Daily math problems for October 5, 2026 — Analytic Geometry, Calculus, Trigonometry & more

$\mathrm{\parallel }\vec{v}\mathrm{\parallel }=\sqrt{1+4+4}=\sqrt{9}=3$.

Cylinder of radius $2$ and height $5$: $V=\pi \cdot 4\cdot 5=20\pi$.

In the second quadrant ($\pi \mathrm{/}2<x<\pi$), $\sin x>0$.

${\displaystyle \frac{7\pi }{4}}\in \text{\hspace{0.17em}⁣}({\displaystyle \frac{3\pi }{2}},2\pi )$, which is the fourth quadrant.

${\displaystyle \frac{72}{7}}\approx 10.3$ years.

$\bullet$ Differentiate by the chain rule:

$$f^{\mathrm{\prime }}(x)=6\pi \cos (2\pi x+\frac{\pi }{4})$$

$\bullet$ Differentiate again:

$$f^{\mathrm{\prime }\mathrm{\prime }}(x)=-12{\pi }^2\sin (2\pi x+\frac{\pi }{4})$$

$\bullet$ Evaluate at $x=\pi \mathrm{/}3$ and $x=\pi \mathrm{/}6$:

$$f^{\mathrm{\prime }}\text{\hspace{0.17em}⁣}\left(\frac{\pi }{3}\right)=6\pi \cos \text{\hspace{0.17em}⁣}(\frac{2{\pi }^2}{3}+\frac{\pi }{4})$$

$$f^{\mathrm{\prime }\mathrm{\prime }}\text{\hspace{0.17em}⁣}\left(\frac{\pi }{6}\right)=-12{\pi }^2\sin \text{\hspace{0.17em}⁣}(\frac{{\pi }^2}{3}+\frac{\pi }{4})$$

$\bullet$ Add the two:

$$6\pi \cos \text{\hspace{0.17em}⁣}(\frac{2{\pi }^2}{3}+\frac{\pi }{4})-12{\pi }^2\sin \text{\hspace{0.17em}⁣}(\frac{{\pi }^2}{3}+\frac{\pi }{4})$$

$x^2+3=(x+1{)}^2=x^2+2x+1\Rightarrow 2=2x\Rightarrow x=1$. Check: $\sqrt{4}=2=1+1$ ✓.

By induction, $A^n=\left(\begin{array}{cc}1& n\\ 0& 1\end{array}\right)$, so $(A^n{)}_{12}=n$.

Let $L$ be the trail length. Define

- $u(t)$: the monk's distance from the base on Monday at time $t\in [7,17]$ - $d(t)$: her distance from the base on Tuesday at the same clock time

Both functions are continuous on $[7,17]$ (a hiker doesn't teleport).

Now consider $f(t)=u(t)-d(t)$. By the problem statement:

- $u(7)=0$ and $d(7)=L$, so $f(7)=-L<0$. - $u(17)=L$ and $d(17)=0$, so $f(17)=+L>0$.

Since $f$ is continuous and changes sign on $[7,17]$, the Intermediate Value Theorem guarantees some $t^{\ast }\in (7,17)$ with $f(t^{\ast })=0$, i.e. $u(t^{\ast })=d(t^{\ast })$. At that clock time, the monk stands at the same point on the trail on both days. $\mathrm{■}$

The physical intuition ("imagine two monks: one ascending Monday, one descending Tuesday at the same time — they must meet") collapses into a one-line IVT argument once you let $f$ do the work.

${\displaystyle \frac{n(n-1)}{2}}=45\Rightarrow n(n-1)=90\Rightarrow n=10$.