Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 12

Table of contents
- a 12
  - Solution a
- b 12
  - Solution b
- c 12
  - Solution c
- d 12
  - Solution d
- e 12
  - i e
    - Solution i
  - ii e
    - Solution ii
  - iii e
    - Solution iii

a

Solution

For

r > 0

, differentiate

r = x^{2} + y^{2}

while keeping the other coordinate fixed in the Cartesian coordinate system:

r_{x} = \frac{x}{r} = cos ϕ, r_{y} = \frac{y}{r} = sin ϕ .

(1)

The origin is excluded from this polar coordinates chart.

b

 0  0

Solution

 0  0

Differentiate the angular relation locally, or use

d ϕ = (x d y - y d x) / (x^{2} + y^{2})

. Holding each other coordinate fixed gives

ϕ_{x} = - \frac{y}{r ^{2}} = - \frac{sin ϕ}{r}, ϕ_{y} = \frac{x}{r ^{2}} = \frac{cos ϕ}{r} .

(1)

These are local partial derivatives on a consistent angular branch, independent of where its branch cut is placed.

c

 0  0

Solution

 0  0

Apply the multivariable chain rule to

f (x, y) = g (r (x, y), ϕ (x, y))

. Using the coordinate partial derivatives above gives

f_{x} = g_{r} cos ϕ - \frac{g _{ϕ}}{r} sin ϕ, f_{y} = g_{r} sin ϕ + \frac{g _{ϕ}}{r} cos ϕ .

(1)

Equivalently,

\partial_{x} = cos ϕ \partial_{r} - (sin ϕ / r) \partial_{ϕ}

and

\partial_{y} = sin ϕ \partial_{r} + (cos ϕ / r) \partial_{ϕ}

. Their coefficients vary with position and must also be differentiated when applying them twice.

d

 0  0

Solution

 0  0

Write

c = cos ϕ

and

s = sin ϕ

. The needed coefficient derivatives are

c_{x} = s^{2} / r

s_{x} = - sc / r

c_{y} = - sc / r

and

s_{y} = c^{2} / r

. Apply the product rule to both terms of

f_{x} = c g_{r} - (s / r) g_{ϕ}

and use the chain rule for the derivatives of

g

. Collecting terms gives

f_{xx} = c^{2} g_{rr} - \frac{2 sc}{r} g_{r ϕ} + \frac{2 sc}{r ^{2}} g_{ϕ} + \frac{s ^{2}}{r} g_{r} + \frac{s ^{2}}{r ^{2}} g_{ϕϕ} .

(1)

Similarly, differentiating

f_{y} = s g_{r} + (c / r) g_{ϕ}

gives

f_{yy} = s^{2} g_{rr} + \frac{2 sc}{r} g_{r ϕ} - \frac{2 sc}{r ^{2}} g_{ϕ} + \frac{c ^{2}}{r} g_{r} + \frac{c ^{2}}{r ^{2}} g_{ϕϕ} .

(2)

On addition, both the mixed partial derivative terms and the first angular derivative terms cancel. Using

s^{2} + c^{2} = 1

, the Laplacian in polar coordinates is

\nabla^{2} f = g_{rr} + \frac{1}{r} g_{r} + \frac{1}{r ^{2}} g_{ϕϕ}, r > 0.

(3)

e

 0  0

i

 0  0

Solution

 0  0

The double-angle formula gives

r^{2} sin 2 ϕ = 2 (r cos ϕ) (r sin ϕ) = 2 x y

. Therefore

f (x, y) = 2 x y, \nabla f = (2 y, 2 x) .

(1)

ii

 0  0

Solution

 0  0

In the positive quadrant, both components of the gradient are positive. Along

y = x

the arrows point along

(1, 1)

; near the horizontal axis they point mainly upwards, and near the vertical axis mainly rightwards. The level sets are the hyperbolas

x y = f /2

, including the coordinate axes for level zero. The gradient is orthogonal to each regular level set and points towards increasing values of

f

Figure 1.
Gradient vectors and hyperbolic level sets of twice the coordinate product
. **Details**.

iii

 0  0

Solution

 0  0

(1, 2)

, the gradient is

(4, 2)

. The requested direction must be normalized to the unit vector

(1, 1) / 2

. The directional derivative is therefore

D_{(1, 1) / 2} f (1, 2) = \nabla f (1, 2) \cdot \frac{( 1 , 1 )}{2} = 32 .

(1)

Using

(1, 1)

without normalization would instead measure change per unit of a differently scaled path parameter.

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 12

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