Calculus: applications of differentiation (gradients, maxima and minima) — KCSE Mathematics

In calculus, optimisation involves finding the maximum or minimum values of a function. To solve these problems, we apply differentiation. Here’s a step-by-step approach:

1. Identify the function: Determine the function that represents the situation.
2. Differentiate the function: Find the first derivative to determine the rate of change.
3. Set the derivative to zero: Solve for critical points by setting the first derivative equal to zero.
4. Determine maxima or minima: Use the second derivative test or evaluate the function at critical points to confirm whether each point is a maximum or minimum.

Example Problem: A farmer wants to create a rectangular pen with a fixed perimeter of 100 meters. What dimensions will maximise the area?

Solution:

1. Let the length be x meters and the width be y meters. Then, the perimeter P = 2x + 2y = 100.
2. Rearranging gives y = 50 - x.
3. Area A = x * y = x(50 - x) = 50x - x².
4. Differentiate A: A' = 50 - 2x.
5. Set A' = 0: 50 - 2x = 0 → x = 25.
6. Width y = 50 - 25 = 25.
7. Thus, dimensions that maximise area are 25m by 25m.

To find the area under a curve defined by a function between two limits, we use definite integrals. The definite integral of a function f(x) from a to b is denoted as ( \int_{a}^{b} f(x) , dx ). This integral calculates the net area between the curve and the x-axis from x = a to x = b.

Steps to evaluate a definite integral:

1. Find the antiderivative of the function f(x).
2. Evaluate the antiderivative at the upper limit (b) and lower limit (a).
3. Subtract the value at the lower limit from the value at the upper limit.

Example: Evaluate the area under the curve ( f(x) = 2x^2 + 3 ) from x = 1 to x = 3.

1. Find the antiderivative: ( F(x) = \frac{2}{3}x^3 + 3x + C )
2. Evaluate at the limits:
   • At x = 3: ( F(3) = \frac{2}{3}(3)^3 + 3(3) = 18 + 9 = 27 )
   • At x = 1: ( F(1) = \frac{2}{3}(1)^3 + 3(1) = \frac{2}{3} + 3 = \frac{11}{3} )
3. Calculate the area: ( Area = F(3) - F(1) = 27 - \frac{11}{3} = \frac{81}{3} - \frac{11}{3} = \frac{70}{3} ).

Thus, the area under the curve from x = 1 to x = 3 is ( \frac{70}{3} ).

In calculus, differentiation helps us find gradients, maxima, and minima of functions. The gradient of a function at a point gives the slope of the tangent line at that point. To find the gradient, we differentiate the function.

Maxima and Minima are points where a function reaches its highest or lowest values, respectively. To find these points:

1. Differentiate the function to obtain f'(x).
2. Set f'(x) = 0 and solve for x to find critical points.
3. Use the second derivative test: if f''(x) > 0, it's a local minimum; if f''(x) < 0, it's a local maximum.

For example, consider f(x) = x² - 4x + 4.

• First, find the derivative: f'(x) = 2x - 4.
• Set it to zero: 2x - 4 = 0, giving x = 2.
• Now find f''(x) = 2, which is greater than 0. Thus, x = 2 is a local minimum.