Multivariate Calculus

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Last Updated: Friday, 02 October 2026 22:18
Published: Thursday, 01 October 2026 21:05
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Multivariate Calculus

 

 

Calculus on Vector Functions:

All laws of calculus can be applied on vector functions. For scalar/vectors, see in "scalars and vectors" in Maths.

Vector func r(t) = f(t)i + g(t)j + h(t)k in 3D space, where i,j,k are x,y,z dir, and f(t), g(t), h(t) are component func which are real valued in terms of parameter "t". vector r(t) is also expressed as <f(t),g(t),h(t)>. Curve C is traced by the terminal point of vector r(t), and is known as the plot or graph of vector r(t). r(t) is written in bold or wih a arrow on top to indicare it's a vector.

Domain of a vector func is the intersection of domains of f, g and h. 

Representing any graph by a vector valued func: This can be done by choosing parameter "t", and defining component func in terms of t.

 ex: y = x^2 + 1. Choose x=t (may choose anything). y=t^2+1. So, vector r(t) = ti +(t^2+1)j. The direction of the curve is from t=-inf to t=+inf (left to right). IF we choose x=-t, then dir of curve would reverse, i.e from right to left. To sketch a curve given as parametrized func, it's usually eadier to convert it to c,y,z cartesian coordinates, and see if it coresponds to any of the known shapes. Otherwise, nly a computer program may sketch such curves, as they can get complicated.

Limit of Vector func: Limits follow same rule as sclar as each component i,j,k of vector func is a scalar func, so we just take limits of each scalar component func.

Continuity of vector func: It's cont at t=a, if each of it's cmponent func is cont at t=a.

Differentiation/Integration of Vector func:

Derivative: Dt([r(t)] = r'(t) = Lim Δt->0 r(t+Δt) - r(t) / Δt = f'(t)i + g'(t)j + h'(t)k => i.e take differentiation of each component of vector. Similarly higher order differentials may be found by successively differentiating.

Vector func r(t) is smooth on an open interval I, if f'(t), g'(t), h'(t) are cont on I, and r'(t) ≠ 0 (i.e derivative for all components are not 0) for any value of t in the interval I. There is no such requirement in scalar calculus, but shows up in vector calculus. Inline discussion suggests that having drivative as 0  reverses dir of the curve (te slope of curve doesn't keep going in same dir) which poduces cusps, resulting in non smooth func at those points. However, I couldn't find a formal proof for this. What's interesting is that it's not the curve traced by vector, that determines if it's differentiable, but the way the curve is traced, i.e by changing parameters, the same curve may now become undifferentiable.

Ex:  Take a plane parabola: (t, t^2, 0). According to the definition, it is smooth, and indeed, a parabola is a curve we would intuitively like to call smooth: At (0,0,0), differentiation is (1,2t,0)=(1,0,0), since it's not 0 for all components at origin, it's smooth => it doesn't take a sudden rough turn. However, consider the same parabola, but parametrize slightly differently: (t^3, t^6, 0). This is not smooth at t=0, as it's diff is (3t^2, 6t^5,0) = (0,0,0) at origin. The way the curve is traced made a difference. Below is one of the links explaiing this:

Source: https://www.physicsforums.com/threads/why-must-dr-dt-never-equal-zero-for-a-smooth-curve.180821/

Properties of derivatives of vector func: Same as in real func for given vector func r(t). All easy to prove by writing vector func in terms of components.

  • Dt([f(t)*r(t)] = f'(t)*r(t) + f(t).r'(t). Here f(t) is real func (as simple multiplication operation doesn't exist for 2 vectors).
  • Dt([r((f(t))] = r'(f(t))*f'(t). Here f(t) is real fun.
  • Dot product (of 2 vectors): Dt([u(t).r(t)] = u'(t)*r(t) + u(t).r'(t). This gives a scalar answer (as dot product is always a scalar)
  • Cross product (of 2 vectors): Dt([u(t) X r(t)] = u'(t) X r(t) + u(t) X r'(t). This gives a vector answer (as cross product is always a scalar)
  • Dot product: If r(t).r(t) = c (a constant), then r(t) . r'(t) = 0. This is a very important property, as it states that if r(t) itself is constant, then it's derivative, r'(t) will always be orthogonal to r(t). This is what is observed in Kinematics eqn in Physics. If velocity is constant, then it's derivative, acceleration vector is orthogonal to velocity (obvious as if acceleration was in dir of velocity, then velocity would have changed)

Integration: 

Integration is same as for scalar func, we integrate each component for both definite and indefinite integral. For indefinite integral, we have to add constant C for each component, so constantbecomes C1i + C2j + C3k = C vector. So, r(t)dt = R(t) +  C, where R'(t) = r(t) and both R(t) and C are vector func. If initial cond is given, then we can find vector C.

Displacement, velocity and acceleration:

From eqn above, displacement is curve C plotted by tip of r(t). The derivative of r(t) is velocity vector, v(t), and derivative of v(t) is acceleration vector, a(t). 

Vector r(t) can be from anywhere (not necessarily from origin), the curve plotted by r(t) remains same as C, as that's what the moving object is tracing.

 v(t) = r'(t) = x'(t)i + y'(t)k + z'(t)k. As Δt -> 0, v(t) becomes tangent to the curve C [as v(t) is pointing in same dir as r(t+Δt) - r(t) ]. 

a(t) = v'(t) = r''(t) = x''(t)i + y''(t)k + z''(t)k. (double differentiation).

We can also do reverse of diff, i.e integration to determine v(t) and r(t) from a(t) given some initial conditions to eliminate constant vector C.

Projectile motion: This is eqn of a particle thrown at an angle from height h1 and hitting a target at heohgt h2. The curve traced by this motion is a parabola (i.e quadratic eqn). 

a(t) = -gj, where g=9.8m/s^2 (or 32 ft/s^2), and t is time

v(t) = ∫ a(t)dt = -gtj + C1

r(t) = ∫ v(t)dt =  ∫ [-gtj + C1] dt = -gt^2/2j + C1t + C2

To find C1 and C2, you may use initial cond: v(0) = v0 and r(0) = r0. These are both vectors.

Solving, we get: r(t) = -gt^2/2j + v0t + r0, where v0 and r0 are both vectors

In most cases, initial height and speed along with angle of projection are given, which gives following:

r(0) = +h1j

v(0) = ucos(θ)i + usin(θ)j, where u is the initial speed, and theta is angle of projection measured from horizontal surface.

So,  r(t) = -gt^2/2j + v0t + r0, =  -gt^2/2j +  utcos(θ)i + utsin(θ)j + h1j =  [ucos(θ)]t i + [h1 + [usin(θ)]t -gt^2/2] j

v(t) = -gtj + [ucos(θ)]i + [usin(θ)]j =  [ucos(θ)] i + [usin(θ) - gt] j

From rqn above, we can very easily find all the stats w/o knowing any physics, just calculus info is enough:

  • Max height: Max height is achieved when velocity vector has no vertical component, i.e V(t1) = x i + 0j => u*sin(θ) - gt1 =0 => t1 = u*sin(θ)/g. Substitute t1 in r(t) to get max height.
  • Total time to travel: It's when j component of r(t) =0 =>  [h1 + [usin(θ)]t -gt^2/2] = 0 => gives time t. Ignre the value with -ve sign before the sq root.
  • Total distance traveled: It's when j component of r(t) =0 => We got time from above eqn, we subtitute in r(t) =  [ucos(θ)]t with t from above eqn.
  • If projectile motion hits object at height h2, then r(t1) = Di + h2j => h2 = [h1 + [usin(θ)]t -gt^2/2] => [(h1-h2) + [usin(θ)]t -gt^2/2] =0 => we get t here, with h1 replaced by (h1-h2).

 

 

Functions of more than one variable:

Chapter 13:

If we have a func like z=f(x,y)= x^2+y^2, it's a func of 2 var. We can graph this in 3D space with points (x,y,z), which will be a curved surface. The domain is all possible values of (x,y) that the func can take, and range is all possible values of z.

Level curve or contour lines are a 2D map of multiple lines where along each line, the value of f(x,y) is constant. Drawing these lines on 2D map gives an idea of how the 3D surface looks like. Similar to Level curve, we've level surface when we extend the func to 3 var, i.e f(x,y,z). It's a graph where f(x,y,z) = contant along those surfaces (this shows in a 3D plot).

ex: z=f(x,y) = √ (16 - 4*x^2 - y^2) 

  • Domain: Here the eqn under sq root ≥ 0 for it to be real number. Hence domain is   (16 - 4*x^2 - y^2) ≥ 0. The boundary of this is an ellipse x^2/4 + y^2/16 = 1. All points on or inside this ellipse are in the domain, i.e x^2/4 + y^2/16 ≤ 1. 
  • Range: Range of this func is all possible z values. Smallest z value is 0, which happens on the boundary of the ellipse. The eqn under sq root can't be > 16, as the smallest possible value of x^2 and y^2 is 0.  So, range of z is 0  ≤ z  ≤ 4.
  • To plot the graph of (x,y,z), we get z^2 = 16 - 4*x^2 - y^2 => x^2/4 + y^2/16 + z^16 = 1, with 0  ≤ z  ≤ 4 which is eqn of ellipsoid with only upper half included due to militaion of z being +ve number only.

 

Limits and Continuity:

Limits for 2 var func follows same principle as that for 1 var, For limit to exist for 1 var func, f(x) at x=a, we calc value of f(x) as x approaches "a" from left and right, and if they are both equal to L, then limit exists and is equal to L. In func of 2 var, f(x,y), for limit to exist at f(x0,y0), the value of func should equal to L from all directions, not just from 2 axis.