Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2007 / Paper 2 / Section A

The cross product is $\mathbf{a}\times \mathbf{b}=(-2,0,2)$, which is perpendicular to both vectors. Any nonzero scalar multiple works, so a simple answer isIts dot product with each given vector is zero.

The product-to-sum identity for the sine gives $\sin 2x\sin 4x=\frac{1}{2}(\cos 2x-\cos 6x)$. Each cosine has zero integral on this interval, soThis is also an instance of Fourier orthogonality.

Using $d({\sin }^{2}x)/dx=2\sin x\cos x$, the definite integral is

The gradient is $(f_x,f_y)=(2-2xy,1-x^2)$. Both partial derivatives vanish at $(1,1)$, so it is a stationary point.

The second component of the gradient forces $x=\pm 1$. The first then requires $xy=1$. Apart from $(1,1)$, the only stationary point is

Taking the negative gradient of the potential energy gives the force

The divergence is the sum of the three diagonal partial derivatives:

Each component depends only on its own coordinate, so all cross partial derivatives vanish. Therefore the curl isThis also follows from the identity that the curl of a gradient vanishes for a twice continuously differentiable potential energy.

The matrix product of the transposed first vector with the second is their dot product:

The reverse arrangement is an outer product, with entries $a_i{b}_j$:

The two factors have moduli $\sqrt{2}$ and $2$, and complex arguments $\pi /4$ and $\pi /6$. Multiplication multiplies the moduli and adds the complex arguments, givingThe displayed angle is the principal argument in radians; equivalently the argument is $75^{\circ }$ modulo $360^{\circ }$.

The logarithmic series gives the first two nonzero terms:

For $x_0>0$, separation of variables gives $1/x=kt+1/x_0$, hence $x(t)=x_0/(1+k{x}_{0}t)$. Setting $x(t)=x_0/2$ gives the half-lifeThis is the same reciprocal decay law as a second-order reaction with the specified disappearance rate coefficient. A positive initial value is needed for a positive half-time; $x_0=0$ instead gives the identically zero solution.

There are $\left(\genfrac{}{}{0ex}{}{7}{2}\right)=21$ equally likely unordered pairs, and $3\cdot 4=12$ contain one ball of each colour. Thus the hypergeometric distribution gives

Let $I_1,I_2$ be indicator random variables for the two draws being red. Each draw has red probability $3/7$, even though the draws are dependent. By linearity of expectation, the number $R=I_1+I_2$ satisfies