Sine Problem

In this 'mesh' of sine graphs, one of the graphs is the graph of the sine function. Find the equations of the other graphs to reproduce the pattern.

Cocked Hat

Sketch the graphs for this implicitly defined family of functions.

Folium of Descartes

Investigate the family of graphs given by the equation x^3+y^3=3axy for different values of the constant a.

Quartics

Stage: 5 Challenge Level:

Congratulations to Aleksander Twarowski from Gdynia Bilingual High School No 3, Poland for solving this Tough Nut. Well done!

We can observe that for $t=2$ it has 3 stationary points (derivative is equal to 0), 2 minima and 1 maximum. It is an even function symmetrical about the $y$-axis. The function and its graph are the same for $t=-1/2$ as for $t=1/2$.

Let's expand this formula: \eqalign{ y&=[1+(x-t)^2][1+(x+t)^2] \cr &= [x^2 + (1+t^2) - 2tx][x^2 +(1+t^2)+2tx]\cr &= x^4 +2x^2(1+t^2)+(1+t^2)^2 - 4t^2x^2\cr &= x^4 +2(1-t^2)x^2 +(1+t^2)^2} When we find its derivative with respect to $x$ we obtain $$y'=4x^3+4x(1-t^2)= 4x[x^2+(1-t)(1+t)].$$ Now we can investigate stationary points when $$4x[x^2+(1-t)(1+t)]=0.$$ It is important to observe that $[x^2+(1-t)(1+t)]$ can be factorized only when $(1-t^2)$ is negative or zero, that is $t^2\geq 1$. So there are three stationary points when $t$ belongs to interval from negative infinity to $-1$ and from $1$ to positive infinity. Hence, we can state that when $t$ belongs to $[-1,1]$ the graph looks like it does for $t=1/2$ above, and when $t$ belongs to $(-\infty,-1)$ or $(1,+\infty)$ the graphs have three stationary points and look like the one for $t=2$ above. Because these three intervals include all real values of $t$, the graphs cannot have other shapes.