Mixed review R3: approximation, geometry, and local change

Alternation and absolute convergence require separate arguments

The magnitudes $b_n=n/(n^2+1)$ tend to zero. Since $b'(x)=(1-x^2)/(x^2+1)^2\le0$ for $x\ge1$, they are nonincreasing. The alternating-series theorem proves convergence. For absolute values, $b_n/(1/n)=n^2/(n^2+1)\to1$, so limit comparison with the harmonic series gives divergence. Thus the original series is conditionally convergent. The alternating remainder satisfies $\boxed{|R_{10}|\le b_{11}=11/122}$. This is a bound, not a claim that the actual error equals that number.

Use a vanished coefficient to improve a bound

Substitute $2x$ in $1-u^2/2!+u^4/4!$ to get $\boxed{P_4(x)=1-2x^2+2x^4/3}$. The fifth derivative at zero vanishes, so $P_5=P_4$. The sixth derivative has absolute value at most $64$. Taylor’s theorem therefore gives

The estimate is uniform over the stated interval. Using the degree-four remainder directly would be valid but less sharp with a crude fifth-derivative bound.

Divide output change per time by input change per time

At $t=\pi/4$ the point is $(\sqrt2,\sqrt2/2)$. Derivatives are $x'=-2\sin t$ and $y'=\cos t$, so the slope is $-1/2$. Thus $\boxed{y-\sqrt2/2=-(x-\sqrt2)/2}$. The chain rule says $\dd y/\dd t=(\dd y/\dd x)(\dd x/\dd t)$ where $x'\ne0$, so dividing by $x'$ converts change per parameter to change per horizontal input. Here $x'=-\sqrt2\ne0$.

Two independent in-plane directions give a normal

The vectors $Q-P=(1,-1,0)$ and $R-P=(0,1,-1)$ have cross product $(1,1,1)$. It is nonzero, so the three points are not collinear and determine one plane. Using $P$ gives $x+(y-1)+(z-2)=0$, or $\boxed{x+y+z=3}$. Each of $P,Q,R$ has coordinate sum $3$, independently checking the result.

A curve’s speed and curvature are not interchangeable

Velocity is $(-2\sin2t,2\cos2t,1)$, with norm $\boxed{\sqrt5}$. Acceleration is $(-4\cos2t,-4\sin2t,0)$. Their cross product is $(4\sin2t,-4\cos2t,8)$, with norm $4\sqrt5$. Thus $\boxed{\kappa=(4\sqrt5)/(\sqrt5)^3=4/5}$. The speed is constant and nonzero, so the curvature formula applies at every parameter value. Curvature measures bending per arc length, not distance traveled per unit time.

A single bound handles every approach

For $(x,y)\ne(0,0)$,

The right side tends to zero as the point approaches the origin, so the squeeze theorem gives the claimed limit. Unlike checking a few lines, this bound applies to every point in a punctured disk.

Normalize the direction before taking the dot product

Away from the origin, $\grad f=(2x/(x^2+y^2),2y/(x^2+y^2))$. At $(3,4)$ it is $(6/25,8/25)$. The unit direction is $(-4/5,3/5)$, so the directional derivative is $(-24+24)/125=\boxed0$. The greatest rate is the gradient norm $\boxed{2/5}$, in direction $(3/5,4/5)$. The zero-rate direction is tangent to the circular level curve $x^2+y^2=25$; first-order change vanishes along that tangent direction, not along every possible displacement.

Treat zero coordinates as legitimate candidates

Let $f=x^2+y^2$, $g=4x^2+y^2$. The ellipse is compact and smooth, so extrema exist and $\grad g\ne0$ on it. The multiplier equations are $2x=8\lambda x$, $2y=2\lambda y$, and $4x^2+y^2=4$. If both coordinates were nonzero, the first two equations would give incompatible values $\lambda=1/4$ and $\lambda=1$. Thus one coordinate vanishes. At $(\pm1,0)$, $f=1$; at $(0,\pm2)$, $f=4$. Therefore the minimum is $\boxed1$ and maximum $\boxed4$. Alternatively, the constraint gives $f=4-3x^2$ with $0\le x^2\le1$, proving the same global bounds directly.