Algebra, Year 10: gradients, areas under graphs and circles

A tangent to the curve

HintThe same word is used for a line that touches a circle once.

The whyA curve has a different steepness at every point, so it has no single gradient. This line matches the curve at the point of contact, and its gradient is taken as the gradient of the curve there.

Draw the tangent at that point and work out the gradient of the tangent

HintA ruler is needed, laid so it just touches the curve at the chosen place.

The whyLay a ruler against the curve so it touches at the point and follows the direction of the curve there, then draw the line. Pick two well-separated points on that line and divide the change in y by the change in x.

3

HintThe steepness of the curve at P is taken from the straight line that touches it there.

The whyThe tangent rises 14 − 2 = 12 while going across 5 − 1 = 4, so its gradient is 12 ÷ 4. The curve has the same gradient as its tangent at the point where they touch.

The speed of the sprinter at that instant

HintGradient on this graph is metres divided by seconds.

The whyOn a distance–time graph, gradient is change in distance divided by change in time. A curve means this is changing all the time, and the tangent gives its value at one moment: the instantaneous rate of change.

A chord of the curve

HintA line joining two points on a circle has the same name.

The whyIf a distance–time curve passes through (2, 6) and (6, 30), with time in seconds and distance in metres, the line joining them has gradient (30 − 6) ÷ (6 − 2) = 6. So the average speed over those 4 seconds is 6 m/s, even though the speed was changing.

The acceleration of the car at that instant

HintThink about the units: metres per second, divided by seconds.

The whyGradient is change in velocity divided by change in time, which is how quickly the velocity is changing. A steep tangent means the velocity is changing fast; a tangent sloping downwards means the car is slowing.

distance

HintMultiply the units of the two axes: metres per second times seconds.

The whyAt a steady 10 m/s for 4 s the graph is a rectangle of height 10 and width 4, with area 40, and 40 m is how far the object goes. The same idea holds when the velocity changes: the area still adds up how far it has gone.

20

HintFind the area of the shape between the line and the time axis.

The whyThe shape under the line is a triangle with base 4 and height 10, so its area is 1/2 × 4 × 10. The object speeds up steadily from rest, so on average it moves at 5 m/s for the 4 seconds.

Split it into vertical strips, treat each strip as a trapezium, and add up their areas

HintReplace the curved top with a few short straight edges.

The whyJoining the points on the curve with straight lines turns each strip into a shape whose area has a formula: half the sum of the two parallel sides, times the width. More, narrower strips follow the curve more closely and give a better estimate.

About 22 m

HintJoin the three points with straight lines and find the area of each of the two shapes underneath.

The whyThe first strip is a triangle: 1/2 × 2 × 6 = 6. The second is a trapezium with parallel sides 6 and 10: 1/2 × (6 + 10) × 2 = 16. Together 6 + 16 = 22, and area under a velocity–time graph is distance.

Too small, because the straight top of each trapezium lies below the curve

HintPicture the thin sliver left between the ruled edge and the arc above it.

The whyEach trapezium misses the sliver between its straight top edge and the curve bulging above it, so the total is an underestimate. For a curve that sags below its chords, like the bottom of a valley, the trapezia include extra area and the estimate is too big.

x² + y² = r²

HintUse the theorem about right-angled triangles on a point (x, y) of the circle.

The whyFor any point (x, y) on the circle, the distance across (x), the distance up (y) and the radius form a right-angled triangle with the radius as hypotenuse. Pythagoras gives the equation, and every point at distance r from the origin fits it.

7

HintThe number on the right-hand side is not the radius itself.

The whyThe equation has the form x² + y² = r², so r² = 49 and r = 7. The circle is centred on the origin and passes through (7, 0), (0, 7), (−7, 0) and (0, −7).

x² + y² = 25

HintThe right-hand side is found from the radius, but it is not the radius.

The whyThe equation is x² + y² = r², and r² = 5² = 25. A quick check: the point (3, 4) is 5 units from the origin, and 3² + 4² = 9 + 16 = 25.

Inside, because 2² + 3² = 13, which is less than 16

HintSubstitute the coordinates into the left-hand side and compare with the right-hand side.

The whyx² + y² is the square of the distance of a point from the origin. For (2, 3) that is 4 + 9 = 13, and 13 is less than the radius squared, 16, so the point is closer to the centre than the circle is. Equal would mean on the circle; greater would mean outside.

8

HintA point on the circle must make the equation true.

The whySubstituting x = 6 and y = k gives 36 + k² = 100, so k² = 64 and k = 8 (the positive root). The point (6, −8) is on the circle too, directly below.