IB Maths AI HL · Paper 2

The spending in year $n$ (with 2021 as $n=1$) is $625\times 1.064^{n-1}$.

The model is not realistic for a long time: a constant yearly growth rate of 6.4% would make the cost very large over many years, and inflation, the dog's age and its appetite will not stay the same.

Expand $C=\frac{k^{2}x}{10}-\frac{3x^3}{1000}$ and differentiate.

The cafe does not lose money while $C\ge 0$, so the most coffee is at the right-hand intercept.

Midpoint of $B(6,0)$ and $D(2,2)$.

Use the distance formula with $Y(-1,3)$ and $Z(7,7)$.

Find XZ and use the cosine rule in triangle XYZ.

Use $\frac{1}{2}ab\sin C$ with the two sides next to the angle at Y.

A larger cell only means D is the closest shop to more land. It does not mean more people live there, and people choose where to shop for reasons other than distance.

Convert each pair to $\left(\frac{1}{T},\ln k\right)$, for example $(1.695\times 10^{-3},-7.601)$, and run linear regression.

Solve $\det (M-\lambda I)=0$.

For $\lambda =1$: $-0.06x+0.02y=0$, so $y=3x$. For $\lambda =0.92$: $0.02x+0.02y=0$, so $y=-x$.

Initial velocity is $(8,10)$.

The height is $y=10t-5t^2$, with maximum at $t=1$ (halfway between the roots 0 and 2).

Landing is when $y=0$.

Eliminate $t$ using $x=5+8t$.

The system matrix is $\left(\begin{array}{cc}0& 1\\ -2& -3\end{array}\right)$.

The graph starts at the origin, rises to one maximum and then decays towards the $t$-axis. For the maximum set $\frac{dx}{dt}=-e^{-t}+2e^{-2t}=0$, so $e^{-t}=\frac{1}{2}$ and $t=\ln 2$.

Solve $e^{-t}-e^{-2t}=0.1$ on a graph. The two solutions are $t\approx 0.1196$ and $t\approx 2.183$.

Stopping fishing for a little longer builds in a safety margin: the concentration may not be the same all along the river, and the cost of allowing fishing while it is unsafe is greater than the cost of stopping it slightly too long.