Unit 9
8 marks · EasyA function is defined by the curve . The diagram below shows part of this curve.
Find the gradient of the curve at the point A.
Draw a tangent at point A and estimate its gradient. Show your working.
A cuboid has an edge OA of 6 cm and OB of 8 cm. OC is perpendicular to both OA and OB and has a length of 5 cm. O, A, B, and C are adjacent vertices joining at one corner.
Use Pythagoras’ theorem in 3D to find the length of the diagonal OC.
Calculate the length OC, giving your answer correct to one decimal place.
Explain how the ideas of gradient (from part a) and using Pythagoras’ theorem in three dimensions (from part b) are similar when working with curves and 3D shapes.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) Gradient is 4 at A, using tangent or differentiation. · must show Correct tangent drawn or explained, Use of difference in y and x for two points near A, Approximate answer close to 4
(b) OC = 10.2 cm · must show Correct 3D Pythagoras' application
(c) Both involve finding the rate of change (gradient) or length by using geometric reasoning with right-angled triangles, whether on a curve or within a cuboid. · must show Appropriate comparison
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Draws or describes an appropriate tangent at A.
A1 (a) · 1 mark · Correctly calculates rise/run for two suitable points on the tangent.
A2 (a) · 1 mark · Obtains an answer of approximately 4 for the gradient at A.
M2 (b) · 1 mark · Uses Pythagoras' theorem in three dimensions: .
A3 (b) · 1 mark · Correctly calculates cm to 1 d.p.
B1 (c) · 3 marks · Explains the connection between gradients (as rates of change) and using Pythagoras' theorem in 3D geometry, relating both to geometric reasoning.
Worked solution
a) To estimate the gradient at A (2, 4), draw a tangent at A. Choose two points close to A on the tangent, say (1.5, 2.25) and (2.5, 6.25).
So, the gradient at A is approximately 4.
b) OA = 6 cm, OB = 8 cm, OC = 5 cm. Length OC (the diagonal from O through A and then up to C) uses:
But OC is 5 cm (so if OC is the diagonal from O, not an edge): = 10, then include OC up, = cm.
So, the 3D diagonal from O to opposite corner is 11.2 cm (rounded as needed).
c) Both parts involve finding how a value changes – the slope of a curve or a length in space. In curves, gradient is found by tangent (just as in a triangle, gradient = change in y/change in x). In space, the 'change' happens in three directions (x, y, z), and Pythagoras' theorem is extended into 3D. Both use geometric concepts of rate of change or distance, connected by squares and roots.