5.2 & 5.4 Tangents, normals and the second derivative

Tangents, Normals, and the next level

In this unit we look at find the equations of tangents and normals to curves, increasing and decreasing functions and the second derivative. All this builds on the fundamental notion that calculus tells us about the rate of of change of a function.

Key Concepts

In this unit you should learn to…

find the equations of straight lines
find the equations of tangents and normals
recognise when functions are increasing and decreasing
find the rate of change of the rate of change.

Essentials

Slides Gallery

Use these slides to review the material and key points covered in the videos.

1. Finding equations of straight lines

This video covers the technique required to find the equation of a straight line given its gradient and a point on the line. This is key for finding the equations of tangents and normals.

2. The equation of the tangent

Here we look at how you can use differentiation to find the gradient of a curve at a given point and then work out the equation of the tangent at that point.

3. The equation of the normal

This is the same technique as we used for the tangent. They key here is the relationship between gradients of perpendicular lines.

4. Increasing and decreasing functions

Definition and understanding of these key terms.

5. The second derivative

Here we look at the 'rate of change' of the 'rate of change'. A key idea that helps us identify local maximums and minimums.

Summary

This section of the page can be used for quick review. The flashcards help you go over key points and the quiz lets you practice answering questions on this subtopic.

Revision Cards

Review these condensed 'key point' flashcards to help you check and keep ideas fresh in your mind.

Test Yourself

Self Checking Quiz

Practice your understanding on these quiz questions. Check your answers when you are done and read the hints where you got stuck. If you find there are still some gaps in your understanding then go back to the videos and slides above.

The equation of the straight line that goes through the point (3, 4) with a gradient of 2 is given as y = ax + b. Find the values of a and b

a = , b =

You are given the gradient. So you know y = 2x + b, now substitute x = 4 and y = 4 (from the given point) to get the value of b.

......

Given the function $f left parenthesis x right parenthesis equals x cubed plus 3 x squared minus 2 x plus 4$ , what is the gradient when x = -2

Gradient =

$f apostrophe left parenthesis x right parenthesis equals 3 x squared plus 6 x minus 2 comma space f apostrophe left parenthesis negative 2 right parenthesis space equals space minus 2$ $f apostrophe left parenthesis x right parenthesis equals 3 x squared plus 6 x minus 2 comma space f apostrophe left parenthesis negative 2 right parenthesis space equals space minus 2$

Consider the function $f left parenthesis x right parenthesis equals x cubed plus 3 x squared minus 2 x plus 4$ , The tangent to this curve at x = -2 is given by y = ax +b. Find the values of a and b

a = , b =

$f apostrophe left parenthesis x right parenthesis equals 3 x squared plus 6 x minus 2 comma space s o space f apostrophe left parenthesis negative 2 right parenthesis space equals space minus 2$ Therefore the gradient, a = -2. So you know y = -2x + b, $f left parenthesis negative 2 right parenthesis space equals space 12$ which means the line goes through the point (-2, 12). Now substitute x = -2 and y = 12 (from the gievn point) to get the value of b.

Consider the function $f left parenthesis x right parenthesis equals 2 over 5 x squared plus 1 over x squared$ , The tangent to this curve at x = 2 is given by y = ax +b. Find the values of a and b.

a = , b =

$f left parenthesis x right parenthesis equals 0.4 x squared plus x to the power of negative 2 end exponent comma space s o space f apostrophe left parenthesis x right parenthesis equals 0.8 x squared minus 2 x to the power of negative 3 end exponent space a n d space f apostrophe left parenthesis 2 right parenthesis space equals space 1.35$ You can read this from the table function on your GDC. $f left parenthesis 2 right parenthesis equals 1.85$ , so the line goes through the point (2, 1.85). Use this to work out the value of c when y = 1.35x + c (since you know that the gradient is 1.35)

Consider the function $f left parenthesis x right parenthesis equals x cubed plus 3 x squared minus 2 x plus 4$ , The normal to this curve at x = -2 is given by y = ax +b. Find the values of a and b

a = , b =

$f(x)={ \quad x }^{ 3 }\quad +\quad 3{ x }^{ 2 }\quad -2x\quad +\quad 4\\ f(-2)=\quad -8\quad +\quad 12\quad +\quad 4\quad +\quad 4\quad =\quad 12\\ f'(x)={ \quad 3x }^{ 2 }\quad +\quad { 6x }\quad -2\\ f'(-2)\quad =\quad 12\quad -\quad 12\quad -\quad 2\quad =\quad -\quad 2\\ So\quad at\quad x\quad =\quad -\quad 2\quad the\quad gradient\quad of\quad the\quad function\quad is\quad -\quad 2\\ The\quad gradient\quad of\quad the\quad normal,\quad { m }_{ n }\times -2\quad =\quad -1\\ so\quad { m }_{ n }\quad =\quad 0.5\\ The\quad equation\quad of\quad the\quad normal\quad will\quad there\quad for\quad be\\ y\quad =\quad 0.5x\quad +\quad c\quad and\quad goes\quad through\quad the\quad point\quad (-2,\quad 12)\\ so,\quad 12\quad =\quad 0.5\times -2\quad +\quad c\\ therefore,\quad c\quad =\quad 13\\ Equation\quad of\quad Normal\quad is\quad y\quad =\quad 0.5x\quad +\quad 13$

......

Consider the function $f(x)\quad =\quad 3{ x }^{ 2 }+\frac { 5 }{ { x }^{ 2 } } $ at the point x = 2.

a) The function can be re written in this form $f(x)\quad =\quad 3{ x }^{ 2 }+\quad p{ x }^{ q }$

What are the values of p and q?

b) The differential is given as $f'(x)\quad =\quad r{ x }+s{ x }^{ t }$

What are the values of r, s and t?

c) What is the gradient of the function at x = 2? (give the exact answer)

The equation of the normal is given by y = ax + b.

d) Use your GDC or otherwise ind the values of a and b (correct to 3sf)

a) p = , q =

b) r = , s = , t =

c) Gradient at x = 2 is

d) a = , b =

$f(x)\quad =\quad 3{ x }^{ 2 }+\frac { 5 }{ { x }^{ 2 } } \\ f(x)\quad =\quad 3{ x }^{ 2 }+\quad 5{ x }^{ -2 }\\ f(2)\quad =\quad 12\quad +\quad 1.25\quad =\quad 13.25\\ f'(x)\quad =\quad 6{ x }-10{ x }^{ -3 }\\ f'(2)\quad =\quad 12\quad -\quad 1.25\quad =\quad 10.75\\ So\quad at\quad x\quad =\quad 2\quad the\quad gradient\quad of\quad the\quad function\quad is\quad 10.75\\ The\quad gradient\quad of\quad the\quad normal,\quad { m }_{ n }\times 10.75\quad =\quad -1\\ so\quad { m }_{ n }\quad =\quad -\frac { 1 }{ 10.75 } \\ The\quad equation\quad of\quad the\quad normal\quad will\quad there\quad for\quad be\\ y\quad =\quad -\frac { 1 }{ 10.75 } x\quad +\quad c\quad and\quad goes\quad through\quad the\quad point\quad (2,\quad 13.25)\\ so,\quad 13.25\quad =\quad -\frac { 1 }{ 10.75 } \times 2\quad +\quad c\\ therefore,\quad c\quad =\quad 13.44\\ Equation\quad of\quad Normal\quad is\quad y\quad =\quad -\frac { 1 }{ 10.75 } x\quad +\quad 13.44$

Consider the function, $f(x)\quad =\quad { x }^{ 3 }+\frac { 3 }{ 2 } { x }^{ 2 }-18x$shown in the diagram below

The derivative of the function is given as $f'(x)\quad =\quad a{ x }^{ 2 }+{ bx }+c$

a) What are the values of a, b and c?

b) What is the gradient of the function at x = -3

c) What is the gradient of the function at x = 2

At any given value of x on the function we can say it is either A - increasing or B - decreasing.

d) For the intervals and x values below enter either A or B as appropriate

a) a = , b = , c =

b) Gradient at x = -3 is

c) Gradient at x = -3 is

d) For the intervals and x values below enter either A (increasing) or B (decreasing) as appropriate

$x=-2$

$x=4$

$x=-4$

$x<-3$

\(-3

$x>2$

a) $f(x)\quad =\quad { x }^{ 3 }+\frac { 3 }{ 2 } { x }^{ 2 }-18x\\ f'(x)\quad =\quad 3{ x }^{ 2 }+3{ x }-18$

b) $f'(x)\quad =\quad 3{ x }^{ 2 }+3{ x }-18\\ f'(-3)\quad =\quad 27\quad -9\quad -18\quad =\quad 0$

c) $f'(x)\quad =\quad 3{ x }^{ 2 }+3{ x }-18\\ f'(2)\quad =\quad 12\quad +6\quad -18\quad =\quad 0$

The diagram below shows the function $f(x)\quad =\quad { x }^{ 4 }+8{ x }^{ 3 }+6{ x }^{ 2 }-40x$

The derivative of the function is given as $f'(x)\quad =\quad { ax }^{ 3 }+b{ x }^{ 2 }+c{ x }-d$

a) What are the values of a, b, c and d?

a = , b = , c = , d =

b) What is the gradient where x = -5, x = -2 and x = 1

Gradient at x = -5 is

Gradient at x = -2 is

Gradient at x = 1 is

c) What is the equation of the tangent at x = -2? y =

d) What is the equation of the normal at x = 1, x =

The function is said to be decreasing for the following intervals reading from left to right...

\(x what are the values of a, b and c?

a = , b = , c =

a) $f(x)\quad =\quad { x }^{ 4 }+8{ x }^{ 3 }+6{ x }^{ 2 }-40x\\ f'(x)\quad =\quad { 4x }^{ 3 }+24{ x }^{ 2 }+12{ x }-40$

b) $f'(x)\quad =\quad { 4x }^{ 3 }+24{ x }^{ 2 }+12{ x }-40\\ f'(-5)\quad =\quad -500\quad +\quad 600\quad -60\quad -40\quad =\quad 0\\ f'(-2)\quad =\quad -32\quad +\quad 96\quad -24\quad -40\quad =\quad 0\\ f'(1)\quad =\quad 4+24+12-40\quad =\quad 0$ or using the table function on the GDC

c) at x = -2, tangent has gradient 0 and as such the equation y = 56

d) Since the gradient at x = 1 is 0, the normal is a vertical line with equation x = 1

Consider the function $f(x)\quad =\quad { x }^{ 3 }+\frac { 3 }{ 2 } { x }^{ 2 }-18x$ that has two points at which the gradient is zero. (-3, 40.5) and ( 2, 22).

The second derivative is given as $f''(x)\quad =\quad ax\quad +\quad b$

a) What are the values of a and b?

b) Now consider the value of the second derivative at x = -3 and use this to tell you if this point is a local maximum (max) or a local minimum (min)

c) Now consider the value of the second derivative at x = -2 and use this to tell you if this point is a local maximum (max) or a local minimum (min)

a) a = , b =

b) at x = -3, f''(x) = so point is (enter max/min)

c) at x = 2, f''(x) = so point is (enter max/min)

a) $f(x)\quad =\quad { x }^{ 3 }+\frac { 3 }{ 2 } { x }^{ 2 }-18x\\ f'(x)\quad =\quad 3{ x }^{ 2 }+3{ x }-18\\ f''(x)\quad =\quad 6x\quad +\quad 3$

b) $f'(x)\quad =\quad 3{ x }^{ 2 }+3{ x }-18\\ f''(x)\quad =\quad 6x\quad +\quad 3\\ f''(-3)\quad =\quad -15$

Point is a maximum since the second derivative is negative, we unerstand that the gradient is changing from positive to negative as occurs at local maximum points.

c) $f''(x)\quad =\quad 6x\quad +\quad 3\\ f''(2)\quad =\quad 15$

Point is a minimum since the second derivative is positive, we unerstand that the gradient is changing from negative to positive as occurs at local minimum points.

Consider the function $f(x)\quad =\quad \frac { { x }^{ 4 } }{ 4 } -\frac { 4{ x }^{ 3 } }{ 3 } +\frac { { x }^{ 2 } }{ 2 } +6x$ which has gradient = 0 at x = -1, 2 and 3. By finding the second derivative determine the nature of each of those points.

f''(-1) = so point is (enter max/min)

f''(2) = so point is (enter max/min)

f''(3) = so point is (enter max/min)

$f(x)\quad =\quad \frac { { x }^{ 4 } }{ 4 } -\frac { 4{ x }^{ 3 } }{ 3 } +\frac { { x }^{ 2 } }{ 2 } +6x\\ f'(x)\quad =\quad { x }^{ 3 }-{ 4{ x }^{ 2 } }+x+6\\ f''(x)\quad =\quad 3{ x }^{ 2 }-8x+1\\ f''(-1)\quad =\quad 3+8+1=12\quad (positive\quad so\quad min)\\ f''(2)\quad =\quad 12-16+1=-3\quad (negative\quad so\quad max)\\ f''(3)\quad =\quad 27-24+1=4\quad (positive\quad so\quad min)$

Exam Style Questions

The following questions are based on IB exam style questions from past exams. You should print these off (from the document at the top) and try to do these questions under exam conditions. Then you can check your work with the video solution.

Question 1