What is the difference between a stationary point and a turning point?

At a stationary point the derivative is zero, so the tangent there is horizontal. A turning point needs more: the derivative must also change sign , so the curve really does turn around. Every turning point is therefore a stationary point, but not every stationary point is a turning point.

The normal to a curve at a point is the straight line through that point which is perpendicular to the tangent there. Every point on a curve has both a tangent and a normal, and they cross at right angles at that point.

IGCSEFurther MathsEdexcelFlashcardsCalculusApplications of Differentiation

Applications of Differentiation (Edexcel IGCSE Further Pure Maths): Flashcards

Exam code: 4PM1

1/29

0Still learning

Know0

FrontFinding Gradients
How do you find the gradient of a curve at a given value of $x$ ?

Was this flashcard helpful?

Cards in this collection (29)

How do you find the gradient of a curve at a given value of $x$ ?
Differentiate the equation of the curve to get $f' (x)$ , then substitute that value of $x$ into it.
The first step produces a function, and only the second step turns it into a number.
Given $f (x) = x^{2} + 3 x - 4$ , find the gradient at $x = 1$ and at $x = - 2$ .
Differentiating gives $f' (x) = 2 x + 3$ .
Then $f' (1) = 5$ and $f' (- 2) = - 1$ , so the same curve has a different gradient at each point.
Which function do you substitute into to show that a point lies on a curve?
Into $f (x)$ itself, not into the derivative.
For $f (x) = x^{3} - 6 x^{2} + 5 x - 12$ , $f (2) = - 18$ , which shows that $(2, - 18)$ is on the curve.
What does the sign of the gradient at a point tell you?
A positive gradient means $y$ is increasing there, and a negative gradient means $y$ is decreasing.
The size of the number says how steeply, so a gradient of $- 7$ is far steeper than one of $- 1$ .
True or False?
A curve can have the same gradient at two different points.
True.
For $f (x) = x^{3} - 6 x^{2} + 5 x - 12$ the derivative is $f' (x) = 3 x^{2} - 12 x + 5$ .
Both $f' (0)$ and $f' (4)$ come to $5$ , so the curve is equally steep at those two places.
What is the difference between a stationary point and a turning point?
At a stationary point the derivative is zero, so the tangent there is horizontal.
A turning point needs more: the derivative must also change sign, so the curve really does turn around.
Every turning point is therefore a stationary point, but not every stationary point is a turning point.
What are the three steps for finding the stationary points of $y = f (x)$ ?
Three steps, the last of which is often not needed:
- differentiate to find $f' (x)$
- solve $f' (x) = 0$ for the $x$ -coordinates
- substitute each one back into $f (x)$ for the $y$ -coordinates
Find the stationary points of $f (x) = 2 x^{3} - 3 x^{2} - 36 x + 25$ .
Setting $f' (x) = 6 x^{2} - 6 x - 36 = 0$ gives $(x - 3) (x + 2) = 0$ , so $x = - 2$ or $x = 3$ .
Substituting each back into $f (x)$ gives the points $(- 2, 69)$ and $(3, - 56)$ .
What is the second derivative, and how do you find it?
It is the derivative of the derivative, written $\frac{d^{2} y}{d x^{2}}$ or $f'' (x)$ .
Simply differentiate twice: $2 x^{3}$ gives $6 x^{2}$ , and differentiating again gives $12 x$ .
Complete the second derivative test at a stationary point:
$if f'' (x)______0 it is a local minimum, if f'' (x)______0 it is a local maximum$
The completed test is:
$if f'' (x) > 0 it is a local minimum, if f'' (x) < 0 it is a local maximum$
The pairing looks backwards to many people, so it is worth learning as it stands: positive goes with minimum.
What do you do if $f'' (x) = 0$ at a stationary point?
Nothing follows from it: the point may be a local maximum, a local minimum or neither.
The test has simply failed, so fall back on the first derivative test instead.
How does the first derivative test classify a stationary point?
Work out $f' (x - h)$ and $f' (x + h)$ for some small $h$ , and see how the sign changes.
Negative to positive means a local minimum, and positive to negative a local maximum.
Choose $h$ small enough that you do not step over a second stationary point on the way.
Classify the stationary points $(- 2, 69)$ and $(3, - 56)$ of $f (x) = 2 x^{3} - 3 x^{2} - 36 x + 25$ .
Differentiating twice gives $f'' (x) = 12 x - 6$ .
Since $f'' (- 2) = - 30$ , which is negative, $(- 2, 69)$ is a local maximum.
Since $f'' (3) = 30$ , which is positive, $(3, - 56)$ is a local minimum.
True or False?
Every function has a global maximum.
False.
A global maximum is the greatest value a function reaches anywhere, and plenty of functions never reach one.
The cubic $y = x^{3}$ climbs without limit as $x$ grows, so it has no greatest value at all.
Define normal.
The normal to a curve at a point is the straight line through that point which is perpendicular to the tangent there.
Every point on a curve has both a tangent and a normal, and they cross at right angles at that point.
What two things do you need for the equation of a tangent, and where does each come from?
A point and a gradient.
The point comes from substituting $x_{1}$ into $f (x)$ , and the gradient comes from substituting the same $x_{1}$ into $f' (x)$ .
What is the gradient of the normal to $y = f (x)$ where $x = x_{1}$ ?
It is $- \frac{1}{f' (x_{1})}$ .
The derivative gives the tangent's gradient, and the normal's is the negative reciprocal of that, never of anything read straight off the curve.
Find the tangent to $y = 2 x^{4} + \frac{3}{x^{2}}$ at $x = 1$ , as $y = m x + c$ .
Since $f (1) = 5$ the point is $(1, 5)$ , and $f' (x) = 8 x^{3} - \frac{6}{x^{3}}$ gives $f' (1) = 2$ .
Then $y - 5 = 2 (x - 1)$ , which rearranges to $y = 2 x + 3$ .
Find the normal to $y = 2 x^{4} + \frac{3}{x^{2}}$ at $x = 1$ , as $a x + b y = c$ .
The tangent's gradient there is $2$ , so the normal's is $- \frac{1}{2}$ , through the same point $(1, 5)$ .
Then $y - 5 = - \frac{1}{2} (x - 1)$ , and doubling to clear the fraction gives $x + 2 y = 11$ .
True or False?
A tangent to a curve can meet that curve again somewhere else.
True.
A tangent only has to touch without cutting through at its own point, and it is free to cross the curve elsewhere.
The tangent to $y = x^{3}$ at $(1, 1)$ is $y = 3 x - 2$ , and that line also passes through $(- 2, - 8)$ on the curve.
Complete the approximation for a small change in $x$ :
$______\approx \frac{d y}{d x}______$
The completed approximation is:
$d y \approx \frac{d y}{d x} d x$
A small change in $x$ produces a change in $y$ of roughly the rate of change multiplied by that small change.
Why is $d y \approx \frac{d y}{d x} d x$ only an approximation?
Because $\frac{d y}{d x}$ is the rate of change at one particular value of $x$ , and it starts to change as soon as $x$ moves.
The smaller the change in $x$ , the closer the rate stays to its original value and the better the estimate.
How do you find $\frac{d x}{d y}$ when you know $\frac{d y}{d x}$ ?
Take the reciprocal, so that $\frac{d x}{d y} = \frac{1}{(\frac{d y}{d x})}$ .
This is needed constantly, because the formula you can differentiate is often the wrong way up for the rate you actually want.
A sphere of radius $5 cm$ has its surface area increased by $15 {cm}^{2}$ . Estimate the increase in radius.
From $A = 4 π r^{2}$ , $\frac{d A}{d r} = 8 π r$ , so the reciprocal gives $\frac{d r}{d A} = \frac{1}{8 π r}$ .
Then $d r \approx \frac{1}{8 π (5)} \times 15 = \frac{3}{8 π} = 0.12 cm$ to two significant figures.
What are connected rates of change?
Two or more rates of change linked by a shared variable, most often time $t$ .
As water runs into a bowl, both the height and the volume are changing with time, and time is what connects them.
How do you build the chain rule equation for a connected rates problem?
Write down the rate you are given and the rate you want as derivatives, then join them through a third.
Choose that third so the letters appear to cancel, as in $\frac{d x}{d t} = \frac{d x}{d V} \times \frac{d V}{d t}$ .
What tells you which rate of change a piece of information gives?
The units it is quoted in.
A rate of ${cm}^{3} s^{- 1}$ is a volume per unit time, so it is $\frac{d V}{d t}$ , while $cm s^{- 1}$ would be a length per unit time.
A cuboid of height $5 cm$ has a square cross-section of side $x$ . Its volume grows at $20 {cm}^{3} s^{- 1}$ ; find $\frac{d x}{d t}$ when $x = 3$ .
Here $V = 5 x^{2}$ , so $\frac{d V}{d x} = 10 x$ and therefore $\frac{d x}{d V} = \frac{1}{10 x}$ .
Then $\frac{d x}{d t} = \frac{d x}{d V} \times \frac{d V}{d t} = \frac{1}{30} \times 20 = \frac{2}{3} cm s^{- 1}$ .
True or False?
A quantity decreasing at $3 cm s^{- 1}$ has a rate of change of $- 3$ .
True.
A negative rate of change means the quantity is getting smaller, and a positive one means it is growing.
Words like decreasing, leaking or cooling are the signal to put the minus sign in before substituting anything.

Sign up to unlock flashcards

By signing up you agree to our Terms and Privacy Policy

Previous:DifferentiationNext:Integration