MATH2010 revision route

Complementary Mock B: curves, coordinates and equality conditions · Complete worked solutions and marking

Use the notes to retrieve a method, the mock paper to choose it independently, and the marking guide to find the step that needs work. Start by explaining a definition from memory before doing a calculation. A correct numerical answer should follow from a stated geometric or analytic reason.

Mock A: 6 questions, 100 marks · Solutions and marking

1. Turn geometry into equations

A plane in $\mathbb R^3$ needs both a normal and a point. From independent directions $u,v$, form $n=u\times v$, then write $n\cdot(X-P)=0$. Check that the supplied point satisfies the final equation. The normal determines a perpendicular line; its intersection with the plane gives the projection. This also checks a distance computed by the plane-distance formula.

$$\begin{aligned} d(Q,\Pi)&=\frac{|n\cdot(Q-P)|}{\|n\|},\\ B&=Q-\frac{n\cdot(Q-P)}{\|n\|^2}n. \end{aligned}$$

The denominator differs between the distance and the projection vector. A scalar triple product has a sign because it records orientation; take its absolute value for volume. For coordinate conversion, fix the meaning and range of every angle. An upper-half-plane point has a polar angle in $(0,\pi)$; squared radius alone does not determine the point's position.

In the formulas, $Q$ is the point being projected, $P$ is a known point of the plane, and $B$ is the resulting foot. The displacement $Q-P$ and the normal $n$ each have three components. Their dot product is one scalar. Dividing that scalar by the normal's squared length and multiplying by $n$ produces a vector again: the part of the displacement perpendicular to the plane. Subtracting that part gives $B$. A distance, by comparison, is one nonnegative scalar, so its formula uses an absolute value and one power of the normal's length. Q1 walks through both computations and checks that they agree.

Revisit 1.1 vectors and inner products, 1.2 planes and volumes, and 1.3 coordinates. Test this with Q1 and Q2(c).

2. Describe the set before testing a function

For a point $a$, take every ball below to be centred at $a$. Interior means that some whole open ball stays inside the set. Closure means that every open ball meets the set. Boundary means that every open ball meets both the set and its complement. Drawing an outline is useful, but the inequalities decide which points belong to the set and which belong only to its closure.

$$\partial S=\overline S\setminus\operatorname{Int}S.$$

A set described by several inequalities may acquire several boundary pieces. For a half-annulus, there are circular arcs and straight segments on the axis. Check their endpoints too. To prove a set is not closed, identify a missing limit point and a sequence or path approaching it from the set. To prove it is not open, exhibit a point that has no contained open ball.

The ball tests take place in the surrounding plane. Intersecting every test ball with the set first would lose the very outside points that reveal a boundary. A point in a hole may be an exterior point rather than a boundary point: it can have an entire small ball disjoint from the set. Q2 uses this observation to decide exactly which portions of the axis belong to the boundary.

For path-connectedness, construct a continuous path between arbitrary points and verify containment along the whole path. A straight segment may cross a hole even when the set is path-connected. Polar interpolation can work where Cartesian interpolation fails.

Revisit 2.1 domains and point sets. Test this with Q2(a–b).

3. Separate evidence against a limit from proof of a limit

Two approaches with different limiting values disprove a joint limit. Many approaches with the same value do not prove it. Even agreement on every line leaves curved approaches untested. Iterated limits answer another question: they approach one coordinate at a time, with the other temporarily fixed.

To prove a joint limit, seek a bound that depends only on distance from the target. With $r=\sqrt{x^2+y^2}$, suppose that whenever $0<r<r_0$, the bound $|F(x,y)-L|\le Cr^\alpha$ holds, where $C>0$, $\alpha>0$ and $r_0>0$. An explicit epsilon–delta choice is:

$$\delta=\min\{r_0,(\varepsilon/C)^{1/\alpha}\}.$$

The choice comes from solving the error requirement. First prescribe the allowed output error $\varepsilon$. To make $Cr^\alpha<\varepsilon$, require $r<(\varepsilon/C)^{1/\alpha}$. The estimate itself is valid only when $r<r_0$, so both restrictions must hold; taking the smaller radius achieves that. The constants must be fixed independently of the approaching direction. A slope-dependent radius for each line would not control all nearby points together.

The estimate must hold for every domain point sufficiently near the target. In a polar calculation, a bound on the angular factor must be uniform in the angle. When extending a function continuously to a missing point, first establish the joint limit; the extension value is then forced.

When searching for a counterexample, inspect the sizes of the denominator's terms. In Q3, balancing its two terms identifies a curve on which the numerator has the same size. This supplies a reason for choosing a path, rather than a list of guesses. Agreement of two iterated limits is a weaker type of evidence because each one freezes a coordinate during its inner limit.

Revisit 2.2 sequences and multivariable limits and the continuous-extension parts of 2.3 continuity. Test this with Q3.

4. Keep derivatives and their hypotheses distinct

A partial derivative holds the other coordinates fixed. A gradient collects those partials at one point. Differentiability asks whether one linear map predicts all sufficiently small increments with an error small relative to their length:

$$\begin{aligned} f(a+h)&=f(a)+A h+E(h),\\ \frac{|E(h)|}{\|h\|}&\longrightarrow0. \end{aligned}$$

When a scalar function is differentiable, $A h=\nabla f(a)\cdot h$. The affine linearization includes the constant $f(a)$; the derivative map itself sends zero to zero. Continuous first partials on an open neighbourhood give a convenient sufficient differentiability condition. Existence of partials only at the point is much weaker.

Here $a$ is the base input, $h$ is the input displacement, and $E(h)$ is the error in the output. For a scalar function with two input coordinates, $A:\mathbb R^2\to\mathbb R$ combines the two displacement components into one predicted scalar change; it can be written as one row of two coefficients. The input length is $\|h\|$, while $A h$ and $E(h)$ are output quantities. The condition asks the error to become negligible compared with that input length. An error merely tending to zero is too weak. Q5 derives an exact error and then verifies the stronger ratio condition.

For mixed derivatives, compute in the indicated order. A $C^2$ function on an open neighbourhood has equal mixed second partials there. Merely obtaining equal values at a point does not check those hypotheses. Keep logarithm and denominator restrictions visible while identifying an open neighbourhood.

A sufficient condition provides one route to a proof. Failing to verify continuous partials is not itself a proof that a function is not differentiable. Return to the definition or find another sufficient argument. To disprove differentiability, one failed path for the uniquely forced linear approximation is decisive; to prove it by the definition, the normalized error must be controlled for all nearby increments.

Revisit 3.1 partial and mixed derivatives and 3.2 differentiability and linearization. Test this with Q4 and Q5. In Q5, rationalization gives an exact remainder, so an error estimate can be established directly.

5. Choose a direction and check the scale

For a differentiable scalar function and a unit vector $u$,

$$D_u f(a)=\nabla f(a)\cdot u.$$

This formula follows by applying the linear approximation to the displacement $h=su$. Differentiability gives $f(a+su)-f(a)=s\nabla f(a)\cdot u+E(su)$. Divide by the nonzero scalar $s$ to form the directional difference quotient. Because $\|u\|=1$, we have $\|su\|=|s|$, and therefore $|E(su)/s|=|E(su)|/\|su\|\to0$. The quotient's limit is the displayed dot product. Without the differentiability remainder condition, that error term has not been controlled, so this derivation is unavailable. Q6(c) shows why substituting partials into the formula prematurely can give a wrong answer.

If the question gives a nonzero direction vector $v$, use $u=v/\|v\|$ for a rate per unit distance. The dot product with $v$ itself corresponds to motion at parameter velocity $v$ and includes its speed. Cauchy–Schwarz identifies the extreme rates. When the gradient is nonzero, the increasing and decreasing unit directions are the normalized gradient and its negative. When it is zero, every unit directional rate is zero.

The unit restriction makes the comparison fair: every proposed direction uses the same travel distance. Otherwise speeding up in an increasing direction could make the rate arbitrarily large. A zero gradient says that all first-order directional rates vanish at that point; it does not say that nearby function values are constant.

All directional derivatives may exist while the function is not differentiable. In a candidate counterexample, compute the coordinate partials, test another direction, and compare with the linear map those partials would force. Alternatively, test the normalized remainder along a carefully chosen path.

Use the unit-directional-derivative and fastest-direction material in 3.3 directional derivatives. Test this with Q6.

Choose a method from the requested conclusion

These are different logical tasks. Q3(d) and Q6(c) ask you to correct an inference, so a numerical output alone does not complete the answer. State which implication fails and supply the missing or contradicting evidence.

Work, mark, repair

After repairing a part, explain why a tempting wrong method fails. That explanation is the final check that the method can be transferred to a new problem.