Test of Mathematics for University Admission (TMUA) Calculus Syllabus

State the equation of the normal to the curve $y = f(x)$ at the point $(a, f(a))$ (assuming $f'(a)\neq 0$).

$$y - f(a) = -\frac{1}{f'(a)}\,(x - a).$$

Find the gradient of the normal to $y = x^{2}$ at the point $(1, 1)$.

$\dfrac{dy}{dx} = 2x$, so the tangent gradient at $x=1$ is $2$. The normal gradient is $-\dfrac{1}{2}.$

What is a stationary point of a curve, defined in terms of the derivative?

A point where $\dfrac{dy}{dx} = 0$, i.e. the tangent is horizontal (gradient zero).

List the three types of stationary point.

Local maximum, local minimum, and point of inflection (horizontal/stationary inflection).

How do you find the coordinates of the stationary points of $y = f(x)$?

Solve $f'(x) = 0$ for $x$, then substitute each solution back into $y = f(x)$ to find the corresponding $y$-values.

State the second derivative test for classifying a stationary point at $x = a$ where $f'(a) = 0$.

If $f''(a) > 0$ it is a local minimum; if $f''(a) < 0$ it is a local maximum; if $f''(a) = 0$ the test is inconclusive (examine further).

When the second derivative test gives $f''(a) = 0$, what method can be used instead to classify the stationary point?

The first derivative (sign) test: examine the sign of $f'(x)$ just to the left and right of $x=a$ to see how the gradient changes.

Using the first derivative test, what sign pattern of $f'(x)$ around $x=a$ indicates a local maximum?

$f'(x)$ changes from positive (increasing) to negative (decreasing) as $x$ passes through $a$: $+ \to 0 \to -$.

Using the first derivative test, what sign pattern of $f'(x)$ indicates a local minimum?

$f'(x)$ changes from negative to positive: $- \to 0 \to +$.

What does $f'(x) > 0$ tell you about the function on an interval?

The function is increasing on that interval (the curve slopes upward).

What does $f'(x) < 0$ tell you about the function on an interval?

The function is decreasing on that interval (the curve slopes downward).

Find and classify the stationary point of $y = x^{2} - 4x + 1$.

$\dfrac{dy}{dx} = 2x - 4 = 0 \Rightarrow x = 2$, giving $y = -3$. Since $\dfrac{d^{2}y}{dx^{2}} = 2 > 0$, it is a local minimum at $(2, -3)$.

What does the concavity of a curve relate to in terms of the second derivative?

$f''(x) > 0$ means the curve is concave up (convex); $f''(x) < 0$ means concave down. A change of concavity occurs at a point of inflection.

Define a point of inflection.

A point where the curve changes concavity (from concave up to concave down or vice versa); $f''(x) = 0$ and changes sign there. It is stationary only if $f'(x)=0$ too.

Indefinite integration is the reverse of what operation?

Differentiation — it is antidifferentiation, finding a function whose derivative is the given integrand.

State the rule for indefinite integration of $x^{n}$.

$$\int x^{n}\,dx = \frac{x^{n+1}}{n+1} + c, \quad n \neq -1$$ where $c$ is the constant of integration.

Why must the constant of integration $c$ always be included in an indefinite integral?

Because differentiating any constant gives $0$, so infinitely many functions differing by a constant share the same derivative; $c$ accounts for all of them.

Evaluate $\displaystyle\int (6x^{2} - 4x + 3)\,dx$.

$2x^{3} - 2x^{2} + 3x + c.$

Evaluate $\displaystyle\int \sqrt{x}\,dx$.

$\displaystyle\int x^{1/2}\,dx = \dfrac{x^{3/2}}{3/2} + c = \dfrac{2}{3}x^{3/2} + c.$

How do you find the particular constant $c$ when given a point through which the integral curve passes?

Integrate to get the general solution, substitute the known point's coordinates to form an equation, and solve for $c$.

State the Fundamental Theorem of Calculus used to evaluate a definite integral.

$$\int_{a}^{b} f(x)\,dx = \big[F(x)\big]_{a}^{b} = F(b) - F(a),$$ where $F'(x) = f(x)$.

What key difference distinguishes a definite integral from an indefinite integral?

A definite integral has limits $a$ and $b$ and evaluates to a number; the constant $c$ cancels out. An indefinite integral yields a function (plus $c$).

Evaluate $\displaystyle\int_{1}^{3} 2x\,dx$.

$\big[x^{2}\big]_{1}^{3} = 9 - 1 = 8.$

How is the area under the curve $y = f(x)$ between $x = a$ and $x = b$ (with $f(x)\geq 0$) found?

$$A = \int_{a}^{b} f(x)\,dx.$$

What does a definite integral give when the curve lies below the $x$-axis over the interval?

A negative value. The actual area is the modulus (absolute value) of that integral; split the integral at the roots to handle regions above and below separately.

How do you find the area enclosed between two curves $y = f(x)$ (upper) and $y = g(x)$ (lower) from $x=a$ to $x=b$?

$$A = \int_{a}^{b} \big(f(x) - g(x)\big)\,dx,$$ integrating the upper curve minus the lower curve.

Outline the general method for solving an optimisation problem using calculus.

Express the quantity to optimise as a function of one variable (using any constraint to eliminate others), differentiate and set the derivative to zero, solve for the variable, then use the second derivative to confirm a maximum or minimum.

In the context of rates of change, what does $\dfrac{dV}{dt}$ represent?

The rate of change of volume $V$ with respect to time $t$ (how fast the volume is changing per unit time).

State the chain rule connection used to relate two rates of change, e.g. $\dfrac{dV}{dt}$ via $\dfrac{dV}{dr}$.

$$\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}.$$

How do you find the gradient of a graph at a point, and how is this connected to the area under its gradient function?

The gradient at a point is the derivative $f'(x)$; conversely, the area under the gradient function $f'(x)$ between $a$ and $b$ recovers the change in $f$: $\int_a^b f'(x)\,dx = f(b)-f(a)$. Differentiation and integration are inverse processes connecting a graph and its gradient/area.