Derivatives of Vector Valued Functions

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Derivatives of Vector-Valued Functions

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

Key Ideas

A vector-valued derivative is a limit involving the vector-valued function, exactly analogous to the limit definition of a scalar function. You can find a vector-valued derivative without explicitly taking a limit by finding the derivative of each component of the function. This works exactly the same in 2 dimensions as in 3 (and more).

The limit definition leads to a definition of what it means for a vector-valued function to be differentiable over a closed interval.

Derivatives of vector-valued functions have properties similar to those of scalar functions, with the addition of dot and cross product rules (which are analogous to the product rule for scalar functions).

The derivative of a vector-valued function gives a vector tangent to the function’s curve; scaled to a unit vector, this is the “principal unit tangent vector.”

Questions

An example of finding a derivative?

Mostly, taking derivatives of vector-valued functions boils down to applying differentiation rules from earlier courses to each component of the vector function. For example…

An example involving the dot or cross product rules? A good way to start such problems is to find the derivatives of each vector in the product:

Then plug these derivatives and the original vectors into the appropriate product rule:

Let’s calculate the product of these vectors and then differentiate, to see that dot product rule gives the right results:

This demonstration also illustrates why you don’t need the dot or cross product rules all that often: you can usually evaluate the product first, and then differentiate if you want.

Beware that the order of terms in the products in the cross product rule matters, because it matters in the cross product.

An example of tangent and unit tangent vectors?

Spacey the Space Traveler is flying along the path r(t) = ⟨t² - 1, t, (2t+3)³  ⟩. At point (-1, 0, 27), when t = 0, they accidentally drop their space baseball out the window. Find an equation for the line the baseball subsequently moves along (remember there’s no gravity or air friction).

Once dropped, the ball will move in a straight line in the direction it was traveling at the instant it was dropped. Direction (and speed) at some instant is what the derivative of a vector-valued function captures, and is in turn the tangent to the function. So find a direction vector for the ball by finding the derivative of Spacey’s path, and evaluating it at t = 0:

This direction vector is kind of big, and any vector in that direction will do. Maybe use a unit vector instead. This unit vector in the tangent direction is the principal unit tangent vector:

Finally, we can give an equation for the line the ball follows. We used the unit tangent vector, since we had it, but could have used the unscaled tangent vector too.

Next

Integrals (both definite and indefinite) of vector-valued functions.

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

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