Mathematics Specification A - 4MA1 PearsonEdexcel

Calculus

Gbogbo ọrọ náà

Calculus is the mathematics of change. It gives you a precise way to find the gradient of a curve at any point, locate maximum and minimum values, and describe how quantities such as speed and acceleration are related.

This Higher-tier topic introduces differentiation of polynomial functions, finding stationary points, and applying calculus to kinematics and optimisation problems. It is one of the most powerful tools in the specification.

Ebumnobi

[Higher] Understand the concept of a variable rate of change
[Higher] Differentiate integer powers of x
[Higher] Determine gradients, rates of change, stationary points, turning points (maxima and minima) by differentiation and relate these to graphs
[Higher] Distinguish between maxima and minima by considering the general shape of the graph only
[Higher] Apply calculus to linear kinematics and to other simple practical problems

Akọmọ Ojú-ẹkọ

On a straight line, the gradient is constant. On a curve, the gradient changes from point to point. Calculus lets you calculate that changing gradient exactly, without drawing a tangent by hand.

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Ayẹwo Ẹkọ

Ekele diri gi maka imecha ihe karịrị na Calculus. Ugbu a na ị na-enyochakwa isi echiche na echiche ndị dị mkpa, ọ bụ oge iji nwalee ihe ị ma. Ngwa a na-enye ụdị ajụjụ ọmụmụ dị iche iche emebere iji kwado nghọta gị wee nyere gị aka ịmata otú ị ghọtara ihe ndị a kụziri.

Ị ga-ahụ ngwakọta nke ụdị ajụjụ dị iche iche, gụnyere ajụjụ chọrọ ịhọrọ otu n’ime ọtụtụ azịza, ajụjụ chọrọ mkpirisi azịza, na ajụjụ ede ede. A na-arụpụta ajụjụ ọ bụla nke ọma iji nwalee akụkụ dị iche iche nke ihe ọmụma gị na nkà nke ịtụgharị uche.

Jiri akụkụ a nke nyocha ka ohere iji kụziere ihe ị matara banyere isiokwu ahụ ma chọpụta ebe ọ bụla ị nwere ike ịchọ ọmụmụ ihe ọzọ. Ekwela ka nsogbu ọ bụla ị na-eche ihu mee ka ị daa mba; kama, lee ha anya dị ka ohere maka ịzụlite onwe gị na imeziwanye.

Differentiate y = 5x^3. A) 15x^2 B) 5x^2 C) 15x^3 D) 3x^2 Answer: A
What is the gradient of y = x^2 - 4x at x = 3? A) -3 B) 2 C) 6 D) 0 Answer: B
A stationary point occurs when: A) y = 0 B) x = 0 C) dy/dx = 0 D) dy/dx = 1 Answer: C
If s = 4t^2 + 3t, what is the velocity? A) 4t + 3 B) 8t + 3 C) 8t D) 4t^2 Answer: B
Differentiate y = 7. A) 7 B) 0 C) 1 D) 7x Answer: B