Calculus and Differential Equations: Derivatives, Optimization, Areas, Volumes, and First-Order ODEs

Calculus questions on the MSTE paper are graded purely on whether your final number is right, and almost every one reduces to a short list of rules. This lesson walks through differentiation, the two classic applications (optimization and related rates), integration for area and volume, and first-order differential equations, the exact sequence the board tests.

Differentiation rules

Rule	Formula
Power rule	d/dx[x^n] = n x^(n-1)
Product rule	d/dx[uv] = u'v + uv'
Quotient rule	d/dx[u/v] = (u'v - uv') / v^2
Chain rule	d/dx[f(g(x))] = f'(g(x)) x g'(x)

Worked example (product rule): Differentiate f(x) = x^2 e^x. f'(x) = (2x)(e^x) + (x^2)(e^x) = e^x(x^2 + 2x). Worked example (chain rule): Differentiate f(x) = (3x^2 + 1)^4. Treat (3x^2 + 1) as the inner function: f'(x) = 4(3x^2 + 1)^3 x 6x = 24x(3x^2 + 1)^3. The same idea handles trig compositions: d/dx[cos(2x^2)] = -sin(2x^2) x 4x = -4x sin(2x^2).

Maxima and minima

A critical point occurs where f'(x) = 0. The second-derivative test then classifies it: f''(x) > 0 means a relative minimum, f''(x) < 0 means a relative maximum. Optimization word problems follow a fixed routine: write the quantity to optimize as a function of one variable, differentiate, set the derivative to zero, and check the second derivative or the domain endpoints.

Worked example: A rectangular field is fenced on three sides with 200 m of fencing, the fourth side runs along a river and needs no fence. Let x be each side perpendicular to the river and y the side parallel to it, so 2x + y = 200 and the area is A = xy = x(200 - 2x) = 200x - 2x^2. Setting dA/dx = 200 - 4x = 0 gives x = 50 m, y = 100 m, for a maximum area of 5,000 m^2.

Related rates

Related-rates problems connect two changing quantities through an equation, then use implicit differentiation with respect to time. The routine: write the governing equation, differentiate both sides with respect to t, substitute the known values, and solve for the unknown rate.

Worked example: A 10 m ladder leans against a wall, its base sliding away at 2 m/s. When the base is 6 m from the wall (so the top is at sqrt(100 - 36) = 8 m), the governing equation is x^2 + y^2 = 100. Differentiating: 2x(dx/dt) + 2y(dy/dt) = 0, so dy/dt = -(x/y)(dx/dt) = -(6/8)(2) = -1.5 m/s. The top slides down at 1.5 m/s.

Integral calculus: area and volume

The definite integral of f(x) from a to b gives the area under the curve (above the x-axis). For a region revolved about the x-axis, the disk method gives the volume: V = pi x integral from a to b of [f(x)]^2 dx.

Worked example (area): The area under y = x^2 from x = 0 to x = 3 is the integral of x^2, evaluated from 0 to 3: (3^3)/3 - 0 = 9. Worked example (volume): Revolve the region under y = x from x = 0 to x = 2 about the x-axis: V = pi x integral from 0 to 2 of x^2 dx = pi x [x^3/3] from 0 to 2 = pi(8/3) = 8 pi/3, about 8.38 cubic units. As a check, the region under y = 2x from x = 0 to x = 3 revolved about the x-axis forms a cone of radius 6 and height 3; the disk-method integral gives 36 pi, which matches the cone formula (1/3) pi r^2 h exactly.

First-order differential equations

A separable equation can be rearranged so all y-terms sit with dy and all x-terms sit with dx, then both sides are integrated. A linear first-order equation, dy/dx + P(x)y = Q(x), is solved with an integrating factor mu = e^(integral of P(x) dx).

Worked example (separable): Solve dy/dx = 2xy. Separate: dy/y = 2x dx. Integrating both sides: ln|y| = x^2 + C1, so y = C e^(x^2). This same pattern models exponential growth and decay: for dP/dt = kP with P(0) = 200 and P(5) = 400, the solution is P(t) = 200 e^(kt); since e^(5k) = 400/200 = 2, the population at t = 10 is P(10) = 200(e^(5k))^2 = 200(4) = 800.

Exam-day strategy

• Differentiate a composite function from the outside in: identify the outer function first, then multiply by the derivative of what is inside it.
• On optimization problems, always reduce to a function of a single variable before differentiating; a leftover second variable means you skipped the constraint equation.
• For related rates, write down every given rate and the target rate before touching the formula; a clear list prevents plugging a value into the wrong variable.
• Volume-of-revolution answers commonly carry pi; leave pi in the answer unless the item explicitly asks for a decimal approximation.
• For a differential equation, always apply the initial condition last, after integrating, to solve for the constant C.