Rates Of Change And Small Increments
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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.
Awọn Afojusun
- Apply differentiation to connected rates of change, small increments and approximations.
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Oriire fun ipari ẹkọ lori Rates Of Change And Small Increments. Ni bayi ti o ti ṣawari naa awọn imọran bọtini ati awọn imọran, o to akoko lati fi imọ rẹ si idanwo. Ẹka yii nfunni ni ọpọlọpọ awọn adaṣe awọn ibeere ti a ṣe lati fun oye rẹ lokun ati ṣe iranlọwọ fun ọ lati ṣe iwọn oye ohun elo naa.
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- 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
- 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
- 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
- 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
- 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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