Define the normal to a curve.

The normal at a point is the line through that point which is perpendicular to the tangent there. Where there is no tangent, as at a sharp corner, there is no normal either.

True or False? A function that is increasing on one interval is increasing everywhere.

False. Increasing and decreasing are properties of an interval , not of the function as a whole. Most curves rise over some intervals and fall over others.

What is the second derivative mainly used for?

Determining the nature of a stationary point , that is whether it is a maximum or a minimum. Its sign at that point is what distinguishes the two.

Define stationary point .

A point on a curve where the gradient is zero . It may be a local minimum, a local maximum, or a point of inflection.

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

Every turning point is stationary, but a point of inflection is stationary without being a turning point. A turning point is one where the curve actually changes direction, from rising to falling or the other way round.

How do you use the first derivative to determine the nature of a stationary point?

Check its sign just to the left and just to the right of the point. Negative then positive is a minimum, positive then negative is a maximum, and the same sign on both sides is a point of inflection.

Which test for the nature of a stationary point should you reach for first?

The second derivative , because it is usually much quicker. Fall back on the first-derivative test only when the second derivative turns out to be zero.

What has to happen before you can optimise a quantity by differentiating?

It has to be written as a formula in a single variable. These problems usually give two variables plus a constraint linking them, and the constraint is what lets you eliminate one.

Why might a stationary point of a model not be a sensible answer?

Because the context may restrict the variable: a length cannot be negative, and a number of items has to be a whole number. A mathematically valid stationary point can fall outside the range the situation allows.

A LevelMathsEdexcelPureFlashcardsDifferentiationApplications of Differentiation

Applications of Differentiation (Edexcel A Level Maths: Pure): Flashcards

Exam code: 9MA0

1/34

0Still learning

Know0

FrontGradients, Tangents & Normals
How do you find the gradient of a curve at a particular point?

Was this flashcard helpful?

Cards in this collection (34)

How do you find the gradient of a curve at a particular point?
Differentiate to get $f' (x)$ , then substitute the $x$ -coordinate of the point into it.
The derivative is a formula for the gradient, so it has to be evaluated at the point you actually want.
Define the normal to a curve.
The normal at a point is the line through that point which is perpendicular to the tangent there.
Where there is no tangent, as at a sharp corner, there is no normal either.
The tangent to $y = f (x)$ at the point $(a, f (a))$ is:
$y - f (a) =______(x - a)$
$y - f (a) = f' (a) (x - a)$
It is simply $y - y_{1} = m (x - x_{1})$ , with the gradient supplied by the derivative.
If the tangent at a point has gradient $f' (a)$ , what is the gradient of the normal there?
$- \frac{1}{f' (a)}$ .
The normal is perpendicular to the tangent, so the two gradients must multiply to $- 1$ .
You are told only the $x$ -coordinate of the point where a tangent touches. What else do you need, and how do you get it?
The $y$ -coordinate, found by substituting that $x$ into the original function.
Two different substitutions are needed: into $f' (x)$ for the gradient, and into $f (x)$ for the point.
True or False?
Where the tangent to a curve is horizontal, the normal is vertical.
True.
If $f' (a) = 0$ the tangent is horizontal, so the normal must be vertical.
Its equation is then $x = a$ , because $- \frac{1}{f' (a)}$ has no value when the derivative is zero.
A function is increasing on an interval where $f' (x)______0$ , and decreasing where $f' (x)______0$ .
It is increasing where $f' (x) \geq 0$ , and decreasing where $f' (x) \leq 0$ .
With the strict inequalities $> 0$ and $< 0$ instead, the function is called strictly increasing or decreasing.
How do you find the intervals on which a function is increasing?
Differentiate, then solve the inequality $f' (x) \geq 0$ .
The solution set of that inequality is exactly the interval you are looking for.
Why does the sign of $f' (x)$ tell you whether a function is increasing?
Because $f' (x)$ is the gradient, and a positive gradient means the curve rises as $x$ increases.
A negative gradient means it falls.
True or False?
A function that is increasing on one interval is increasing everywhere.
False.
Increasing and decreasing are properties of an interval, not of the function as a whole.
Most curves rise over some intervals and fall over others.
Does an increasing function have to have a positive gradient at every single point?
Not quite: the definition allows $f' (x) = 0$ at isolated points.
That is exactly why strictly increasing, with $f' (x) > 0$ throughout, is a stronger statement.
Define second derivative.
The result of differentiating a function twice, written $f'' (x)$ or $\frac{d^{2} y}{d x^{2}}$ .
It does not mean squaring the first derivative: the superscripts are part of the notation, not powers.
Complete the second derivative notation:
$\frac{d^{2} y}{______}$
The completed notation is:
$\frac{d^{2} y}{d x^{2}}$
Note the positions: the $2$ sits on the $d$ on top and on the $x$ underneath, never on the $y$ .
What does the second derivative measure?
The rate of change of the gradient.
The first derivative says how fast $y$ is changing; the second says how fast that rate is itself changing.
What is the second derivative mainly used for?
Determining the nature of a stationary point, that is whether it is a maximum or a minimum.
Its sign at that point is what distinguishes the two.
True or False?
The second derivative of a straight line is zero.
True.
A straight line has a constant gradient, so the rate at which that gradient changes is zero.
Differentiating $y = m x + c$ gives $m$ , and differentiating again gives $0$ .
Define stationary point.
A point on a curve where the gradient is zero.
It may be a local minimum, a local maximum, or a point of inflection.
How do you find the stationary points of $f (x)$ ?
Differentiate, solve $f' (x) = 0$ for the $x$ -coordinates, then substitute those back into $f (x)$ for the $y$ -coordinates.
The derivative gives the positions; the original function gives the heights.
What is the difference between a stationary point and a turning point?
Every turning point is stationary, but a point of inflection is stationary without being a turning point.
A turning point is one where the curve actually changes direction, from rising to falling or the other way round.
Complete the second derivative test for the nature of a stationary point:
$f'' (x) > 0 \Rightarrow local______$
$f'' (x) < 0 \Rightarrow local______$
A positive second derivative means a local minimum, and a negative one means a local maximum.
Substitute the stationary point's $x$ -coordinate into $f'' (x)$ to find out which.
How do you use the first derivative to determine the nature of a stationary point?
Check its sign just to the left and just to the right of the point.
Negative then positive is a minimum, positive then negative is a maximum, and the same sign on both sides is a point of inflection.
True or False?
If $f'' (x) = 0$ at a stationary point, the point is a point of inflection.
False.
A zero second derivative tells you nothing: the point could still be a maximum or a minimum.
The first-derivative test has to be used instead, and that one always works.
Which test for the nature of a stationary point should you reach for first?
The second derivative, because it is usually much quicker.
Fall back on the first-derivative test only when the second derivative turns out to be zero.
How many stationary points does a quadratic have, and what kind?
Exactly one, and it is always a turning point rather than a point of inflection.
Its $y$ -value is therefore the minimum or maximum value the whole quadratic can take.
What does a stationary point on $y = f (x)$ become on the graph of $y = f' (x)$ ?
A point where the gradient graph meets the $x$ -axis, since the gradient there is zero.
The stationary points of $f$ are exactly the roots of $f'$ .
Where $f (x)$ is increasing, $f' (x)$ lies $______$ the $x$ -axis; where $f (x)$ is decreasing, it lies $______$ it.
Where $f$ is increasing, $f'$ lies above the axis; where $f$ is decreasing, it lies below it.
That is simply because the gradient is positive on a rising section and negative on a falling one.
True or False?
Where $f (x)$ cuts the $x$ -axis, $f' (x)$ does something notable.
False.
Where $f$ crosses the axis tells you nothing at all about $f'$ .
Only the gradient of $f$ matters, and a curve can cross the axis at any gradient whatever.
If $f (x)$ is a smooth curve, what does that tell you about $f' (x)$ ?
It will be a smooth curve as well.
So once the key features have been marked, the rest of the gradient graph is drawn by joining them smoothly.
What can you not work out about $f' (x)$ from the graph of $f (x)$ alone?
The exact coordinates of its $y$ -intercept, or of its own stationary points.
A sketch of $f$ gives you the shape of $f'$ , not its precise values.
Can you differentiate variables other than $y$ and $x$ ?
Yes: a derivative can be taken with respect to any variable.
$\frac{d V}{d r}$ , for instance, gives the rate of change of a volume with respect to a radius.
What has to happen before you can optimise a quantity by differentiating?
It has to be written as a formula in a single variable.
These problems usually give two variables plus a constraint linking them, and the constraint is what lets you eliminate one.
Why is the answer to an optimisation question often not just the value of $x$ ?
Because the question usually asks for the maximum or minimum value itself, not for where it occurs.
Substituting the stationary point's $x$ back into the original formula is what gives that value.
True or False?
A derivative in a modelling question always represents a gradient on a graph.
False.
It represents a rate of change of one quantity with respect to another, whatever those quantities happen to be.
The gradient of a graph is simply the case where they are $y$ and $x$ .
Why might a stationary point of a model not be a sensible answer?
Because the context may restrict the variable: a length cannot be negative, and a number of items has to be a whole number.
A mathematically valid stationary point can fall outside the range the situation allows.

Sign up to unlock flashcards

By signing up you agree to our Terms and Privacy Policy

Previous:DifferentiationNext:Further Differentiation