In this video I go further into vector functions and this time show how to obtain their derivatives and integrals. The definition of derivative for a vector function is much the same as with a real valued function, but the difference is that we apply this definition to each component of the vector function. I also show this geometrically as a tangent vector to a point on a 3D surface and the limit as the second vector between that point and another point on the surface approach each other. The differentiation rules, such as the chain rule and product rule of dot products and cross products, have their similar counterparts for vector functions as well. Likewise, integration is also similar to real valued functions but the integral of each component is solved. This will be the basis for future videos when I cover velocity and acceleration of planetary bodies so stay tuned for that!
#math #science #calculus #vectorfunctions #education
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Later in this chapter we are going to use vector functions to describe the motion of planets and other objects through space.
Here we prepare the way by developing the calculus of vector functions.
The derivatives r' of a vector function r is defined in much the same way as for real-valued functions:
if this limit exists.
The geometric significance of this definition is shown in the figure below.
If the points P and Q have position vectors r(t) and r(t + h), then (PQ) represents the vector r(t + h) - r(t), which can therefore be regarded as a secant vector.
(a) The secant vector (PQ)
If h > 0, the scalar multiple (1/h)(r(t + h) - r(t)) has the same direction as r(t + h) - r(t).
As h → 0, it appears that this vector approaches a vector that lies on the tangent line.
For this reason the vector r'(t) is called the tangent vector to the curve defined by r at the point P, provided that r'(t) exists and r'(t) ≠ 0.
The tangent line to C at P is defined to be the line through P parallel to the tangent vector r'(t).
(b) The Tangent vector r'(t)
Here is a Secant / Tangent Line animation someone made with GeoGebra! https://www.geogebra.org/m/sVCRDDmA
As the points P and Q are closer together, the Secant and Tangent Lines become the same, i.e. the derivative is the instantaneous rate of change (i.e. rise over run) of the curve.
We will also have occasion to consider the unit tangent vector, which is:
The following theorem gives us a convenient method for computing the derivative of a vector function r: just differentiate each component of r.
Theorem
If r(t) = <f(t), g(t), h(t)> = f(t) i + g(t) j + h(t) k, where f, g, and h are differentiable functions, then:
r'(t) = <f'(t), g'(t), h'(t)> = f'(t) i + g'(t) j + h'(t) k
Proof:
(a) Find the derivative of r(t) = (1 + t3) i + te-t j + sin 2t k.
(b) Find the unit tangent vector at the point where t = 0.
Solution to (a):
According to the earlier theorem, we can differentiate each component of r:
Solution to (b):
Thus the unit tangent vector at the point (1, 0, 0) is:
Recall the distance formula in 3D (and 2D):
For the curve listed below, find r'(t) and sketch the position vector r(1) and the tangent vector r'(1).
Solution:
We have:
The curve r(t) is a plane curve and elimination of the parameter from the corresponding parametric equations gives:
In the figure below, we draw the position vector r(1) = i + j starting at the origin and the tangent vector r'(1) starting at the corresponding point (1, 1).
Notice from the above figure that the tangent vector points in the direction of increasing t (See Exercise 1 at the end of this video).
Here is how the plane curve looks like in 3D, using GeoGebra's 3D graphing calculator: https://www.geogebra.org/3d/qn5zkfvy
Find parametric equations for the tangent line to the helix with parametric equations:
at the point (0, 1, π/2).
Solution:
The vector equation of the helix is:
The parameter value corresponding to the point (0, 1, π/2) is t = π/2, so the tangent vector there is:
The tangent line is the line through (0, 1, π/2) parallel to the vector <-2, 0, 1>, so by my earlier video, the parametric equations can be obtained:
In our case:
Thus the parametric equations of the tangent line to the helix are:
The helix and the tangent line are shown in the figure below.
Here is the same graph using the GeoGebra 3D graphing calculator, including the tangent vector across the entire Helix: https://www.geogebra.org/3d/na9yk6pn
Just as for real-valued functions, the second derivative of a vector function r is the derivative of r', that is, r'' = (r')'.
For instance the second derivative of the function in Example 3 is:
Note that in a future video, we will see how r'(t) and r''(t) can be interpreted as the velocity and acceleration vectors of a particle moving through space with position vector r(t) at time t.
The next theorem shows that the differentiation formulas for real-valued functions have their counterparts for vector-valued functions.
Theorem:
Suppose u and v are differentiable vector functions, c is a scalar, and f is a real-valued function. Then:
Here is a screenshot of the list from my calculus book:
This theorem can be proved either directly by the definition of a vector function derivative or from the theorem on differentiating vector functions by components.
The proof of Formula 4 follows; the remaining formulas are left as exercises.
Let's expand the Dot Product and write as a summation.
The ordinary Product Rule gives:
Show that if |r(t)| = c (a constant), then r'(t) is orthogonal to r(t) for all t.
Solution:
Recall the Geometric interpretation of the Dot Product.
Let's consider the dot product of r(t) with itself, and apply Formula 4 of the Vector Functions differentiation rules.
Thus by Formula 4, we have:
Thus r'(t) is orthogonal to r(t) when |r(t)| = c (a constant).
Geometrically, this result says that if a curve lies on a sphere with center the origin, then the tangent vector r'(t) is always perpendicular to the position vector r(t).
To better illustrate this, I have made a sphere in GeoGebra using Grok AI to obtain the parametric equations of the sphere and the tangent vectors: https://www.geogebra.org/3d/qjahrjrw
The definite integral of a continuous vector function r(t) can be defined in much the same way as for real-valued functions except that the integral is a vector.
But then we can express the integral of r in terms of the integrals of its component functions f, g, and h as follows (using the notation of my earlier video).
This means that we can evaluate an integral of a vector function by integrating each component function.
We can extend the Fundamental Theorem of Calculus to continuous vector functions as follows:
where R is an antiderivative of r, that is, R'(t) = r(t).
We use the notation ∫ r(t) dt for indefinite integrals (antiderivatives).
where C is a vector constant of integration, and:
Show that the tangent vector to a curve defined by a vector function r(t) points in the direction of increasing t.
Hint: Refer to the earlier figure, also shown below, and consider the cases h > 0 and h < 0 separately.
Solution:
The tangent vector r'(t) is defined as:
Here we assume that this limit exists and r'(t) ≠ 0; then we know that this vector lies on the tangent line to the curve.
As in the above figure, let points P and Q have position vectors r(t) and r(t + h).
The vector r(t + h) - r(t) points from P to Q, so r(t + h) - r(t) = (PQ).
If h > 0, then t < t + h, so Q lies "ahead" of P on the Curve.
If h is sufficiently small (we can take h to be as small as we like since h → 0) then (PQ) approximates the curve from P to Q and hence points approximately in the direction of the curve as t increases.
Since h is positive, 1/h (PQ) = [r(t+h)−r(t)]/h points in the same direction.
If h < 0, then t > t + h so Q lies "behind" P on the curve.
For h sufficiently small, (PQ) approximates the curve but points in the direction of decreasing t.
However, h is negative, so 1/h (PQ) = [r(t+h)−r(t)]/h points in the opposite direction, that is, in the direction of increasing t.
In both cases, the difference quotient [r(t+h)−r(t)]/h points in the direction of increasing t.
The tangent vector r'(t) is the limit of this difference quotient, so it must also point in the direction of increasing t.
Prove Formula 1 of the Differentiation Rules for Vector Functions.
Solution:
Let u(t) = <u1(t), u2(t), u3(t)> and v(t) = <v1(t), v2(t), v3(t)>.
Taking the derivative of the components, we obtain:
Prove Formula 2 and 3 of the Differentiation Rules for Vector Functions.
Solution:
Note that Formula 2 is just Formula 3 for the case f(t) = c, where c is a constant.
Let's now prove Formula 3.
Let u(t) = <u1(t), u2(t), u3(t)>.
Taking the derivative of the components, we obtain:
Prove Formula 5 of the Differentiation Rules for Vector Functions.
Let u(t) = <u1(t), u2(t), u3(t)> and v(t) = <v1(t), v2(t), v3(t)>.
Taking the derivative of the components, we obtain:
Alternatively, we can prove Formula 5 via the definition of derivatives of vector functions.
Also note the properties of the cross product from my earlier video, namely Property 3 and 4 (be careful with the order of the cross product here).
Dividing through by h and taking the limit as h → 0 we have:
Note that I used Properties (a) and (d) of Vector Functions Limits from my earlier video in this derivation.
Prove Formula 6 of the Differentiation Rules for Vector Functions.
Solution:
Let u(t) = <u1(t), u1(t), u1(t)>.
Taking the derivative of the components, we obtain:
