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Calculus - Derivatives as a rate of change

Grade 11ICSE

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

🔑Concepts

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The derivative dydx\frac{dy}{dx} represents the instantaneous rate of change of a dependent variable yy with respect to an independent variable xx. Geometrically, this is visualized as the slope of the tangent line to the curve y=f(x)y = f(x) at any specific point. If the tangent is steep and pointing upwards, the rate of change is high and positive.

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The Average Rate of Change is defined over a specific interval [x1,x2][x_1, x_2] and is calculated as ΔyΔx=f(x2)−f(x1)x2−x1\frac{\Delta y}{\Delta x} = \frac{f(x_2) - f(x_1)}{x_2 - x_1}. Visually, this corresponds to the slope of the secant line passing through two distinct points on a graph, whereas the instantaneous rate of change is the limit as the secant points merge into a single tangent point.

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In Kinematics, if displacement ss is a function of time tt, the first derivative v=dsdtv = \frac{ds}{dt} represents the instantaneous velocity. On a displacement-time graph, a horizontal tangent indicates the object is momentarily at rest (v=0v = 0), while the second derivative a=dvdt=d2sdt2a = \frac{dv}{dt} = \frac{d^2s}{dt^2} represents acceleration, showing how the velocity's slope is changing.

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Related Rates involve equations where multiple variables change with respect to time. By using the Chain Rule, such as dAdt=dAdr⋅drdt\frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt}, we can find the rate of change of one quantity (like the area of a ripple in a pond) based on the rate of change of another (like the radius). Visually, imagine a circle expanding; the rate at which the area 'fills' depends on how fast the boundary moves outward.

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Marginal Functions in Economics utilize derivatives to describe the rate of change of cost, revenue, or profit. Marginal Cost MC=dCdxMC = \frac{dC}{dx} represents the instantaneous rate of change of total cost CC with respect to the number of units xx produced. On a cost curve, the marginal cost at a point xx is the slope of the curve, representing the approximate cost of producing the next single unit.

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The derivative can also be used to find small errors or approximations. The change in yy, denoted Δy\Delta y, can be approximated using the formula Δy≈f′(x)Δx\Delta y \approx f'(x) \Delta x. This is visually represented by moving along the tangent line instead of the curve for a very small step Δx\Delta x, where the tangent line provides a linear approximation of the function's value.

📐Formulae

dydx=lim⁡Δx→0f(x+Δx)−f(x)Δx\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}

v=dsdtv = \frac{ds}{dt}

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

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

Rate of change of Area of a Circle: dAdt=2πrdrdt\text{Rate of change of Area of a Circle: } \frac{dA}{dt} = 2\pi r \frac{dr}{dt}

Rate of change of Volume of a Sphere: dVdt=4πr2drdt\text{Rate of change of Volume of a Sphere: } \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}

Δy≈f′(x)Δx\Delta y \approx f'(x) \Delta x

💡Examples

Problem 1:

The displacement ss of a particle at time tt is given by s=2t3−15t2+36t+10s = 2t^3 - 15t^2 + 36t + 10. Find the velocity of the particle when the acceleration is zero.

Solution:

  1. Find velocity vv by differentiating ss with respect to tt: v=dsdt=ddt(2t3−15t2+36t+10)=6t2−30t+36v = \frac{ds}{dt} = \frac{d}{dt}(2t^3 - 15t^2 + 36t + 10) = 6t^2 - 30t + 36
  2. Find acceleration aa by differentiating vv with respect to tt: a=dvdt=ddt(6t2−30t+36)=12t−30a = \frac{dv}{dt} = \frac{d}{dt}(6t^2 - 30t + 36) = 12t - 30
  3. Set acceleration to zero to find the time tt: 12t−30=0  ⟹  12t=30  ⟹  t=3012=2.5 seconds12t - 30 = 0 \implies 12t = 30 \implies t = \frac{30}{12} = 2.5 \text{ seconds}
  4. Substitute t=2.5t = 2.5 into the velocity equation: v=6(2.5)2−30(2.5)+36v = 6(2.5)^2 - 30(2.5) + 36 v=6(6.25)−75+36=37.5−75+36=−1.5v = 6(6.25) - 75 + 36 = 37.5 - 75 + 36 = -1.5 The velocity is −1.5-1.5 units/sec.

Explanation:

We first find the general expressions for velocity and acceleration using derivatives. We then solve for the specific time tt where acceleration is null and use that timestamp to calculate the specific velocity.

Problem 2:

The radius of a circular blot of ink is increasing at the rate of 0.50.5 cm/s. Find the rate of increase of its area when the radius is 44 cm.

Solution:

  1. Let rr be the radius and AA be the area of the circle. We are given drdt=0.5\frac{dr}{dt} = 0.5 cm/s.
  2. The formula for the area of a circle is A=πr2A = \pi r^2.
  3. Differentiate AA with respect to time tt using the chain rule: dAdt=dAdr⋅drdt\frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt} dAdt=ddr(πr2)⋅drdt=2πrdrdt\frac{dA}{dt} = \frac{d}{dr}(\pi r^2) \cdot \frac{dr}{dt} = 2\pi r \frac{dr}{dt}
  4. Substitute the given values r=4r = 4 and drdt=0.5\frac{dr}{dt} = 0.5: dAdt=2⋅π⋅4⋅0.5\frac{dA}{dt} = 2 \cdot \pi \cdot 4 \cdot 0.5 dAdt=4π cm2/s\frac{dA}{dt} = 4\pi \text{ cm}^2/\text{s} The area is increasing at a rate of 4π4\pi (or approx 12.5712.57) cm2/s\text{cm}^2/\text{s}.

Explanation:

This is a related rates problem. We identify the geometric relationship between area and radius, differentiate with respect to time using the chain rule, and substitute the known rates and dimensions.