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Calculus - Rate of change problems

Grade 11IB_AA

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

🔑Concepts

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The average rate of change of a function f(x)f(x) over the 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)), calculated as f(b)−f(a)b−a\frac{f(b) - f(a)}{b - a}.

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The instantaneous rate of change of yy with respect to xx at a specific point x=ax = a is the derivative f′(a)f'(a). Geometrically, this represents the gradient of the tangent to the curve at that point.

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In kinematics, if displacement is s(t)s(t), then the instantaneous velocity is v(t)=s′(t)=dsdtv(t) = s'(t) = \frac{ds}{dt} and the instantaneous acceleration is a(t)=v′(t)=s′′(t)=dvdta(t) = v'(t) = s''(t) = \frac{dv}{dt}.

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Related rates problems involve finding the rate at which one quantity changes by relating it to other quantities whose rates of change are known. This typically requires the use of the Chain Rule: dydt=dydx×dxdt\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}.

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To solve related rates problems: 1) Identify the given variables and their rates. 2) Find an equation relating the variables. 3) Differentiate both sides with respect to time tt using the chain rule. 4) Substitute the known values to find the required rate.

📐Formulae

Average Rate of Change=f(x+h)−f(x)h\text{Average Rate of Change} = \frac{f(x+h) - f(x)}{h}

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

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

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

dydt=dydx×dxdt\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt}

💡Examples

Problem 1:

The radius rr of a circular oil spill is increasing at a constant rate of 2 m/s2 \text{ m/s}. Find the rate at which the area AA of the spill is increasing when the radius is 10 m10 \text{ m}.

Solution:

Let AA be the area and rr be the radius. We know A=πr2A = \pi r^2. Differentiating both sides with respect to time tt: dAdt=dAdr×drdt\frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt} dAdt=(2πr)×drdt\frac{dA}{dt} = (2\pi r) \times \frac{dr}{dt} Given drdt=2 m/s\frac{dr}{dt} = 2 \text{ m/s} and r=10 mr = 10 \text{ m}: dAdt=2π(10)×2=40π m2/s\frac{dA}{dt} = 2\pi(10) \times 2 = 40\pi \text{ m}^2\text{/s}

Explanation:

We use the area of a circle formula and apply the chain rule because the radius is a function of time. Multiplying the derivative of the area with respect to the radius by the rate of change of the radius gives the rate of change of the area.

Problem 2:

A particle moves along a straight line such that its displacement ss in meters at time tt seconds is given by s(t)=t3−6t2+9ts(t) = t^3 - 6t^2 + 9t. Find the velocity of the particle when t=2t = 2.

Solution:

The velocity v(t)v(t) is the derivative of the displacement s(t)s(t). v(t)=s′(t)=3t2−12t+9v(t) = s'(t) = 3t^2 - 12t + 9 Substituting t=2t = 2: v(2)=3(2)2−12(2)+9v(2) = 3(2)^2 - 12(2) + 9 v(2)=12−24+9=−3 m/sv(2) = 12 - 24 + 9 = -3 \text{ m/s}

Explanation:

To find the instantaneous velocity at a specific time, we differentiate the displacement function and evaluate it at that time. A negative velocity indicates the particle is moving in the opposite direction to the positive displacement.

Problem 3:

The volume VV of a cube is increasing at a rate of 12 cm3/s12 \text{ cm}^3\text{/s}. Find the rate of change of the side length xx when x=2 cmx = 2 \text{ cm}.

Solution:

The volume of a cube is V=x3V = x^3. Differentiating with respect to tt: dVdt=dVdx×dxdt\frac{dV}{dt} = \frac{dV}{dx} \times \frac{dx}{dt} dVdt=3x2×dxdt\frac{dV}{dt} = 3x^2 \times \frac{dx}{dt} Given dVdt=12\frac{dV}{dt} = 12 and x=2x = 2: 12=3(2)2×dxdt12 = 3(2)^2 \times \frac{dx}{dt} 12=12×dxdt12 = 12 \times \frac{dx}{dt} dxdt=1 cm/s\frac{dx}{dt} = 1 \text{ cm/s}

Explanation:

By relating the volume of the cube to its side length and differentiating with respect to time, we can solve for the unknown rate of change of the side length.