Mathematics - Additional - 0606 CIE

Rates Of Change And Small Increments

Gbogbo ọrọ náà

Differentiation measures how one quantity changes as another changes. But often two or more quantities change together over time, and we want to link their rates: if a balloon's radius grows, how fast does its volume grow? Connected rates of change answer exactly these questions using the chain rule.

In this lesson you will use the chain rule to connect rates, with \(\frac{dy}{dt}=\frac{dy}{dx}\times\frac{dx}{dt}\), and learn how a derivative estimates a small change through \(\delta y \approx \frac{dy}{dx}\,\delta x\). Together they show how calculus handles change that ripples from one quantity to another.

Ebumnobi

  1. Apply differentiation to connected rates of change, small increments and approximations.

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Akwụkwọ Ọmụmụ

Most real change is connected. As a tank fills, the depth rises and the surface area changes; as a price shifts, demand responds. Connected rates of change let you find one rate from another using a relationship between the quantities. Small increments then let you estimate the effect of a tiny change quickly, a powerful approximation used across science and engineering.

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Ekele diri gi maka imecha ihe karịrị na Rates Of Change And Small Increments. 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.

  1. The chain rule for connected rates states that dy/dt equals: A. dy/dx + dx/dt B. dy/dx x dx/dt C. dx/dt / dy/dx D. dy/dx - dx/dt Answer: B
  2. The area of a square is A = x^2. If x increases at 0.5 cm/s, what is dA/dt when x = 4? A. 2 cm^2/s B. 4 cm^2/s C. 8 cm^2/s D. 16 cm^2/s Answer: B
  3. The small-increment approximation is: A. delta y = dy/dx + delta x B. delta y ~ (dy/dx) delta x C. delta y ~ (dx/dy) delta x D. delta y = delta x / (dy/dx) Answer: B
  4. For y = x^3, estimate the change in y when x increases from 2 by delta x = 0.01. A. 0.03 B. 0.06 C. 0.12 D. 1.2 Answer: C
  5. In connected rates, dy/dx is found by: A. Integrating the relationship B. Differentiating the relationship between the quantities C. Dividing the two rates D. Adding the two rates Answer: B

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