krit.club logo

Application of Derivatives - Rate of change of quantities

Grade 12CBSE

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

🔑Concepts

•

Definition of Rate of Change: If a variable yy varies with another variable xx such that y=f(x)y = f(x), then the derivative dydx\frac{dy}{dx} represents the instantaneous rate of change of yy with respect to xx. Visually, this is represented as the slope of the tangent line to the curve y=f(x)y = f(x) at a specific point.

•

Rate of Change with Respect to Time: In many physical applications, quantities change with respect to time tt. If two variables xx and yy are both functions of tt, the rate of change of yy with respect to xx can be calculated using the chain rule as dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt}, provided dx/dt≠0\frac{dx/dt} \neq 0.

•

Increasing and Decreasing Quantities: A positive rate of change (dydt>0\frac{dy}{dt} > 0) indicates that the quantity yy is increasing over time, which corresponds to an upward-sloping graph. Conversely, a negative rate (dydt<0\frac{dy}{dt} < 0) indicates the quantity is decreasing, represented by a downward-sloping graph.

•

Geometric Interpretations: For geometric shapes, the rate of change of one dimension often relates to another. For example, for a circle where A=πr2A = \pi r^2, the rate dAdr=2πr\frac{dA}{dr} = 2\pi r is numerically equal to the circumference. This visualizes as the thin 'ring' added to the boundary as the circle expands.

•

Marginal Cost (MC): In economics, the marginal cost is the instantaneous rate of change of the total cost C(x)C(x) with respect to the number of items produced xx. It is mathematically expressed as MC=dCdxMC = \frac{dC}{dx}. On a graph of cost versus units, this is the steepness of the cost curve at a production level xx.

•

Marginal Revenue (MR): Marginal revenue is the instantaneous rate of change of total revenue R(x)R(x) with respect to the number of items sold xx, expressed as MR=dRdxMR = \frac{dR}{dx}. It represents the additional income generated by selling one additional unit at a specific sales volume.

•

Related Rates: When two related quantities (like the height and radius of water in a conical tank) change over time, their rates are linked by the derivative of the equation relating them. Visually, as the water level rises in a cone, the surface area of the top circle also increases, and their rates are coupled by the geometry of the container.

📐Formulae

dydx=lim⁡Δx→0ΔyΔx\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}

dydt=dydx⋅dxdt\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}

Area of Circle: A=πr2  ⟹  dAdt=2πrdrdtA = \pi r^2 \implies \frac{dA}{dt} = 2\pi r \frac{dr}{dt}

Volume of Sphere: V=43πr3  ⟹  dVdt=4πr2drdtV = \frac{4}{3}\pi r^3 \implies \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}

Surface Area of Sphere: S=4πr2  ⟹  dSdt=8πrdrdtS = 4\pi r^2 \implies \frac{dS}{dt} = 8\pi r \frac{dr}{dt}

Volume of Cube: V=x3  ⟹  dVdt=3x2dxdtV = x^3 \implies \frac{dV}{dt} = 3x^2 \frac{dx}{dt}

Marginal Cost: MC=ddx[C(x)]MC = \frac{d}{dx}[C(x)]

Marginal Revenue: MR=ddx[R(x)]MR = \frac{d}{dx}[R(x)]

💡Examples

Problem 1:

Find the rate of change of the area of a circle with respect to its radius rr when r=5r = 5 cm.

Solution:

Let AA be the area of the circle. We know that A=πr2A = \pi r^2. Differentiating AA with respect to rr, we get: dAdr=ddr(πr2)=2πr\frac{dA}{dr} = \frac{d}{dr}(\pi r^2) = 2\pi r. When r=5r = 5 cm, dAdr=2π(5)=10π\frac{dA}{dr} = 2\pi(5) = 10\pi. Thus, the rate of change of area is 10π10\pi cm2^2/cm.

Explanation:

Since the question asks for the rate of change with respect to the radius (not time), we differentiate the area formula directly with respect to rr and substitute the given value.

Problem 2:

The volume of a cube is increasing at the rate of 99 cubic centimeters per second. How fast is the surface area increasing when the length of an edge is 1010 centimeters?

Solution:

Let xx be the length of the side, VV be the volume, and SS be the surface area of the cube. Given dVdt=9\frac{dV}{dt} = 9 cm3^3/s. We know V=x3V = x^3. Differentiating with respect to tt: dVdt=3x2dxdt  ⟹  9=3x2dxdt  ⟹  dxdt=3x2\frac{dV}{dt} = 3x^2 \frac{dx}{dt} \implies 9 = 3x^2 \frac{dx}{dt} \implies \frac{dx}{dt} = \frac{3}{x^2}. Now, surface area S=6x2S = 6x^2. Differentiating SS with respect to tt: dSdt=12xdxdt\frac{dS}{dt} = 12x \frac{dx}{dt}. Substitute dxdt\frac{dx}{dt}: dSdt=12x(3x2)=36x\frac{dS}{dt} = 12x \left(\frac{3}{x^2}\right) = \frac{36}{x}. When x=10x = 10, dSdt=3610=3.6\frac{dS}{dt} = \frac{36}{10} = 3.6 cm2^2/s.

Explanation:

This is a related rates problem. First, find the rate of change of the side length (dxdt\frac{dx}{dt}) using the volume rate, then use that result to find the rate of change of the surface area.