C.2 · Multivariable calculus and vector analysis

Undergraduate GRE preparation. Local objectives within the reviewed ETS scope; this is original teaching, not an official test or score predictor.

Prerequisites: Single-variable derivatives, dot products and iterated integration.

• Compute partial derivatives, gradients and directional derivatives
• Use multiple integrals and coordinate changes
• Use derivative conditions to distinguish planes from curved surfaces

gradient 梯度: Vector of partial derivatives.

Jacobian 雅可比行列式: The local area or volume scaling in a coordinate change.

A partial derivative changes one coordinate while fixing the others. For f=x²+3xy, f_x=2x+3y and f_y=3x. The gradient collects these derivatives; if f is differentiable, the directional derivative along a unit vector v is grad f·v. Normalise the direction before taking this dot product. A direction vector of length two would double the answer if used without normalisation.

Differentiability means a valid linear approximation, not merely the existence of some partial derivatives at one point. Continuous first partials in a neighbourhood are a sufficient condition. For a composition f(x(t),y(t)), the chain rule gives f_x x′+f_y y′. Second derivatives can introduce both direct and mixed terms; a zero mixed partial alone places no restriction on the pure second partials.

A multiple integral sums contributions over a specified region. Describe the region before choosing iterated limits; nonrectangular bounds may change when the order changes. In polar coordinates area is r dr dtheta, not just dr dtheta. A general coordinate change uses the absolute Jacobian determinant. Sign belongs to oriented vector quantities, while area and volume scaling use a nonnegative factor.

If both first partials of a globally defined function on R² are constant, f_x=a and f_y=b, integrating successively gives f=ax+by+c, a plane. This cannot be inferred from parallel straight level sets alone: e^x has vertical parallel level lines but a curved graph. Nor do f_xy=f_yx=0 force a plane: x²+y² is a counterexample. Distinguish first-derivative constancy from absent mixed dependence and from the shape of selected level sets.

For f(x,y)=x²+3y², gradient f=(2x,6y). At (1,1) this is (2,6). In direction (3,4), the unit vector is (3/5,4/5), giving directional derivative 2·3/5+6·4/5=6.

A non-unit direction vector gives a scaled directional rate, not the derivative per unit distance.

For $f(x,y)=x^2+xy+2y^2$, find the directional derivative at $(1,-1)$ towards $(3,4)$. Evaluate $\iint_D(x^2+y^2)\,dA$ for the unit disk $D$.

Worked solution

The polynomial is differentiable. Normalise the direction: $u=(3/5,4/5)$. The gradient is $(2x+y,x+4y)$, hence $(1,-3)$ at the point.

Polar coordinates give $x^2+y^2=r^2$ and $dA=r\,dr\,d\theta$:

Define $g(x,y)=xy/\sqrt{x^2+y^2}$ away from the origin and $g(0,0)=0$. Both partial derivatives at the origin are zero. Is $g$ differentiable there? Is it continuous?

Check after attempting

Along each axis $g=0$, so both partials are zero. Continuity follows from $|xy|\le(x^2+y^2)/2$, which gives $|g(x,y)|\le\sqrt{x^2+y^2}/2\to0$. If differentiable, its linear derivative would be zero. Along $x=y=t\ne0$,

The required remainder ratio does not tend to zero. Thus continuity and existing partial derivatives do not establish differentiability.