Derivatives of Vector Valued Functions

Return to Course Outline

Previous Lecture

Anything You Want to Talk About?

(No.)

Derivatives of Vector-Valued Functions

Based on “Derivatives of Vector-Valued Functions” and “Tangent Vectors and Unit Tangent Vectors” in section 2.2 of the textbook.

Key Ideas

There is a limit definition of the derivative of a vector-valued function, which is essentially the same as for a scalar function.

For computational purposes, the derivative of a vector-valued function is just the vector of the derivatives of the function’s components.

Vector-valued functions have differentiation rules that are exactly like the rules for scalar functions, plus product rules for dot and cross products that follow the same pattern as the regular product rule.

The unit tangent vector is a standard way of talking about the direction a vector-valued or parametric curve is heading.

Definition

Suppose u(t) = ⟨ 2t + 1, t² - t , √t ⟩.

Use the limit definition of the vector-valued derivative to find u′(t).

We plugged u(t+h) and u(t) into the definition of derivative, then expanded out the resulting expressions and simplified. The trickiest part was probably multiplying by a conjugate in order to simplify the term involving a square root:

Derivative Examples

Let v(t) =⟨ sin t, e^2t, 3 ln t ⟩. Without necessarily using the limit definition, find v′(t)

This time we just applied derivative rules for scalar functions to each component of v:

Derivatives in Mathematica

Mathematica can take derivatives, using the D function. The general form is

D[ expression, var ]

where expression is the expression (possibly a vector, in curly braces) to take the derivative of, and var is the variable to differentiate with respect to.

We demonstrated this by taking the derivative of v(t) from above. You can download the notebook from Canvas to view and experiment with.

Tangent Vectors

A not-implausible function for the position of a ball thrown into the air at the origin of some coordinate system might be s(t) = ⟨ t, t/2, 3t - 5t² ⟩.

Find the ball’s velocity function (which happens to be tangent to the position curve at all points). Velocity is the derivative of position, so we just used standard differentiation rules to find s’(t):

Find a function for the ball’s unit tangent vector. To do this, we just divided the derivative we’d just found by its magnitude. The result is ugly, which often happens with unit tangent vectors, but unit tangent vectors are nonetheless an important part of some things we’ll see later.

Problem Set

Problem set 4, mainly on derivatives of vector-valued functions, is ready.

Work on it (most of) next week, and grade it the week after.

See the handout for more information.

Next

Integrals of vector-valued functions.

Please read “Integrals of Vector-Valued Functions” in section 2.2 of the textbook.

Next Lecture