Multivariable Calculus Flashcards

Spaced repetition cards for multivariable calculus concepts — focused on the “why” and common traps, not rote definitions.

Directional Derivativesmultivariable-calculus

Why must you normalize the direction vector before computing a directional derivative? ? The formula ${\frac{dh}{ds}}_{\hat{u}}=\nabla h\cdot \hat{u}$ requires a unit vector $\hat{u}$. If you use $\hat{i}+\hat{j}$ directly (magnitude $\sqrt{2}$), your answer is off by a factor of $\sqrt{2}$. Always divide by the magnitude first: $\hat{u}=\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})$.

How do you estimate partial derivatives from a topographic map? ? Approximate with finite differences: $\frac{\partial h}{\partial x}\approx \frac{\Delta h}{\Delta x}$. The numerator is the contour interval (fixed by the map, e.g. 100 ft). The denominator is the physical distance between adjacent contour lines in that direction, measured using the scale bar. Valid because $h$ is smooth and contour spacing is small relative to the scale of variation.

On a topographic map, how do you determine the sign of $\frac{\partial h}{\partial x}$ at a point $P$? ? Look at which direction elevation increases. If $h$ increases as $x$ increases at $P$, then $\frac{\partial h}{\partial x}>0$. Read the contour labels — the side with higher numbers tells you the direction of increasing elevation.

Parametric Curvesmultivariable-calculus

What is a parametric curve in 3D, and what does the parameter $t$ represent? ? A curve is a 1D object (one degree of freedom) described by $(x(t),y(t),z(t))$. The parameter $t$ is just a label you assign to points along the curve — it does not have to be time. The curve (set of points) is fixed by geometry; the labeling is your free choice. Different parametrizations give tangent vectors with different magnitudes but the same direction.

Why is $x(t)=t$ a valid parametrization choice for finding tangent planes, and when does it fail? ? For curves moving in the pure $x$-direction on a surface, setting $x(t)=t$ means “use $x$ itself as the parameter.” This is the laziest valid choice. It fails when the curve doubles back — e.g., the unit circle visits the same $x$ twice, so $x$ cannot be a global parameter. For tangent planes we choose curves in pure $x$ and pure $y$ directions specifically because those never double back.

Tangent Planesmultivariable-calculus

For $z=f(x,y)$, where does the tangent vector $(1,0,\frac{\partial f}{\partial x})$ come from? ? Take the curve “walk in pure $x$-direction on the surface”: position = $(t,y_0,f(t,y_0))$. Differentiate with respect to $t$:

• $dx/dt=1$
• $dy/dt=0$
• $dz/dt=\frac{\partial f}{\partial x}$ (chain rule)

So the tangent vector is $(1,0,f_x)$. Similarly $(0,1,f_y)$ for the $y$-direction. Their cross product gives the normal $\mathbf{n}=(-f_x,-f_y,1)$.

Why does the explicit surface approach $z=f(x,y)$ break down for implicit surfaces like $x^{3}y+z^2=2$? ? Three problems:

1. Branch ambiguity — solving gives $z=\pm \sqrt{2-x^{3}y}$, you must choose a branch
2. Blow-ups — $\frac{\partial f}{\partial x}=\frac{-3x^{2}y}{2\sqrt{2-x^{3}y}}$ has denominator zero where $2-x^{3}y=0$, even though the surface is smooth there
3. Impossibility — for surfaces like $x^3+\sin (xyz)+z^5=7$, you cannot isolate any variable at all

Why is $\nabla F$ always normal to the level surface $F(x,y,z)=c$? ? Any curve on the surface satisfies $F(x(t),y(t),z(t))=c$ for all $t$. Differentiate both sides (chain rule): $\frac{\partial F}{\partial x}\frac{dx}{dt}+\frac{\partial F}{\partial y}\frac{dy}{dt}+\frac{\partial F}{\partial z}\frac{dz}{dt}=0$ This is $\nabla F\cdot \mathbf{v}=0$ where $\mathbf{v}$ is the tangent vector. Since this holds for every curve on the surface, $\nabla F$ is perpendicular to every tangent vector — making it the normal.

What is the general tangent plane equation for an implicit surface $F(x,y,z)=c$ at point $(x_0,y_0,z_0)$? ? $F_x(x_0,y_0,z_0)(x-x_0)+F_y(x_0,y_0,z_0)(y-y_0)+F_z(x_0,y_0,z_0)(z-z_0)=0$ No need to isolate variables. Works for any smooth implicit surface. The vector $(F_x,F_y,F_z)$ evaluated at the point is the normal vector.

What is the relationship between “level curve” and “level surface”? ? Same idea, one dimension up. A level curve is $f(x,y)=c$ — a 1D curve in 2D space (like contour lines on a topographic map). A level surface is $F(x,y,z)=c$ — a 2D surface in 3D space. In both cases, the gradient of the function is perpendicular to the level set.

Multivariable Chain Rulemultivariable-calculus

Why does the multivariable chain rule have a SUM and not just “outside times inside”? ? The single-variable chain rule is just one multiplication: $\frac{d}{dx}f(g(x))=f^{\prime }(g(x))\cdot g^{\prime }(x)$. But if $w=f(u,v)$ where $u=u(x,y)$ and $v=v(x,y)$, changing $x$ affects $w$ through two channels simultaneously (via $u$ and via $v$). The sum comes from isolating each channel separately — it is genuinely new, not a consequence of the single-variable rule.

How does the add-and-subtract trick prove the multivariable chain rule from the limit definition? ? Start from $\frac{\partial w}{\partial x}={\lim }_{h\to 0}\frac{f(u(x+h,y),v(x+h,y))-f(u(x,y),v(x,y))}{h}$. Insert a middle term $f(u(x+h,y),v(x,y))$: Term 1 (only $u$ changes): $\frac{f(u+\Delta u,v)-f(u,v)}{\Delta u}\cdot \frac{\Delta u}{h}\to f_u\cdot \frac{\partial u}{\partial x}$ Term 2 (only $v$ changes): $\frac{f(u,v+\Delta v)-f(u,v)}{\Delta v}\cdot \frac{\Delta v}{h}\to f_v\cdot \frac{\partial v}{\partial x}$ Adding: $\frac{\partial w}{\partial x}=f_u\frac{\partial u}{\partial x}+f_v\frac{\partial v}{\partial x}$. The sum is not a postulate — it came from $x$ affecting $w$ through two separate channels.

Why does the differential $dw=f_{u}du+f_{v}dv$ make the sum feel “obvious” while the chain rule does not? ? The differential is built to handle simultaneous changes in all directions — it decomposes the total change into independent contributions by construction. The chain rule derived from the limit definition requires the add-and-subtract trick to isolate contributions. The differential packages that trick into its definition upfront, which is why dividing $dw$ by $dx$ gives the chain rule “for free.” The chain rule is actually the harder way to see it.

Implicit Partial Differentiationmultivariable-calculus

Why is $\frac{\partial w}{\partial z}$ meaningless without specifying what is held fixed? ? If $w=f(x,y)$ and a constraint $g(x,y,z)=c$ links all three variables, the constraint has two degrees of freedom. When $z$ moves, you must choose: hold $x$ fixed (forcing $y$ to adjust) or hold $y$ fixed (forcing $x$ to adjust)? These are different scenarios with different answers. Example: $w=x^2+y$, constraint $x+y+z=1$.

• Hold $y$ fixed: $\partial x/\partial z=-1$, so $\partial w/\partial z=2x(-1)=-2x$
• Hold $x$ fixed: $\partial y/\partial z=-1$, so $\partial w/\partial z=-1$ Same point, same function — two different answers.

In implicit partial differentiation, why does the problem reduce to the ordinary (one-variable) chain rule? ? Once you choose which variable is held fixed, only one variable is left free to adjust. For example: $w=f(x,y)$, constraint $g(x,y,z)=c$, hold $y$ fixed. Then $x=x(z)$ is a function of $z$ alone (the constraint forces it). So $w=f(x(z),y_0)$ is a one-variable composition, and the ordinary chain rule gives: $\frac{\partial w}{\partial z}{|}_y=\frac{\partial w}{\partial x}\cdot \frac{\partial x}{\partial z}$ No multivariate chain rule sum needed — the fixed-variable choice killed the second channel.

What does the implicit function theorem guarantee, and when does it fail? ? It guarantees that a constraint $g(x,y,z)=c$ locally defines one variable as a smooth function of the others — specifically, $x=x(y,z)$ exists near a point provided $\frac{\partial g}{\partial x}\ne 0$ there. When $\frac{\partial g}{\partial x}=0$, the constraint surface is “vertical” in the $x$-direction at that point — $x$ is not responding to changes in the other variables, so you cannot invert the relationship. Note: the condition is sufficient, not necessary — solutions can sometimes exist even when the partial vanishes (e.g. $y^3=0$ at the origin). See The Implicit Function Theorem.

How do you compute $\frac{\partial x}{\partial z}$ from a constraint $g(x,y,z)=c$ with $y$ held fixed? ? Differentiate the constraint implicitly with $dy=0$: $\frac{\partial g}{\partial x}\frac{\partial x}{\partial z}+\frac{\partial g}{\partial z}=0$ Solve: $\frac{\partial x}{\partial z}{|}_y=-\frac{\partial g/\partial z}{\partial g/\partial x}$ This requires $\frac{\partial g}{\partial x}\ne 0$ (implicit function theorem). The formula is just “differentiate the constraint, solve for the ratio.”

What is the geometric meaning of $F_y(p)\ne 0$ on the level curve $F(x,y)=c$? ? Two equivalent readings:

1. The tangent to the level curve is not vertical at $p$ — since the tangent direction is $(-F_y,F_x)$, having $F_y\ne 0$ means nonzero $x$-component, so nearby points have distinct $x$-values. This is what makes $y=\varphi (x)$ well-defined.
2. $F$ is strictly monotonic in $y$ near $p$ — the map $y\mapsto F(x_0,y)$ crosses $c$ exactly once near $p$, giving a unique solution for each $x$.

Why does implicit differentiation actually require the IFT as justification? ? When you differentiate $F(x,y)=c$ to get ${\varphi }^{\prime }=-F_x/F_y$, you apply the chain rule to $F(x,\varphi (x))=c$. But the chain rule requires that:

1. $\varphi$ exists — $y$ is actually a function of $x$ near $p$
2. $\varphi$ is differentiable — not just continuous The IFT is exactly what guarantees both. Without it, you’re computing the derivative of a function that might not exist.

Where does the differentiability of $\varphi$ come from (not just continuity)? ? Continuity comes from monotonicity (topological). Differentiability comes from inverting the linear approximation. Since $F$ is differentiable: $0=F_{x}h+F_{y}k+o(\sqrt{h^2+k^2})$ where $k=\varphi (x+h)-\varphi (x)$. Solving: $k=-\frac{F_x}{F_y}h+\text{remainder}$. A Lipschitz bound $|k|\le C|h|$ (from contraction) kills the remainder, giving $\varphi (x+h)-\varphi (x)=-\frac{F_x}{F_y}h+o(|h|)$. This is differentiability. The $C^1$ hypothesis on $F$ ensures $F_y\ne 0$ on a neighborhood (not just at $p$), which is needed for the contraction to work uniformly.

Change of Variables and the Jacobianmultivariable-calculus

Why can’t you substitute $dx=\cos \theta dr-r\sin \theta d\theta$ directly into $\iint fdxdy$? ? The formula $dx=\cos \theta dr-r\sin \theta d\theta$ is the correct total differential (how $x$ changes when you wiggle $r$ and $\theta$). But $dxdy$ in a double integral is an area element, not a product of two independent length changes. Multiplying two differentials naively produces cross-terms ($drd\theta$ and $d\theta dr$) with no rule for simplifying them. The correct substitution is $dxdy=|\det J|drd\theta$, which accounts for how areas (not lengths) transform.

Where does the factor $r$ in $dxdy=rdrd\theta$ come from geometrically? ? A small rectangle $[r,r+dr]\times [\theta ,\theta +d\theta ]$ in polar parameter space maps to a thin wedge in the $xy$-plane. Its sides are:

• Radial side: length $dr$
• Arc side: length $rd\theta$ (arc of a circle of radius $r$ subtending angle $d\theta$) Area $\approx dr\times rd\theta =rdrd\theta$. The $r$ factor is the arc-length scaling — further from the origin, the same $d\theta$ sweeps a larger arc.

What does $\det J=0$ mean geometrically for a coordinate transformation? ? The two column vectors of the Jacobian matrix become parallel — the parallelogram they span collapses to a line segment with zero area. The transformation crushes a 2D region to a 1D one at that point. It is not locally invertible there (Inverse Function Theorem fails). Example: at $r=0$ in polar coordinates, all $\theta$ values map to the origin — the map is not injective and the “wedge” has zero area.

How does the wedge product $dx\wedge dy$ automatically give the Jacobian determinant? ? Substitute $dx=\cos \theta dr-r\sin \theta d\theta$ and $dy=\sin \theta dr+r\cos \theta d\theta$, then expand using $dr\wedge dr=0$, $d\theta \wedge d\theta =0$, and $d\theta \wedge dr=-dr\wedge d\theta$: $dx\wedge dy=(r{\cos }^2\theta +r{\sin }^2\theta )dr\wedge d\theta =rdr\wedge d\theta$ The antisymmetry of $\wedge$ kills the “diagonal” terms and produces exactly the determinant. This is why the Jacobian works — the determinant is the wedge product.

Polar Integration Boundsmultivariable-calculus

How do you determine the $\theta$ bounds when converting a double integral to polar coordinates? ? For each angle $\theta$, find where the ray from the origin enters the region ($r_{\text{low}}(\theta )$) and exits ($r_{\text{high}}(\theta )$). The allowed $\theta$ values are exactly those where $r_{\text{low}}(\theta )\le r_{\text{high}}(\theta )$ — i.e., the ray actually hits the region. The boundary angles (where $r_{\text{low}}=r_{\text{high}}$) are where the ray is tangent to the region, just grazing it. Those give your $\theta$ limits.

When does the single-pair $r_{\text{low}}(\theta ),r_{\text{high}}(\theta )$ approach fail for polar bounds? ? When a ray from the origin hits the region in two or more disjoint pieces (e.g., an annulus offset from the origin, or a region with a hole the origin “sees through”). In that case, a single interval $[r_{\text{low}},r_{\text{high}}]$ can’t describe the intersection — you must split the region into sub-regions, each of which every ray hits at most once. Check: “does every ray from the origin meet $D$ in a single interval?” If yes, proceed. If no, split first.

Gradient in Curvilinear Coordinatesmultivariable-calculus

What is the coordinate-free defining property of the gradient? ? The gradient $\nabla w$ is the unique vector field satisfying $dw=\nabla w\cdot d\vec{s}$ for every spatial displacement $d\vec{s}$. Equivalently: it points in the direction of steepest ascent, with magnitude equal to the rate of increase per unit distance. The formula “gradient = vector of partial derivatives” is a special case that works only when scale factors are all 1 (Cartesian).

Why does the polar gradient formula have $\frac{1}{r}$ in the $\theta$-component? ? A displacement of $d\theta$ at radius $r$ corresponds to an arc length of $rd\theta$. The partial derivative $\frac{\partial w}{\partial \theta }$ measures change per radian, but the gradient needs change per meter of arc. Dividing by $r$ converts between the two. At $r=100$, a change of 1 per radian is only 0.01 per meter — the $1/r$ captures this.

State the gradient formula in a general orthogonal coordinate system with scale factors $h_i$. ? $\nabla w=\frac{1}{h_1}\frac{\partial w}{\partial q_1}{\hat{e}}_1+\frac{1}{h_2}\frac{\partial w}{\partial q_2}{\hat{e}}_2+\frac{1}{h_3}\frac{\partial w}{\partial q_3}{\hat{e}}_3$ Scale factors: Cartesian $(1,1,1)$, cylindrical $(1,r,1)$, spherical $(1,\rho ,\rho \sin \varphi )$. Each $h_i$ converts “per unit coordinate” to “per unit distance.” Cartesian is special only because all $h_i=1$.

In what sense is “gradient = vector of partial derivatives” wrong? ? Partial derivatives are rates of change per unit coordinate increment. Gradient components are rates of change per unit distance. These are the same only when one unit of coordinate equals one unit of distance (scale factor = 1). In polar coordinates, 1 radian $\ne$ 1 meter — they differ by a factor of $r$. Writing $\nabla w=(\partial w/\partial r,\partial w/\partial \theta )$ in polar gives a “vector” that does NOT point in the steepest ascent direction and is NOT perpendicular to level curves.

What are the scale factors in spherical coordinates and where do they come from? ? $h_{\rho }=1$, $h_{\varphi }=\rho$, $h_{\theta }=\rho \sin \varphi$.

• $h_{\rho }=1$: moving radially, 1 unit of $\rho$ = 1 unit of distance
• $h_{\varphi }=\rho$: moving in $\varphi$ traces a great circle of radius $\rho$, arc length = $\rho d\varphi$
• $h_{\theta }=\rho \sin \varphi$: moving in $\theta$ traces a circle at colatitude $\varphi$ with radius $\rho \sin \varphi$, arc length = $\rho \sin \varphi d\theta$ At the north pole ($\varphi =0$), $h_{\theta }=0$ — the azimuthal circle has collapsed to a point.