GCSE Maths: Gradients of curves and rates of change (Higher)
Part of the GCSE Maths functions and advanced graphs chapter (Higher). Chapter overview · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)
Gradients of curves and rates of change
For a straight line the gradient is the same everywhere, so it tells us a steady rate of change: for a distance–time graph the gradient is the speed (Chapter 12). For a curve the steepness changes from place to place, so the gradient depends on where you look. Think of a car accelerating: its speed is different at different moments.
The gradient at a point on a curve. To find how steep a curve is at one point, draw a tangent: a straight line that just touches the curve at that point and has the same direction as the curve there. The gradient of the tangent is the gradient of the curve at that point. It is called the instantaneous rate of change. To find it:
- Use a ruler to draw the tangent so that the curve is balanced equally on either side of the point.
- Choose two points that are far apart on the tangent, and read their coordinates.
- Work out rise ÷ run.
A hand-drawn tangent is never perfect, so an answer is an estimate. In an exam, answers within a small tolerance are accepted.
Average rate of change. Often we only know the graph at two points. The average rate of change between them is the gradient of the straight line joining them, called a chord: (change in y) ÷ (change in x). For a distance–time graph this is the average speed over the interval.
Closing in on the tangent. To estimate the gradient at x = 3 for y = x2, work out the gradient of the chord from x = 3 to x = 3 + h for smaller and smaller h. The answers 7, 6.5, 6.1, 6.01 get closer and closer to 6, which is the gradient of the tangent. In algebra the chord gradient is (y-change) ÷ h; for y = x2 between x = 3 and 3 + h it simplifies to 6 + h, and when h is tiny this is almost 6. (At A level you will find such gradients exactly using calculus; here we estimate.)
Units. The gradient has units of (y-unit) per (x-unit). On a distance–time graph in metres and seconds the gradient is in m/s; on a speed–time graph it is m/s2 (acceleration); on a graph of cost in pounds against items bought it is £ per item.
Key ideas
- The gradient of a curve at a point = the gradient of the tangent there = the instantaneous rate of change. It is found by rise ÷ run using two points on the tangent.
- Average rate of change between two points = gradient of the chord = (change in y) ÷ (change in x).
- As the second point moves closer, the chord gradient gets closer to the tangent gradient.
- Gradient units: (y-unit) per (x-unit), e.g. m/s on a distance–time graph.
- Hand-drawn tangents give estimates, not exact answers.
Key words. The gradient measures steepness: rise ÷ run. On a curve the gradient changes from point to point, so we draw a tangent to find it at one point. This is the instantaneous rate of change.
Method. The gradient of a curve at a point is the gradient of the tangent drawn at that point: rise ÷ run.
- Draw the tangent. A tangent is a straight line that just touches the curve at one point, without crossing it there. Place a ruler so it touches the curve at x = 2 and has the curve balanced evenly on both sides of that point; draw along the ruler.
- Choose two points far apart on the tangent line. Points far apart give a more accurate gradient. Choose points where the line crosses grid intersections so that they are easy to read.(0, −2) and (4, 6)
- Draw a right-angled triangle under the tangent and read the rise and the run. The run is the horizontal distance and the rise is the vertical distance between the two points.run = 4 − 0 = 4
rise = 6 − (−2) = 8 - Work out the gradient. Gradient = rise ÷ run.gradient = 8 ÷ 4 = 2
Check. A hand-drawn tangent is only an estimate, so answers within about 10% are accepted (here between 1.8 and 2.2). The exact value for this curve is 2. ✓
Watch out. Do not use the gradient of the line joining two points on the curve (that is a chord). Use two points on the tangent.
Key words. The gradient measures steepness: rise ÷ run. On a curve the gradient changes from point to point, so we draw a tangent to find it at one point. This is the instantaneous rate of change.
Method. The gradient of a curve at a point is the gradient of the tangent drawn at that point: rise ÷ run.
- Draw the tangent. A tangent is a straight line that just touches the curve at one point, without crossing it there. Place a ruler so it touches the curve at t = 4 and has the curve balanced evenly on both sides of that point; draw along the ruler.
- Choose two points far apart on the tangent line. Points far apart give a more accurate gradient. Choose points where the line crosses grid intersections so that they are easy to read.(1, −2) and (7, 10)
- Draw a right-angled triangle under the tangent and read the rise and the run. The run is the horizontal distance and the rise is the vertical distance between the two points.run = 7 − 1 = 6
rise = 10 − (−2) = 12 - Work out the gradient. Gradient = rise ÷ run.gradient = 12 ÷ 6 = 2
- Interpret the answer. The gradient of the tangent is the rate of change at that instant.The gradient of a distance–time graph is speed. So at 4 seconds the car is travelling at 2 m/s.
Check. A hand-drawn tangent is only an estimate, so answers within about 10% are accepted (here between 1.8 and 2.2). The exact value for this curve is 2. ✓
Watch out. Do not use the gradient of the line joining two points on the curve (that is a chord). Use two points on the tangent.
Key words. A chord is a straight line joining two points on a curve. The average rate of change is the gradient of the chord.
Method. The average rate of change over an interval = (change in the output) ÷ (change in the input). It is the gradient of the straight line (chord) joining the two points on the graph.
- Find the value at each end of the interval. To use the rule, substitute each value into the formula.at t = 2: s = 22 = 4
at t = 5: s = 52 = 25 - Find the change in the output and the change in the input. Subtract the first from the second.change in s = 25 − 4 = 21m
change in t = 5 − 2 = 3s - Divide. The average rate of change is the gradient of the chord joining the two points: change in output ÷ change in input.average rate = 21 ÷ 3 = 7 m/s
Check. Picture the chord from (2, 4) to (5, 25): rise 21 over run 3 gives 7. ✓
Watch out. An average rate is only an average over the whole interval. The rate at a single instant can be higher or lower.
Method. Use a chord from the fixed point to a second point very close to it. As the second point gets closer, the gradient of the chord approaches the gradient of the tangent.
- Find the value at x = 3. This is the fixed point where we want the gradient.y = 32 = 9
- For each step size h, find the second point and the gradient of the chord. Chord gradient = (change in y) ÷ (change in x) = (y2 − 9) ÷ h. Use smaller and smaller h so the second point slides towards the first.
h x-value y-value gradient of chord 1 4 16 7 0.5 3.5 12.25 6.5 0.1 3.1 9.61 6.1 0.01 3.01 9.0601 6.01 - Look at the pattern. As h gets smaller the chord gets closer to the tangent, so the chord gradients settle down towards one number.The gradients are getting closer to 6.
Check. The last chord gradient, 6.01, is very close to 6. ✓
Watch out. Use the same fixed point every time, and divide by h, the change in x.
Method. Find the gradient of the chord from x = 3 to x = 3 + h, simplify it, then see what happens when h gets tiny.
- Write the two points on the curve. Take x = 3 and a nearby value x = 3 + h. Substitute into y = x2.(3, 9) and (3 + h, (3 + h)2)
- Find the change in y. Expand (3 + h)2 and take away 9.(3 + h)2 − 9 = h2 + 6h
- Divide by the change in x, which is h. Gradient of the chord = change in y ÷ change in x. Every term has a factor h, so divide each term by h.gradient = h + 6
- Let h become very small (almost 0). Any term containing h becomes almost 0, so only the number is left. This is the gradient of the tangent.gradient of the tangent at x = 3 is 6
Check. Try h = 0.01 numerically: ((3 + 0.01)2 − 9) ÷ 0.01 = 6.0100, close to 6. ✓
Watch out. Do not put h = 0 before dividing; that would give 0 ÷ 0. Simplify first, then let h become small.
Watch out
A tangent touches the curve at one point; it does not cut through it there. Do not use two points on the curve to find the gradient of a tangent: that gives a chord (an average rate). Take two points on the tangent line itself, far apart, and give the units of the gradient.Exercise 25H Gradients of curves and rates of change
Non-calculatorKey words. The gradient measures steepness: rise ÷ run. On a curve the gradient changes from point to point, so we draw a tangent to find it at one point. This is the instantaneous rate of change.
Method. The gradient of a curve at a point is the gradient of the tangent drawn at that point: rise ÷ run.
- Draw the tangent. A tangent is a straight line that just touches the curve at one point, without crossing it there. Place a ruler so it touches the curve at x = 2 and has the curve balanced evenly on both sides of that point; draw along the ruler.
- Choose two points far apart on the tangent line. Points far apart give a more accurate gradient. Choose points where the line crosses grid intersections so that they are easy to read.(0, −2) and (4, 6)
- Draw a right-angled triangle under the tangent and read the rise and the run. The run is the horizontal distance and the rise is the vertical distance between the two points.run = 4 − 0 = 4
rise = 6 − (−2) = 8 - Work out the gradient. Gradient = rise ÷ run.gradient = 8 ÷ 4 = 2
Check. A hand-drawn tangent is only an estimate, so answers within about 10% are accepted (here between 1.8 and 2.2). The exact value for this curve is 2. ✓
Watch out. Do not use the gradient of the line joining two points on the curve (that is a chord). Use two points on the tangent.
Key words. The gradient measures steepness: rise ÷ run. On a curve the gradient changes from point to point, so we draw a tangent to find it at one point. This is the instantaneous rate of change.
Method. The gradient of a curve at a point is the gradient of the tangent drawn at that point: rise ÷ run.
- Draw the tangent. A tangent is a straight line that just touches the curve at one point, without crossing it there. Place a ruler so it touches the curve at x = −2 and has the curve balanced evenly on both sides of that point; draw along the ruler.
- Choose two points far apart on the tangent line. Points far apart give a more accurate gradient. Choose points where the line crosses grid intersections so that they are easy to read.(−4, 6) and (0, −2)
- Draw a right-angled triangle under the tangent and read the rise and the run. The run is the horizontal distance and the rise is the vertical distance between the two points.run = 0 − (−4) = 4
rise = −2 − 6 = −8 - Work out the gradient. Gradient = rise ÷ run.gradient = −8 ÷ 4 = −2
Check. A hand-drawn tangent is only an estimate, so answers within about 10% are accepted (here between −2.2 and −1.8). The exact value for this curve is −2. ✓
Watch out. Do not use the gradient of the line joining two points on the curve (that is a chord). Use two points on the tangent.
Key words. The gradient measures steepness: rise ÷ run. On a curve the gradient changes from point to point, so we draw a tangent to find it at one point. This is the instantaneous rate of change.
Method. The gradient of a curve at a point is the gradient of the tangent drawn at that point: rise ÷ run.
- Draw the tangent. A tangent is a straight line that just touches the curve at one point, without crossing it there. Place a ruler so it touches the curve at x = 2 and has the curve balanced evenly on both sides of that point; draw along the ruler.
- Choose two points far apart on the tangent line. Points far apart give a more accurate gradient. Choose points where the line crosses grid intersections so that they are easy to read.(0, −4) and (4, 8)
- Draw a right-angled triangle under the tangent and read the rise and the run. The run is the horizontal distance and the rise is the vertical distance between the two points.run = 4 − 0 = 4
rise = 8 − (−4) = 12 - Work out the gradient. Gradient = rise ÷ run.gradient = 12 ÷ 4 = 3
Check. A hand-drawn tangent is only an estimate, so answers within about 10% are accepted (here between 2.7 and 3.3). The exact value for this curve is 3. ✓
Watch out. Do not use the gradient of the line joining two points on the curve (that is a chord). Use two points on the tangent.
Key words. The gradient measures steepness: rise ÷ run. On a curve the gradient changes from point to point, so we draw a tangent to find it at one point. This is the instantaneous rate of change.
Method. The gradient of a curve at a point is the gradient of the tangent drawn at that point: rise ÷ run.
- Draw the tangent. A tangent is a straight line that just touches the curve at one point, without crossing it there. Place a ruler so it touches the curve at t = 2 and has the curve balanced evenly on both sides of that point; draw along the ruler.
- Choose two points far apart on the tangent line. Points far apart give a more accurate gradient. Choose points where the line crosses grid intersections so that they are easy to read.(0, −1) and (4, 3)
- Draw a right-angled triangle under the tangent and read the rise and the run. The run is the horizontal distance and the rise is the vertical distance between the two points.run = 4 − 0 = 4
rise = 3 − (−1) = 4 - Work out the gradient. Gradient = rise ÷ run.gradient = 4 ÷ 4 = 1
- Interpret the answer. The gradient of the tangent is the rate of change at that instant.Gradient of a distance–time graph = speed = 1 m/s.
Check. A hand-drawn tangent is only an estimate, so answers within about 10% are accepted (here between 0.9 and 1.1). The exact value for this curve is 1. ✓
Watch out. Do not use the gradient of the line joining two points on the curve (that is a chord). Use two points on the tangent.
Key words. The gradient measures steepness: rise ÷ run. On a curve the gradient changes from point to point, so we draw a tangent to find it at one point. This is the instantaneous rate of change.
Method. The gradient of a curve at a point is the gradient of the tangent drawn at that point: rise ÷ run.
- Draw the tangent. A tangent is a straight line that just touches the curve at one point, without crossing it there. Place a ruler so it touches the curve at t = 4 and has the curve balanced evenly on both sides of that point; draw along the ruler.
- Choose two points far apart on the tangent line. Points far apart give a more accurate gradient. Choose points where the line crosses grid intersections so that they are easy to read.(1, −2) and (7, 10)
- Draw a right-angled triangle under the tangent and read the rise and the run. The run is the horizontal distance and the rise is the vertical distance between the two points.run = 7 − 1 = 6
rise = 10 − (−2) = 12 - Work out the gradient. Gradient = rise ÷ run.gradient = 12 ÷ 6 = 2
- Interpret the answer. The gradient of the tangent is the rate of change at that instant.Gradient of a distance–time graph = speed = 2 m/s.
Check. A hand-drawn tangent is only an estimate, so answers within about 10% are accepted (here between 1.8 and 2.2). The exact value for this curve is 2. ✓
Watch out. Do not use the gradient of the line joining two points on the curve (that is a chord). Use two points on the tangent.
Key words. The gradient measures steepness: rise ÷ run. On a curve the gradient changes from point to point, so we draw a tangent to find it at one point. This is the instantaneous rate of change.
Method. The gradient of a curve at a point is the gradient of the tangent drawn at that point: rise ÷ run.
- Draw the tangent. A tangent is a straight line that just touches the curve at one point, without crossing it there. Place a ruler so it touches the curve at t = 6 and has the curve balanced evenly on both sides of that point; draw along the ruler.
- Choose two points far apart on the tangent line. Points far apart give a more accurate gradient. Choose points where the line crosses grid intersections so that they are easy to read.(4, 3) and (8, 15)
- Draw a right-angled triangle under the tangent and read the rise and the run. The run is the horizontal distance and the rise is the vertical distance between the two points.run = 8 − 4 = 4
rise = 15 − 3 = 12 - Work out the gradient. Gradient = rise ÷ run.gradient = 12 ÷ 4 = 3
- Interpret the answer. The gradient of the tangent is the rate of change at that instant.Gradient of a distance–time graph = speed = 3 m/s.
Check. A hand-drawn tangent is only an estimate, so answers within about 10% are accepted (here between 2.7 and 3.3). The exact value for this curve is 3. ✓
Watch out. Do not use the gradient of the line joining two points on the curve (that is a chord). Use two points on the tangent.
Key words. A chord is a straight line joining two points on a curve. The average rate of change is the gradient of the chord.
Method. The average rate of change over an interval = (change in the output) ÷ (change in the input). It is the gradient of the straight line (chord) joining the two points on the graph.
- Find the value at each end of the interval. To use the rule, substitute each value into the formula.at t = 0: s = 02 = 0
at t = 2: s = 22 = 4 - Find the change in the output and the change in the input. Subtract the first from the second.change in s = 4 − 0 = 4m
change in t = 2 − 0 = 2s - Divide. The average rate of change is the gradient of the chord joining the two points: change in output ÷ change in input.average rate = 4 ÷ 2 = 2 m/s
Check. Picture the chord from (0, 0) to (2, 4): rise 4 over run 2 gives 2. ✓
Watch out. An average rate is only an average over the whole interval. The rate at a single instant can be higher or lower.
Key words. A chord is a straight line joining two points on a curve. The average rate of change is the gradient of the chord.
Method. The average rate of change over an interval = (change in the output) ÷ (change in the input). It is the gradient of the straight line (chord) joining the two points on the graph.
- Find the value at each end of the interval. To use the rule, substitute each value into the formula.at t = 2: s = 22 = 4
at t = 4: s = 42 = 16 - Find the change in the output and the change in the input. Subtract the first from the second.change in s = 16 − 4 = 12m
change in t = 4 − 2 = 2s - Divide. The average rate of change is the gradient of the chord joining the two points: change in output ÷ change in input.average rate = 12 ÷ 2 = 6 m/s
Check. Picture the chord from (2, 4) to (4, 16): rise 12 over run 2 gives 6. ✓
Watch out. An average rate is only an average over the whole interval. The rate at a single instant can be higher or lower.
Key words. A chord is a straight line joining two points on a curve. The average rate of change is the gradient of the chord.
Method. The average rate of change over an interval = (change in the output) ÷ (change in the input). It is the gradient of the straight line (chord) joining the two points on the graph.
- Find the value at each end of the interval. To use the rule, substitute each value into the formula.at t = 4: s = 42 = 16
at t = 8: s = 82 = 64 - Find the change in the output and the change in the input. Subtract the first from the second.change in s = 64 − 16 = 48m
change in t = 8 − 4 = 4s - Divide. The average rate of change is the gradient of the chord joining the two points: change in output ÷ change in input.average rate = 48 ÷ 4 = 12 m/s
Check. Picture the chord from (4, 16) to (8, 64): rise 48 over run 4 gives 12. ✓
Watch out. An average rate is only an average over the whole interval. The rate at a single instant can be higher or lower.
Key words. A chord is a straight line joining two points on a curve. The average rate of change is the gradient of the chord.
Method. The average rate of change over an interval = (change in the output) ÷ (change in the input). It is the gradient of the straight line (chord) joining the two points on the graph.
- Find the value at each end of the interval. To use the rule, substitute each value into the formula.at t = 0: V = 100 × (1110)0 = 100
at t = 2: V = 100 × (1110)2 = 121 - Find the change in the output and the change in the input. Subtract the first from the second.change in V = 121 − 100 = 21pounds
change in t = 2 − 0 = 2year - Divide. The average rate of change is the gradient of the chord joining the two points: change in output ÷ change in input.average rate = 21 ÷ 2 = 10.5 pounds/year
Check. Picture the chord from (0, 100) to (2, 121): rise 21 over run 2 gives 10.5. ✓
Watch out. An average rate is only an average over the whole interval. The rate at a single instant can be higher or lower.
Key words. A chord is a straight line joining two points on a curve. The average rate of change is the gradient of the chord.
Method. The average rate of change over an interval = (change in the output) ÷ (change in the input). It is the gradient of the straight line (chord) joining the two points on the graph.
- Find the value at each end of the interval. To use the rule, substitute each value into the formula.at t = 2: V = 100 × (1110)2 = 121
at t = 4: V = 100 × (1110)4 = 146.41 - Find the change in the output and the change in the input. Subtract the first from the second.change in V = 146.41 − 121 = 25.41pounds
change in t = 4 − 2 = 2year - Divide. The average rate of change is the gradient of the chord joining the two points: change in output ÷ change in input.average rate = 25.41 ÷ 2 = 12.705 pounds/year
Check. Picture the chord from (2, 121) to (4, 146.41): rise 25.41 over run 2 gives 12.705. ✓
Watch out. An average rate is only an average over the whole interval. The rate at a single instant can be higher or lower.
Method. Use a chord from the fixed point to a second point very close to it. As the second point gets closer, the gradient of the chord approaches the gradient of the tangent.
- Find the value at t = 4. This is the fixed point where we want the gradient.s = 422 + 2 × 4 = 16
- For each step size h, find the second point and the gradient of the chord. Chord gradient = (change in s) ÷ (change in t) = (s2 − 16) ÷ h. Use smaller and smaller h so the second point slides towards the first.
h t-value s-value gradient of chord 1 5 22.5 6.5 0.5 4.5 19.125 6.25 0.1 4.1 16.605 6.05 0.01 4.01 16.06005 6.005 - Look at the pattern. As h gets smaller the chord gets closer to the tangent, so the chord gradients settle down towards one number.The gradients are getting closer to 6.
Check. The last chord gradient, 6.005, is very close to 6. ✓
Watch out. Use the same fixed point every time, and divide by h, the change in x.
Method. Find the gradient of the chord from x = 3 to x = 3 + h, simplify it, then see what happens when h gets tiny.
- Write the two points on the curve. Take x = 3 and a nearby value x = 3 + h. Substitute into y = x2.(3, 9) and (3 + h, (3 + h)2)
- Find the change in y. Expand (3 + h)2 and take away 9.(3 + h)2 − 9 = h2 + 6h
- Divide by the change in x, which is h. Gradient of the chord = change in y ÷ change in x. Every term has a factor h, so divide each term by h.gradient = h + 6
- Let h become very small (almost 0). Any term containing h becomes almost 0, so only the number is left. This is the gradient of the tangent.gradient of the tangent at x = 3 is 6
Check. Try h = 0.01 numerically: ((3 + 0.01)2 − 9) ÷ 0.01 = 6.0100, close to 6. ✓
Watch out. Do not put h = 0 before dividing; that would give 0 ÷ 0. Simplify first, then let h become small.
Method. Find the gradient of the chord from x = 1 to x = 1 + h, simplify it, then see what happens when h gets tiny.
- Write the two points on the curve. Take x = 1 and a nearby value x = 1 + h. Substitute into y = x3.(1, 1) and (1 + h, (1 + h)3)
- Find the change in y. Expand (1 + h)3 and take away 1.(1 + h)3 − 1 = h3 + 3h2 + 3h
- Divide by the change in x, which is h. Gradient of the chord = change in y ÷ change in x. Every term has a factor h, so divide each term by h.gradient = h2 + 3h + 3
- Let h become very small (almost 0). Any term containing h becomes almost 0, so only the number is left. This is the gradient of the tangent.gradient of the tangent at x = 1 is 3
Check. Try h = 0.01 numerically: ((1 + 0.01)3 − 1) ÷ 0.01 = 3.0301, close to 3. ✓
Watch out. Do not put h = 0 before dividing; that would give 0 ÷ 0. Simplify first, then let h become small.
Method. Find the gradient of the chord from x = 5 to x = 5 + h, simplify it, then see what happens when h gets tiny.
- Write the two points on the curve. Take x = 5 and a nearby value x = 5 + h. Substitute into y = x2.(5, 25) and (5 + h, (5 + h)2)
- Find the change in y. Expand (5 + h)2 and take away 25.(5 + h)2 − 25 = h2 + 10h
- Divide by the change in x, which is h. Gradient of the chord = change in y ÷ change in x. Every term has a factor h, so divide each term by h.gradient = h + 10
- Let h become very small (almost 0). Any term containing h becomes almost 0, so only the number is left. This is the gradient of the tangent.gradient of the tangent at x = 5 is 10
Check. Try h = 0.01 numerically: ((5 + 0.01)2 − 25) ÷ 0.01 = 10.0100, close to 10. ✓
Watch out. Do not put h = 0 before dividing; that would give 0 ÷ 0. Simplify first, then let h become small.
Method. Find the gradient of the chord from x = −2 to x = −2 + h, simplify it, then see what happens when h gets tiny.
- Write the two points on the curve. Take x = −2 and a nearby value x = −2 + h. Substitute into y = x3.(−2, −8) and (−2 + h, (−2 + h)3)
- Find the change in y. Expand (−2 + h)3 and take away −8.(−2 + h)3 − −8 = h3 − 6h2 + 12h
- Divide by the change in x, which is h. Gradient of the chord = change in y ÷ change in x. Every term has a factor h, so divide each term by h.gradient = h2 − 6h + 12
- Let h become very small (almost 0). Any term containing h becomes almost 0, so only the number is left. This is the gradient of the tangent.gradient of the tangent at x = −2 is 12
Check. Try h = 0.01 numerically: ((−2 + 0.01)3 − −8) ÷ 0.01 = 11.9401, close to 12. ✓
Watch out. Do not put h = 0 before dividing; that would give 0 ÷ 0. Simplify first, then let h become small.
Method. Average speed = total distance ÷ total time. For the instantaneous speed, use a very short interval after 4 seconds.
- (a) Distance after 4 s. s(4) = 42 = 16 m, and s(0) = 0
- (a) Average speed. Distance ÷ time.16 ÷ 4 = 4 m/s
- (b) Distance at 4.01 s. s(4.01) = 4.012 = 16.0801 m
- (b) Chord gradient. Change in distance ÷ change in time.(16.0801 − 16) ÷ 0.01 = 0.0801 ÷ 0.01 = 8.01 ≈ 8 m/s
- (c) Compare. The car is accelerating, so by 4 seconds it is going much faster than its average over the first 4 seconds.average 4 m/s but speed at 4 s is about 8 m/s
Check. The exact gradient of s = t2 at t = 4 is 2 × 4 = 8. ✓
Watch out. Average rate over an interval is not the same as the rate at one instant.
Method. Average rate = change in P ÷ change in t. For the instantaneous rate use a very short interval.
- (a) Population at t = 2 and t = 4. Use the formula.P(2) = 1000 × 1.12 = 1210.0
P(5) = 1000 × 1.15 = 1610.5 - (a) Average rate. Change in P ÷ change in t.(1610.5 − 1210.0) ÷ 3 = 133.50 ≈ 134 per hour
- (b) Population at t = 3 and 3.01. P(3) = 1331.000
P(3.01) = 1332.269 - (b) Chord gradient over 0.01 h. A very short interval gives almost the tangent gradient.(1332.269 − 1331.000) ÷ 0.01 = 126.9 ≈ 127 per hour
Check. The rate at t = 3 should lie between the average rates over (2, 3) and (3, 4): 121 and 133. ✓
Watch out. The rate is increasing all the time, so the average over a long interval lies between the rates at the two ends.
Method. The gradient is the change in y for a change of 1 in x. Its unit is (y-unit) per (x-unit).
- Identify the axes. The horizontal axis shows number of items and the vertical axis shows cost in pounds.gradient unit = £ per item = £/item
- Say what the number means. A gradient of 3.5 means that cost in pounds changes by 3.5 £ for every 1 item.each extra item costs £3.50 (price per item)
Check. Multiply back: 5 item gives a change of 17.5 £. ✓
Watch out. State the units; a gradient without units loses marks.
Method. The gradient is the change in y for a change of 1 in x. Its unit is (y-unit) per (x-unit).
- Identify the axes. The horizontal axis shows time in seconds and the vertical axis shows speed in m/s.gradient unit = m/s per s = m/s/s
- Say what the number means. A gradient of 2 means that speed in m/s changes by 2 m/s for every 1 s.the acceleration is 2 m/s²
Check. Multiply back: 5 s gives a change of 10 m/s. ✓
Watch out. State the units; a gradient without units loses marks.
Method. The gradient is the change in y for a change of 1 in x. Its unit is (y-unit) per (x-unit).
- Identify the axes. The horizontal axis shows time in hours and the vertical axis shows distance in km.gradient unit = km per hour = km/hour
- Say what the number means. A gradient of 60 means that distance in km changes by 60 km for every 1 hour.the speed is 60 km/h
Check. Multiply back: 5 hour gives a change of 300 km. ✓
Watch out. State the units; a gradient without units loses marks.
Method. The gradient is the change in y for a change of 1 in x. Its unit is (y-unit) per (x-unit).
- Identify the axes. The horizontal axis shows distance in km and the vertical axis shows fuel left in litres.gradient unit = litres per km = litres/km
- Say what the number means. A gradient of −0.08 means that fuel left in litres changes by −0.08 litres for every 1 km.the car uses 0.08 litres of fuel per km (the minus sign means the fuel is decreasing)
Check. Multiply back: 5 km gives a change of −0.4 litres. ✓
Watch out. State the units; a gradient without units loses marks.
Method. Imagine the tangent at each marked point. The gradient is positive if the curve rises left to right, negative if it falls, and zero at a turning point.
- Look at each point in turn, from left to right. Picture a ruler touching the curve at the point.A: rising, B: flat, C: falling, D: flat, E: rising
- Pick the points that fit. We want the points where the tangent would be horizontal (flat): the curve is at the top of a hill or the bottom of a valley.points B and D
Check. A turning point is where the curve changes from rising to falling or the other way, so the gradient must be 0 there. ✓
Watch out. Read the graph from left to right when deciding between rising and falling.
Method. Imagine the tangent at each marked point. The gradient is positive if the curve rises left to right, negative if it falls, and zero at a turning point.
- Look at each point in turn, from left to right. Picture a ruler touching the curve at the point.A: rising, B: flat, C: falling, D: flat, E: rising
- Pick the points that fit. We want the points where the tangent would slope upwards to the right (the curve is rising).points A and E
Check. A turning point is where the curve changes from rising to falling or the other way, so the gradient must be 0 there. ✓
Watch out. Read the graph from left to right when deciding between rising and falling.
Method. Imagine the tangent at each marked point. The gradient is positive if the curve rises left to right, negative if it falls, and zero at a turning point.
- Look at each point in turn, from left to right. Picture a ruler touching the curve at the point.A: rising, B: flat, C: falling, D: flat, E: rising
- Pick the points that fit. We want the points where the tangent would slope downwards to the right (the curve is falling).points C
Check. A turning point is where the curve changes from rising to falling or the other way, so the gradient must be 0 there. ✓
Watch out. Read the graph from left to right when deciding between rising and falling.
Method. Compare an average rate with an instantaneous rate.
- Say what 80 km/h represents. It is the total distance divided by the total time: the gradient of the chord from the start to the end of the journey.average speed = total distance ÷ total time
- Say what the half-way speed is. The speed at one instant is the gradient of the tangent at that point of the graph, which can be higher or lower.instantaneous speed = gradient of the tangent
- Give a reason they differ. The train speeds up after leaving the station, may travel faster on open track, and slows down before stopping.The speed at the half-way point could easily be 100 km/h or 60 km/h.
Check. Example: 80 km in 1 hour is an average of 80 km/h even if the train goes at 100 km/h for half an hour and 60 km/h for the other half. ✓
Watch out. Average rate = chord gradient; instantaneous rate = tangent gradient.
Method. Picture the tangent at the lowest point of the curve.
- Look at the shape. To the left of the lowest point the curve falls, and to the right it rises.falling, then rising
- Picture the tangent at the lowest point. Just at the bottom the curve is neither rising nor falling. The tangent lies flat along the bottom.the tangent is horizontal
- Find the gradient of a horizontal line. A horizontal line has no rise.gradient = 0 ÷ run = 0
Check. Chord from x = 2 to 2 + h: gradient = h, which is almost 0 when h is tiny. ✓
Watch out. The gradient is 0 at every turning point.
Method. Work out the chord gradient, then think about the shape of the curve.
- Find the two points. (3, 9) and (5, 25)
- Chord gradient. Rise ÷ run.(25 − 9) ÷ (5 − 3) = 16 ÷ 2 = 8
- Compare with 6. 8 is bigger than 6.8 > 6
- Explain with the graph. The curve gets steeper as x increases, so the chord to a point further along is steeper than the tangent at the start.The chord gradient is larger because the curve is getting steeper.
Check. Using the formula 6 + h from the worked example with h = 2 gives 8. ✓
Watch out. For a curve that is getting steeper, chords to the right have a larger gradient than the tangent.
Method. When h is tiny, the chord gradient 2a + h is almost 2a. So the gradient of the curve at x = a is 2a.
- Write the chord gradient. From the algebra of the chord.2a + h
- Let h become very small. The h disappears, leaving the gradient of the tangent.gradient at x = a is 2a
- Set it equal to 10. 2a = 10
- Solve. a = 5
Check. At x = 5, the chord to 5.01 has gradient (5.012 − 25) ÷ 0.01 = 10.01. ✓
Watch out. The gradient of y = x2 at x = a is 2a. It is not a2.