# Derivatives of vector functions

• tony873004
In summary, the trajectory of a baby fly is represented by the position vector at time t given by <t^2, t>. On a graph, <t^2, t>, <2t, 1>, and <2, 1> are carefully sketched in their appropriate positions. The tangent vector at time t is found by dividing the position vector by its magnitude, and the normal vector, which is orthogonal to the tangent vector, can be found by using the formula <a_2, -a_1> or <-a_2, a_1>. The unit normal vector can also be found by dividing the tangent vector by its magnitude.
tony873004
Gold Member

Above is the trajectory of a baby fly, whose position vector at time t is given by

$$\overrightarrow r \left( t \right) = \left\langle {t^2 ,t} \right\rangle$$.

1. On the graph, carefully sketch , $$\overrightarrow r \left( 1 \right),\,\,\overrightarrow {r'} \left( 1 \right),\,\,\overrightarrow T \left( 1 \right),\,\overrightarrow N \left( 1 \right),{\rm{ and }}\overrightarrow {r''} \left( 1 \right)$$ in their appropriate positions. Use a straight edge and make sure to pay attention to the lengths of the vectors.

My effort so far:

$$\begin{array}{l} \overrightarrow r \left( 1 \right) = \left\langle {1^2 ,\,1} \right\rangle = \left\langle {1,\,1} \right\rangle \\ \overrightarrow {r'} \left( t \right) = \left\langle {2t,\,1} \right\rangle ,\,\,\,\,\overrightarrow {r'} \left( 1 \right) = \left\langle {2 \cdot 1,\,1} \right\rangle \,\, = \,\left\langle {2,\,1} \right\rangle \\ \\ \overrightarrow T \left( 1 \right) = \frac{{\overrightarrow {r'} \left( 1 \right)}}{{\left| {\overrightarrow {r'} \left( 1 \right)} \right|}} = \frac{{\,\left\langle {2,\,1} \right\rangle }}{{\sqrt {2^2 + 1^2 } }} = \frac{{\,\left\langle {2,\,1} \right\rangle }}{{\sqrt {4 + 1} }} = \frac{{\,\left\langle {2,\,1} \right\rangle }}{{\sqrt 5 }} \\ \end{array}$$

Our book doesn't cover N. I imagine that's the normal vector, which is orthogonal to the tangent vector. But how do I find that? I know that if N and T are orthogonal, then their dot product should equal 0. So what times <2,1>/sqrt(5)=0? <0,0> would work, but that's the zero vector. I don't imagine that's what I'm trying to find.

Any clues?

nitpick: There's no such thing as a baby fly. Fly is the adult stage of an insect that was once a maggot.

In general, if a 2-coordinate vector is given by $<a_1,a_2>$ then a normal vector (my best guess is as good as yours) can be given by $<-a_2,a_1>$ or $<a_2,-a_1>$. There might be an actual convention as to which to use, but the normal vector typically points "out" away from the curvature I think. I didn't normalize the result for brevity.

Last edited:
Looking over my notes:

$$\overrightarrow N = \frac{\overrightarrow T'}{|\overrightarrow T' |}$$

N is the Unit Normal Vector, a special one since T and T' are orthogonal

Don't know if that's what your problem asks for, but the T's are defined the exactly the same, so guessing this definition is about right.

## What are derivatives of vector functions?

Derivatives of vector functions are mathematical operations that calculate the instantaneous rate of change of a vector function at a specific point. They are used to analyze the behavior and properties of vector functions.

## Why are derivatives of vector functions important?

Derivatives of vector functions are important because they allow us to understand the direction and rate of change of a vector function at any given point. They are used in various fields such as physics, engineering, and economics for modeling and predicting real-world phenomena.

## How do you calculate derivatives of vector functions?

The process of calculating derivatives of vector functions involves using standard rules of differentiation, such as the power rule, product rule, and chain rule. These rules are applied to each component of the vector function to obtain the derivative of the entire function.

## What is the relationship between derivatives of vector functions and tangent vectors?

The derivative of a vector function at a point is the same as the tangent vector to the curve defined by the vector function at that point. This means that the derivative gives us the direction and magnitude of the instantaneous change at that point, which is the same as the direction and magnitude of the tangent to the curve at that point.

## Can derivatives of vector functions be higher order?

Yes, derivatives of vector functions can be taken to higher orders, just like derivatives of scalar functions. Higher order derivatives can provide more information about the behavior and properties of the vector function, such as curvature and concavity. However, they are less commonly used compared to first and second order derivatives.

• Calculus and Beyond Homework Help
Replies
6
Views
583
• Calculus and Beyond Homework Help
Replies
2
Views
429
• Electromagnetism
Replies
1
Views
477
• Calculus and Beyond Homework Help
Replies
1
Views
643
• Calculus and Beyond Homework Help
Replies
6
Views
1K
• Special and General Relativity
Replies
14
Views
823
• Calculus and Beyond Homework Help
Replies
16
Views
1K
Replies
1
Views
489
• Calculus and Beyond Homework Help
Replies
1
Views
299
• Calculus and Beyond Homework Help
Replies
9
Views
765