SL 2.10—Solving equations graphically and analytically

A particle changes direction when its velocity changes sign, i.e. when $v(t)=0$. We need to find how many positive roots $e^{-2t}-3t^2{e}^{-t}=0$ has.

Factor out $e^{-t}$ (which is always positive): $e^{-t}(e^{-t}-3t^2)=0$. Since $e^{-t}>0$ for all $t$, we need $e^{-t}=3t^2$.

Plot $y_1=e^{-t}$ and $y_2=3t^2$ on a GDC for $t>0$. At $t=0$: $y_1=1$, $y_2=0$, so $y_1>y_2$. As $t\to \infty$: $y_1\to 0$ and $y_2\to \infty$, so $y_2>y_1$. The curves must cross at least once.

Using the GDC, the graphs of $y_1=e^{-t}$ and $y_2=3t^2$ intersect at two points for $t>0$. This means $v(t)=0$ has two positive solutions, so the particle changes direction twice.

Since there are two sign changes in $v(t)$, the particle changes direction twice.

We need the number of times $v(t)$ changes sign for $t>0$, which corresponds to positive roots of $v(t)=0$.

Never is incorrect because $v(0)=1>0$ and as $t\to \infty$, $v(t)\to 0^{-}$ (since $-3t^2{e}^{-t}$ eventually dominates from below zero), so there must be at least one sign change.

Once would be true if there were exactly one crossing, but graphical analysis shows the exponential $e^{-2t}$ and $3t^2{e}^{-t}$ intersect at two values of $t>0$, giving two roots.

Three times would require three positive roots. Graphical inspection confirms only two intersection points, so three changes is an overcount.

By elimination and graphical confirmation, the velocity changes sign exactly twice for $t>0$.

Using the product rule: $r(t)=-e^{-t}\ln (t^2+4)+e^{-t}\cdot \frac{2t}{t^2+4}$. Factor out $e^{-t}$: $r(t)=e^{-t}\left(\frac{2t}{t^2+4}-\ln (t^2+4)\right)$.

Since $e^{-t}>0$ always, $r(t)=0$ requires $\frac{2t}{t^2+4}=\ln (t^2+4)$. This equation mixes rational and logarithmic functions, so a graphical/GDC approach is appropriate.

At $t=0$: LHS $=0$, RHS $=\ln 4\approx 1.39$, so RHS $>$ LHS. As $t\to \infty$: LHS $\to 0$ and RHS $\to \infty$, so RHS $>$ LHS. There is a region in between where LHS may exceed RHS.

Plotting both sides on a GDC shows the curves intersect at two values of $t>0$, giving two zeros of $r(t)$.

There are 2 values of $t>0$ for which $r(t)=0$.

We need the number of positive values of $t$ where the derivative $r(t)=0$, i.e. stationary points of $s(t)$ for $t>0$.

0 solutions is incorrect: $s(t)$ increases initially (since $s^{\prime }(0^{+})>0$... checking: $r(0)=e^0(0-\ln 4)<0$, wait—$r(t)$ is negative at $t=0$ and the function must have turning points). The function is non-trivial, so dismissing all stationary points is unreasonable without full verification.

3 solutions would mean three turning points, which is not supported by a GDC graph of $y=r(t)$, where only two zero-crossings are visible.

1 solution would suggest only one turning point. But the shape of $s(t)$ — rising then falling then levelling — typically produces two zeros of its derivative, as confirmed by technology.

The GDC confirms exactly 2 positive values of $t$ where $r(t)=0$.