/dt = dA/dr × dr/dt) A = 4πr² dA/dr = 8πr dr/dt = 0.8cms¯¹ r = 8 dA/dt = 8πr × 0.8 = 8π[8] × 0.8 = 64π × 0.8 = 51.2πcm²s¯¹ Back to Questions

Similar presentations

OK

Example 1 Find the derivative of Function is written in terms of f(x), answer should be written in terms of f ′ (x) Continue

Example 1 Find the derivative of Function is written in terms of f(x), answer should be written in terms of f ′ (x) Continue

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on credit policy sample Ppt on renewable sources of energy in india Ppt on bluetooth communication standard Numeric display ppt on tv Ppt on field study 1 Atoms for kids ppt on batteries Ppt on 4-stroke petrol engine Ppt on unity in diversity and organic farming Ppt on tamper resistant seals Ppt on area of plane figures geometry