Define the normal to a curve at a point.

The normal is the line through that point which is perpendicular to the tangent there. Finding it is the same process as finding a tangent, with one extra step.

Can the same curve be both increasing and decreasing?

A curve can be both, in different intervals: a cubic typically rises, then falls, then rises again. The intervals are separated by the points where the gradient is zero.

True or False? A curve can be falling while its gradient is increasing.

True. On the left-hand side of a u-shape the gradient is negative but rising towards zero. So a positive second derivative means the gradient is increasing, not that the function is.

Define a stationary point .

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

What does a second derivative of zero at a stationary point tell you?

A second derivative of zero tells you nothing, because the point could still be a minimum, a maximum or an inflection. In that case you have to compare the first derivative a little either side of the point instead.

How many stationary points does a quadratic graph have?

A quadratic has exactly one, and it is the overall minimum or maximum rather than merely a local one. For quadratics this point is usually called the turning point .

True or False? A stationary point need not be a turning point.

True. A point of inflection is stationary, but the curve does not turn there and carries on in the same direction. "Turning point" covers only the maxima and the minima.

How does differentiation help you maximise or minimise something?

A maximum or a minimum is a stationary point, so setting the derivative equal to zero locates it. That turns a problem like maximising a volume into an equation you can solve.

What has to happen before you can differentiate in an optimisation problem?

You need a formula for the quantity being optimised, written in terms of a single variable. Getting the expression down to one variable is what makes differentiation possible at all.

GCSEFurther MathsAQAFlashcards4. CalculusApplications of Differentiation

Applications of Differentiation (AQA GCSE Further Maths): Flashcards

Exam code: 8365

1/36

0Still learning

Know0

FrontTangents & Normals
How do you find the equation of a tangent to a curve at $x = a$ ?

Was this flashcard helpful?

Cards in this collection (36)

How do you find the equation of a tangent to a curve at $x = a$ ?
Differentiate and substitute $x = a$ to get the gradient, then find the full coordinate on the curve.
Put the gradient and that point into $y - y_{1} = m (x - x_{1})$ , or into $y = m x + c$ to find $c$ .
Given only the $x$ -value, how do you find the full coordinate on the curve?
Substitute it into the equation of the curve, not into the derivative.
The derivative would give you another gradient, whereas the curve gives the $y$ -coordinate you need.
Find the tangent to $y = 2 x^{2} - 6 x + 10$ where $x = 1$ .
Differentiating gives $\frac{d y}{d x} = 4 x - 6$ , so at $x = 1$ the gradient is $- 2$ .
The curve gives $y = 6$ there, so $6 = - 2 + c$ and the tangent is $y = - 2 x + 8$ .
What do the tangent and the normal at a point have in common?
They both pass through that same point on the curve.
Only their gradients differ, so the same substitution finds $c$ for either line.
Define the normal to a curve at a point.
The normal is the line through that point which is perpendicular to the tangent there.
Finding it is the same process as finding a tangent, with one extra step.
What is the extra step when finding a normal rather than a tangent?
Take the negative reciprocal of the tangent's gradient, so a gradient $g$ becomes $- \frac{1}{g}$ .
A tangent gradient of 3 therefore gives a normal gradient of $- \frac{1}{3}$ .
True or False?
The normal and the tangent at a point have the same gradient.
False.
They are perpendicular, so their gradients multiply to $- 1$ rather than being equal.
Only the tangent's gradient comes straight from the derivative; the normal's needs the extra step.
Find the normal to $y = \frac{4}{x}$ where $x = 1$ .
Rewriting as $4 x^{- 1}$ gives $\frac{d y}{d x} = - \frac{4}{x^{2}}$ , so the tangent's gradient is $- 4$ and the normal's is $\frac{1}{4}$ .
The curve gives $y = 4$ , so $4 = \frac{1}{4} + c$ and the normal is $y = \frac{1}{4} x + \frac{15}{4}$ .
Complete the conditions:
A function is increasing when $\frac{d y}{d x}______0$ , and decreasing when $\frac{d y}{d x}______0$ .
The completed conditions are:
A function is increasing when $\frac{d y}{d x} > 0$ , and decreasing when $\frac{d y}{d x} < 0$ .
Increasing means the graph goes up as $x$ increases, and decreasing means it goes down.
How do you find the interval where a curve is decreasing?
Differentiate, then solve the inequality $\frac{d y}{d x} < 0$ .
The answer is a range of $x$ values rather than a single value.
For what values of $x$ is $y = 2 x^{3} - 3 x^{2} + 5$ decreasing?
Differentiating gives $\frac{d y}{d x} = 6 x^{2} - 6 x$ , so solve $6 x (x - 1) < 0$ .
The gradient function is a positive quadratic, so it lies below zero between its roots, giving $0 < x < 1$ .
When solving $\frac{d y}{d x} < 0$ , which graph should you picture?
The graph of the gradient function, not the graph of the original curve.
Sketching $\frac{d y}{d x}$ shows which $x$ values make it negative, which is exactly what the inequality asks for.
True or False?
A function is increasing wherever its $y$ -values are positive.
False.
Increasing is about the gradient, not about the sign of $y$ .
A curve can sit well below the $x$ -axis and still be rising steeply.
Can the same curve be both increasing and decreasing?
A curve can be both, in different intervals: a cubic typically rises, then falls, then rises again.
The intervals are separated by the points where the gradient is zero.
What is the second derivative of a function?
The result of differentiating twice, written $\frac{d^{2} y}{d x^{2}}$ .
Where the first derivative is the rate of change of the function, the second is the rate of change of the gradient.
Why is the second derivative written $\frac{d^{2} y}{d x^{2}}$ ?
Because you differentiate twice, which gives the $d^{2}$ , and you do it with respect to $x$ twice, which gives the $x^{2}$ .
The two powers sit in different places on the top and the bottom, which is worth checking each time you write it.
Complete what the sign of the second derivative tells you:
A positive second derivative means the gradient is $______$ .
A negative second derivative means the gradient is $______$ .
The completed statements are:
A positive second derivative means the gradient is increasing.
A negative second derivative means the gradient is decreasing.
A u-shape has an increasing gradient, going from negative to positive; an n-shape has a decreasing one.
Find $\frac{d^{2} y}{d x^{2}}$ for $y = x^{5} - 2 x^{3} + 7 x^{2} + 9 x - 18$ .
Differentiating once gives $\frac{d y}{d x} = 5 x^{4} - 6 x^{2} + 14 x + 9$ .
Differentiating that gives $\frac{d^{2} y}{d x^{2}} = 20 x^{3} - 12 x + 14$ .
How do you start finding the second derivative of $y = \frac{3 x + 7}{x^{4}}$ ?
Split the fraction and rewrite each part with negative powers, giving $y = 3 x^{- 3} + 7 x^{- 4}$ .
Only then can you differentiate term by term, twice over.
True or False?
A curve can be falling while its gradient is increasing.
True.
On the left-hand side of a u-shape the gradient is negative but rising towards zero.
So a positive second derivative means the gradient is increasing, not that the function is.
Define a stationary point.
A stationary point is any 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 a curve?
Differentiate, then solve $\frac{d y}{d x} = 0$ to get the $x$ -coordinates.
Substitute those into the equation of the curve to get the matching $y$ -coordinates.
Complete the second-derivative test at a stationary point:
If $\frac{d^{2} y}{d x^{2}}$ is positive the point is a local $______$ , and if it is negative the point is a local $______$ .
The completed test is:
If $\frac{d^{2} y}{d x^{2}}$ is positive the point is a local minimum, and if it is negative it is a local maximum.
This is usually the quickest way to tell the two apart.
What does a second derivative of zero at a stationary point tell you?
A second derivative of zero tells you nothing, because the point could still be a minimum, a maximum or an inflection.
In that case you have to compare the first derivative a little either side of the point instead.
How do you classify a stationary point using the first derivative?
Work out $\frac{d y}{d x}$ a little to the left and a little to the right of the point.
Negative then positive is a minimum, positive then negative is a maximum, and the same sign both times is an inflection.
Find the stationary points of $y = 3 x^{5} - 20 x^{3}$ .
$\frac{d y}{d x} = 15 x^{4} - 60 x^{2}$ , which factorises to $15 x^{2} (x^{2} - 4) = 0$ , giving $x = 0$ and $x = \pm 2$ .
Substituting into the curve gives the points $(0, 0)$ , $(- 2, 64)$ and $(2, - 64)$ .
For $y = 3 x^{5} - 20 x^{3}$ the second derivative is $60 x^{3} - 120 x$ . Classify the stationary point at the origin.
The second derivative is zero there, so it gives no answer and the first derivative has to be used instead.
It is $- 45$ at both $x = - 1$ and $x = 1$ , the same sign either side, so the origin is a point of inflection.
How many stationary points does a quadratic graph have?
A quadratic has exactly one, and it is the overall minimum or maximum rather than merely a local one.
For quadratics this point is usually called the turning point.
True or False?
A stationary point need not be a turning point.
True.
A point of inflection is stationary, but the curve does not turn there and carries on in the same direction.
"Turning point" covers only the maxima and the minima.
Complete the meaning of the notation:
$\frac{d A}{d r}$ is the rate of change of $______$ with respect to $______$ .
The completed statement is:
$\frac{d A}{d r}$ is the rate of change of $A$ with respect to $r$ .
Derivatives are not limited to $y$ and $x$ : for $A = 4 π r^{2}$ , $\frac{d A}{d r} = 8 π r$ .
How does differentiation help you maximise or minimise something?
A maximum or a minimum is a stationary point, so setting the derivative equal to zero locates it.
That turns a problem like maximising a volume into an equation you can solve.
What has to happen before you can differentiate in an optimisation problem?
You need a formula for the quantity being optimised, written in terms of a single variable.
Getting the expression down to one variable is what makes differentiation possible at all.
A cuboid has length $4 x$ , width $x$ and height $\frac{3}{x} - 5$ . Show that $V = 12 x - 20 x^{2}$ .
Volume is length times width times height, so $V = 4 x \times x \times (\frac{3}{x} - 5)$ .
That gives $4 x^{2} (\frac{3}{x} - 5) = \frac{12 x^{2}}{x} - 20 x^{2} = 12 x - 20 x^{2}$ .
Find the maximum value of $V = 12 x - 20 x^{2}$ .
$\frac{d V}{d x} = 12 - 40 x$ , which is zero when $x = 0.3$ .
Substituting that back gives $V = 12 (0.3) - 20 {(0.3)}^{2} = 1.8$ .
How do you prove that a value you have found is a maximum?
Show that the second derivative is negative there.
For $V = 12 x - 20 x^{2}$ , $\frac{d^{2} V}{d x^{2}} = - 40$ , which is negative, so the value is a maximum.
True or False?
Solving $\frac{d V}{d x} = 0$ gives you the maximum volume.
False.
Solving it gives the value of $x$ at which the maximum occurs, not the volume itself.
Substitute that $x$ back into the formula for $V$ to get the maximum volume.

Sign up to unlock flashcards

By signing up you agree to our Terms and Privacy Policy

Previous:DifferentiationNext:Matrix Multiplication