Exercises — Vector fields — definition, visualization

(a) $P=2,Q=2$ do not contain $x$ or $y$ → the arrow never changes → constant field. At $(1,0)$ the arrow is $(2,2)$: points up-right at $45^{\circ }$, length $\sqrt{2^2+2^2}=2\sqrt{2}$. (b) Output equals the position vector → radial (outward) field. At $(1,0)$ the arrow is $(1,0)$: points straight right, away from the origin. (c) Output is the position vector turned $90^{\circ }$ → rotational field. At $(1,0)$ the arrow is $(-0,1)=(0,1)$: points straight up (tangent to the unit circle, counter-clockwise).

$\mathbf{F}(3,4)=(3,4)$, so $|\mathbf{F}|=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$. Why: the length of the arrow is the length of the position vector here, i.e. the distance $r$ from the origin, which is $5$.

Plug each point into $(-y,x)$:

• $(1,0)\to (-0,1)=(0,1)$ — up
• $(0,1)\to (-1,0)$ — left
• $(-1,0)\to (0,-1)$ — down
• $(0,-1)\to (1,0)$ — right

Reading up → left → down → right as you go counter-clockwise around the circle is exactly a counter-clockwise swirl. See the figure.

Direction $=\frac{\mathbf{F}}{|\mathbf{F}|}$. From L1.2, $|\mathbf{F}|=5$, so $\hat{\mathbf{F}}=\frac{(3,4)}{5}=\left(\frac{3}{5},\frac{4}{5}\right)=(0.6,0.8).$ Why divide by the magnitude? Dividing a vector by its own length scales it to length $1$ while keeping its direction — check: $\sqrt{0.6^2+0.8^2}=\sqrt{0.36+0.64}=\sqrt{1}=1.$ ✅

The gradient $\nabla f=\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right)$ collects the partial derivatives (see Gradient and Directional Derivatives).

• $\frac{\partial }{\partial x}(x^{2}y)=2xy$ (treat $y$ as a constant).
• $\frac{\partial }{\partial y}(x^{2}y)=x^2$ (treat $x$ as a constant).

So $\nabla f=(2xy,x^2)$. This is automatically a conservative (gradient) field — see Conservative Fields and Potential Functions.

Take the dot product with the position vector: $\mathbf{F}\cdot (x,y)=(-y)(x)+(x)(y)=-xy+xy=0.$ Why the dot product? It is the tool that answers "are these two arrows perpendicular?" — its value is $0$ exactly when they meet at $90^{\circ }$. Here it is $0$ for all $x,y$. Geometric meaning: the arrow is always at right angles to the line from the origin (the radius), i.e. tangent to the circle through that point → pure rotation, no outward/inward push.

The zero arrow needs both components zero at once: $x-1=0\text{ and }y+2=0\Rightarrow x=1,y=-2.$ So $\mathbf{F}=0$ only at $(1,-2)$. Rewrite $\mathbf{F}=((x-1),(y+2))$ — this is exactly the radial field $(u,v)$ shifted so its centre sits at $(1,-2)$. Arrows point outward from $(1,-2)$, growing longer with distance. The zero point is a source (an equilibrium where nothing pushes).

Here $\mathbf{r}=(0,2)$ so $|\mathbf{r}|=\sqrt{0^2+2^2}=2$. The field's magnitude is $|\mathbf{F}|=\frac{|\mathbf{r}|}{|\mathbf{r}{|}^3}=\frac{1}{|\mathbf{r}{|}^2}=\frac{1}{2^2}=\frac{1}{4}=\mathrm{0.25.}$ Why $1/r^2$? The numerator $\mathbf{r}/|\mathbf{r}|$ is a unit arrow (length $1$); the extra $1/|\mathbf{r}{|}^2$ is the strength, which falls off as the square of distance — the signature of gravity/electrostatics.

Add the two archetypes with unknown weights $a,b$: $a(x,y)+b(-y,x)=(ax-by,ay+bx).$ Match components with $(x-y,x+y)$:

• $x$-part: $ax-by=x-y\Rightarrow a=1,b=1.$
• $y$-part: $ay+bx=y+x$ ✅ with $a=1,b=1.$

So $\mathbf{F}=(x,y)+(-y,x)$: an outward push plus a counter-clockwise swirl — arrows spiral outward. This "radial + rotational" split previews Divergence and Curl (spreading vs. swirling).

We need $\frac{\partial f}{\partial x}=2xy$. Integrate in $x$ (treat $y$ constant): $f=\int 2xydx=x^{2}y+g(y),$ where $g(y)$ is an unknown "constant of integration" that may depend on $y$. Now demand $\frac{\partial f}{\partial y}=x^2$: differentiating $x^{2}y+g(y)$ gives $x^2+g^{\prime }(y)$. Set equal to $x^2$: $x^2+g^{\prime }(y)=x^2\Rightarrow g^{\prime }(y)=0\Rightarrow g(y)=C.$ So $f(x,y)=x^{2}y+C$. Since a potential exists, $\mathbf{F}$ is conservative — matching L2.3 where we built exactly this field from $f=x^{2}y$. See Conservative Fields and Potential Functions.

Start from the counter-clockwise swirl $(-y,x)$; flip the sign to reverse the spin: $(y,-x)$. Its magnitude is $\sqrt{y^2+x^2}=r$ already, which is exactly what we want. So take $\boxed{\mathbf{F}(x,y)=(y,-x)}.$ Check tangent (perpendicular to radius): $\mathbf{F}\cdot (x,y)=(y)(x)+(-x)(y)=xy-xy=0$ ✅ → tangent to circles. Check clockwise: at $(1,0)$, $\mathbf{F}=(0,-1)$ points down — moving clockwise around the circle ✅. Check magnitude $=r$: $|\mathbf{F}|=\sqrt{y^2+(-x{)}^2}=\sqrt{x^2+y^2}=r$ ✅.

(i) $\mathbf{F}(2,3)=(2,-3)$. (ii) $|\mathbf{F}|=\sqrt{2^2+(-3{)}^2}=\sqrt{4+9}=\sqrt{13}\approx 3.606$. (iii) Zero needs $x=0$ and $-y=0$ → only $(0,0)$. (This is a saddle-type point: arrows flow out along the $x$-axis, in along the $y$-axis.) (iv) Look for $f$ with $\partial f/\partial x=x$: integrate → $f=\frac{1}{2}{x}^2+g(y)$. Then $\partial f/\partial y=g^{\prime }(y)$ must equal $-y$, so $g(y)=-\frac{1}{2}{y}^2$. Thus $f(x,y)=\frac{1}{2}{x}^2-\frac{1}{2}{y}^2+C,$ and $\mathbf{F}$ is conservative. Check: $\nabla f=(x,-y)$ ✅.

From L3.3, $|\mathbf{F}|=1/r^2$.

• $r\to 0^{+}$: $1/r^2\to +\infty$ — the field blows up near the origin. The origin is a singularity (the point mass / charge sits there; the field is undefined at $r=0$).
• $r\to \infty$: $1/r^2\to 0$ — the field fades to nothing far away, but never actually reaches zero. Meaning: gravity/electric pull is enormous up close and weakens with the square of distance — you feel a planet strongly at its surface and barely at all in deep space.