krit.club logo

Calculus - Rates of change

Grade 11IB_AI

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

The average rate of change of a function f(x)f(x) over an interval [a,b][a, b] is the gradient of the secant line passing through (a,f(a))(a, f(a)) and (b,f(b))(b, f(b)).

•

The instantaneous rate of change of a function at a specific point is the gradient of the tangent line at that point, represented by the derivative f′(x)f'(x) or dydx\frac{dy}{dx}.

•

In kinematics, if s(t)s(t) is the displacement at time tt, then the velocity v(t)v(t) is the first derivative s′(t)s'(t) and the acceleration a(t)a(t) is the second derivative s′′(t)s''(t) or v′(t)v'(t).

•

In economics, marginal cost and marginal revenue are the rates of change (derivatives) of the total cost and total revenue functions with respect to the number of items produced/sold.

•

The sign of the rate of change indicates direction: a positive rate means the quantity is increasing, while a negative rate means the quantity is decreasing.

📐Formulae

Average Rate of Change=f(b)−f(a)b−a\text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a}

Instantaneous Rate of Change=lim⁡h→0f(x+h)−f(x)h\text{Instantaneous Rate of Change} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

ddx(xn)=nxn−1\frac{d}{dx}(x^n) = nx^{n-1}

v(t)=dsdtv(t) = \frac{ds}{dt}

a(t)=dvdt=d2sdt2a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2}

💡Examples

Problem 1:

A ball is thrown upwards, and its height hh in meters after tt seconds is given by the function h(t)=−5t2+20t+2h(t) = -5t^2 + 20t + 2. Find the instantaneous velocity of the ball at t=3t = 3 seconds.

Solution:

First, find the derivative of the height function to get the velocity function: v(t)=h′(t)=ddt(−5t2+20t+2)v(t) = h'(t) = \frac{d}{dt}(-5t^2 + 20t + 2) v(t)=−10t+20v(t) = -10t + 20 Now, substitute t=3t = 3 into the velocity function: v(3)=−10(3)+20v(3) = -10(3) + 20 v(3)=−30+20=−10 m/sv(3) = -30 + 20 = -10 \text{ m/s}

Explanation:

The derivative of displacement (height) gives the velocity. A negative velocity indicates the ball is moving downwards at 10 m/s10 \text{ m/s} at that specific moment.

Problem 2:

Find the average rate of change of the function f(x)=x2+3xf(x) = x^2 + 3x over the interval [1,4][1, 4].

Solution:

Calculate the values of the function at the endpoints of the interval: f(1)=(1)2+3(1)=4f(1) = (1)^2 + 3(1) = 4 f(4)=(4)2+3(4)=16+12=28f(4) = (4)^2 + 3(4) = 16 + 12 = 28 Use the average rate of change formula: Average Rate=f(4)−f(1)4−1\text{Average Rate} = \frac{f(4) - f(1)}{4 - 1} Average Rate=28−43\text{Average Rate} = \frac{28 - 4}{3} Average Rate=243=8\text{Average Rate} = \frac{24}{3} = 8

Explanation:

The average rate of change is the slope of the line connecting the points (1,4)(1, 4) and (4,28)(4, 28) on the graph of the function.

Problem 3:

The total cost CC (in dollars) of producing xx units is given by C(x)=0.5x2+10x+100C(x) = 0.5x^2 + 10x + 100. Find the marginal cost when x=20x = 20.

Solution:

The marginal cost is the derivative of the cost function C′(x)C'(x): C′(x)=ddx(0.5x2+10x+100)C'(x) = \frac{d}{dx}(0.5x^2 + 10x + 100) C′(x)=1.0x+10C'(x) = 1.0x + 10 Substitute x=20x = 20: C′(20)=1.0(20)+10=30C'(20) = 1.0(20) + 10 = 30 The marginal cost is 3030 per unit.

Explanation:

The marginal cost represents the approximate cost of producing one additional unit (the 21st unit) when the current production level is 20.