Calculus and Vector Calculus

Tangent plane to level surface

A level surface is given implicitly as F(x,y,z)=c. At any point where \nabla F\neq \mathbf{0}, the gradient \nabla F=(F_x,F_y,F_z) is perpendicular to the surface, so it serves as the normal vector of the tangent plane. The tangent plane at P_0=(x_0,y_0,z_0) is F_x(x-x_0)+F_y(y-y_0)+F_z(z-z_0)=0. A common mistake is to use the position vector or the constant level value as the normal. A point lies on the tangent plane only if it satisfies this equation, so substituting the point is a quick check. For example, for F=x^2+y^2-z at (1,0,1), the gradient is (2,0,-1), and the plane is 2x-z-1=0. Its check point (1,0,1) gives 2-1-1=0.

Line integral by Green's theorem

For a positively oriented (counterclockwise) simple closed curve C bounding a region R, Green's theorem states \oint_C P\,dx+Q\,dy=\iint_R\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dA, provided P and Q have continuous first partial derivatives on R. Reversing the direction of traversal changes the sign of the line integral, so the orientation must be checked first. The theorem converts a line integral into a double integral, which is often easier to evaluate. Choosing P=0 and Q=x gives \oint_C x\,dy=\text{Area}(R). For example, for a circle of radius 2 traversed counterclockwise, \oint x\,dy=\pi\cdot 2^2=4\pi. When the field is given in a plane with a constant curl-type integrand, the double integral reduces to that constant times the area of the region.

Chain rule for partial derivatives

If z=f(x,y) and x, y are themselves functions of u and v, then the partial derivative of z with respect to v is \frac{\partial z}{\partial v}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial v}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial v}. Each intermediate variable contributes one term, and every partial derivative is taken with the other independent variable held fixed. Students often confuse partial and total derivatives, or omit a path through an intermediate variable. A simple check is to substitute the functions first and differentiate directly. For z=x^2y with x=s+t and y=st, the chain rule gives \frac{\partial z}{\partial t}=2xy\cdot 1+x^2\cdot s, which agrees with direct differentiation of (s+t)^2st. Sometimes the expression simplifies to a constant, which can be seen before computing each term.

Directional derivative of scalar field

The directional derivative of a scalar field f at a point in the direction of a vector \mathbf{a} is D_{\mathbf{u}}f=\nabla f\cdot\mathbf{u}, where \mathbf{u}=\mathbf{a}/|\mathbf{a}| is the unit vector. The direction must be normalized before the dot product is taken; using \mathbf{a} directly gives a wrong result. The directional derivative is largest in the direction of \nabla f, where it equals |\nabla f|, and it is zero in directions perpendicular to \nabla f. The partial derivative f_x is only the special case where the direction is \hat{i}. For example, for f=xy at (2,1), \nabla f=(1,2), and along \hat{i}+\hat{j} the unit vector is (1/\sqrt{2},1/\sqrt{2}), so D_{\mathbf{u}}f=3/\sqrt{2}.

Divergence theorem for flux integrals

For a closed surface S enclosing a volume V, the outward flux of a vector field \mathbf{F} satisfies \oint\!\!\oint_S \mathbf{F}\cdot\hat{n}\,dS=\iiint_V \nabla\cdot\mathbf{F}\,dV, where \nabla\cdot\mathbf{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z}. The theorem applies only to closed surfaces, and the normal must point outward. A flux written as \iint(P\,dy\,dz+Q\,dz\,dx+R\,dx\,dy) is the same as the surface integral of \mathbf{F}\cdot\hat{n}. It is useful when \nabla\cdot\mathbf{F} is simple, because the volume integral is often easier. For example, for \mathbf{F}=(x,0,0) over a unit cube, \nabla\cdot\mathbf{F}=1, so the total outward flux is the volume, 1.

Arc length of parametric curve

For a curve given parametrically as x=x(t), y=y(t) on a\le t\le b, the arc length is L=\int_a^b\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt. The integrand is the speed of the point on the curve. Before integrating, it helps to simplify the expression under the square root, often using trigonometric identities such as 1-\cos t=2\sin^2(t/2). A constant speed makes the arc length easy to find: if x=3t and y=4t for 0\le t\le 2, the speed is 5, so L=10. For a standard cycloid of radius a, one arch has length 8a, which is a useful known result.

Critical points of functions of two variables

Critical points of f(x,y) satisfy f_x=0 and f_y=0. To classify them, use the Hessian discriminant D=f_{xx}f_{yy}-f_{xy}^2. If D>0 and f_{xx}>0, the point is a local minimum; if D>0 and f_{xx}<0, it is a local maximum; if D<0, it is a saddle point, where the surface curves up in one direction and down in another; and if D=0, the test is inconclusive. In GATE-style questions, the origin of such a saddle is sometimes described as a point where concavity changes, and the key accepts that wording. To find the minimum value, compare the function values at all critical points. Solving the equations often involves eliminating one variable. For example, f=x^2+y^2-2x+4y has critical point (1,-2), where D=4>0 and f_{xx}>0, so it is a local minimum with value -5.

Triple integral over box region

A box region a\le x\le b, c\le y\le d, e\le z\le f has constant limits, so the triple integral can be written as an iterated integral in any order. For a product of single-variable functions, the integral splits into a product of simpler integrals. Terms that are odd about a symmetric interval vanish: if g(-x)=-g(x), then \int_{-a}^{a}g(x)\,dx=0. Each term of a sum can be integrated separately, and this symmetry check can save much work. For example, \iiint_{[0,1]^3}xyz\,dV=\left(\frac12\right)^3=\frac18. A more involved example is \iiint(xy^2+yz^3)\,dV over a symmetric box in x; the x-odd term xy^2 drops out, leaving only the other term.