Multivariable Lesson 1.2: Derivatives and Integrals of Vector Functions

Differentiating

The derivative of a vector function has much the same definition and meaning as that of scalar functions: r'(t)=lim(h->0) (r(t+h)-r(t))/h. This can be interpreted as the limit of a secant vector to the space curve: as h approaches zero, the secant vector approaches a tangent vector, just as in single-variable calculus a secant line approaches a tangent line: Animation of a secant vector approaching a tangent vector

Because limiting a vector is equivalent to limiting all of its components, it’s very easy to find the derivative of a vector r(t)=(x(t), y(t), z(t)) by simply differentiating each component: r'(t)=(x'(t), y'(t), z'(t)). For similar reasons, it is straightforward to show that standard differentiation rules apply to vector functions. In particular, the product rule applies for the dot and cross products as well as the scalar product (the product of a scalar function and a vector funciton). The chain rule also applies for the composition of a vector function evaluated at a scalar function.

Integrating

As to the integral of a vector function, that too can be found by integrating each component separately. However, this will of course result in a vector, and usually won’t have any geometric meaning for a space curve. (In particular, it certainly doesn’t represent area.) In fact, the only time you’d really directly integrate a vector function is in the case of a particle motion problem such as those you’ve seen before; for example, integrating the vector acceleration of a particle to find its velocity. You already know how to tackle this sort of problem, and it won’t be addressed again.

Arc Length

Though directly integrating doesn’t help us analyze space curves, we still might be able to use integrals on them somehow. Recall our use of integration to find the arc length of parametrized functions in the plane. Since a vector function is simply a parametrization of a space curve, the Pythagorean Theorem can be used on the scalar component functions to derive an expression for the arc length function, analogous to the formula for two-dimensional parametric functions: s(t)=Int(a to t) sqrt ((dx/du)^2+(dy/du)^2+(dz/du)^2)du However, since dx/du, dy/du, and dz/du are the components of the vector function’s derivative, we notice that the integrand can much more simply be written as the magnitude of the function’s derivative: s(t)=Int(a to t) |r'(u)|du

It is often handy to reparametrize the function r(t)=(cos t, sin t, 3t) for t greater than or equal to 0. First, solve for the arc length. s(t)=Int(0 to t) sqrt(sin^2 u + cos^2 u + 9)du = t sqrt(10) Since s=t sqrt(10), then t=s/sqrt(10). If we substitute that expression for t in the original function, then s will be our new parameter, meaning that we have successfully changed the parameter to the arc length: r(s)=(cos(s/root 10), sin(s/root 10), 3(s/root 10))

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