What is meant by the gradient of a curve at a point?

The gradient of a curve at a point is the gradient of the tangent to the curve there. Unlike a straight line, a curve has a different gradient at every point.

True or False? A tangent to a curve can cross that curve somewhere else.

True. A tangent must not cut the curve at the point of contact , but elsewhere it may cross it freely. Only one tangent can be drawn at any given point, however.

What does "rate of change" mean in a calculus question?

Rate of change means the gradient , so a question asking for one is asking you to differentiate. For a car, the way its position changes is its speed, and the way its speed changes is its acceleration.

Why can you not use two points on a curve to find its gradient at a point?

Because two points on a curve give the gradient of the straight line joining them, not the gradient at either one. The curve's gradient keeps changing between them, which is why a tangent is needed instead.

GCSEFurther MathsAQAFlashcards4. CalculusDifferentiation

Differentiation (AQA GCSE Further Maths): Flashcards

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What is meant by the gradient of a curve at a point?

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Cards in this collection (15)

What is meant by the gradient of a curve at a point?
The gradient of a curve at a point is the gradient of the tangent to the curve there.
Unlike a straight line, a curve has a different gradient at every point.
Define the derivative of a function.
The derivative is a function that gives the gradient of the original function for any value of $x$ .
It is also called the gradient function, because it turns an $x$ -value into a gradient.
True or False?
A tangent to a curve can cross that curve somewhere else.
True.
A tangent must not cut the curve at the point of contact, but elsewhere it may cross it freely.
Only one tangent can be drawn at any given point, however.
What does "rate of change" mean in a calculus question?
Rate of change means the gradient, so a question asking for one is asking you to differentiate.
For a car, the way its position changes is its speed, and the way its speed changes is its acceleration.
Why can you not use two points on a curve to find its gradient at a point?
Because two points on a curve give the gradient of the straight line joining them, not the gradient at either one.
The curve's gradient keeps changing between them, which is why a tangent is needed instead.
A curve is drawn with its tangent at $A (2, 1)$ . Which two points should you use to find the gradient?
Two points on the tangent, not on the curve, chosen where the coordinates are whole numbers.
Using $(1, - 2)$ and $(3, 4)$ gives a gradient of $\frac{4 - (- 2)}{3 - 1} = 3$ .
Complete the rule for differentiating a power of $x$ :
If $y = a x^{n}$ then $\frac{d y}{d x} =______$ .
The completed rule is:
If $y = a x^{n}$ then $\frac{d y}{d x} = a n x^{n - 1}$ .
Bring the power down to the front, then subtract one from the power.
What are the derivatives of $y = a x$ and $y = a$ ?
For $y = a x$ the derivative is $a$ , and for $y = a$ it is 0.
A constant has no gradient at all, which is why differentiating it gives zero.
How do you differentiate $y = \frac{4}{x}$ ?
Rewrite it as $y = 4 x^{- 1}$ first, then apply the rule.
That gives $\frac{d y}{d x} = - 4 x^{- 2}$ , which can also be written $- \frac{4}{x^{2}}$ .
True or False?
$y = (2 x - 3) (x^{2} - 4)$ can be differentiated bracket by bracket.
False.
A product cannot be differentiated term by term, so it has to be expanded first.
Expanding gives $y = 2 x^{3} - 3 x^{2} - 8 x + 12$ , which is a sum of powers and can then be differentiated.
How do you differentiate $y = \frac{8 x^{6} - x^{3}}{2 x^{4}}$ ?
Split the fraction over its single denominator and simplify with the index laws, giving $y = 4 x^{2} - \frac{1}{2} x^{- 1}$ .
Then differentiate term by term to get $\frac{d y}{d x} = 8 x + \frac{1}{2} x^{- 2}$ .
Differentiate $y = 5 x^{3} + 2 x + \frac{3}{x^{2}} + 8$ .
Rewrite the fraction as $3 x^{- 2}$ , then differentiate term by term.
That gives $\frac{d y}{d x} = 15 x^{2} + 2 - 6 x^{- 3}$ .
How do you find the gradient of a curve at a particular point?
Differentiate, then substitute the point's $x$ -coordinate into $\frac{d y}{d x}$ .
The answer is a number, whereas the derivative itself is a function.
What do "gradient of the curve", "gradient of the tangent" and "rate of change" have in common?
At a given point all three mean the same thing, and all three are found by substituting into $\frac{d y}{d x}$ .
A question may use any of the three wordings for what is the same calculation.
For $y = \frac{4}{3} x^{3} + 3 x - 8$ , find the values of $x$ where the rate of change is 4.
Differentiating gives $\frac{d y}{d x} = 4 x^{2} + 3$ , so set $4 x^{2} + 3 = 4$ .
That gives $x^{2} = \frac{1}{4}$ , so $x = \pm \frac{1}{2}$ ; the plural "values" is the hint that there is more than one.

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