how do we prove that the velocity vector is perpendicular to the position vector using the dot product?

In general, it is not necessarily true that the velocity vector is perpendicular to the position vector (for example, the particle could be traveling in a straight line through the origin). An additional assumption is needed, such as assuming that the position vector's magnitude is constant. Assume that |r(t)| = c for some constant c. We need to show that r'(t) and r(t) are perpendicular, or equivalently r'(t) dot r(t) is zero. Since the square of the magnitude of any vector is the dot product of the vector and itself, we have r(t) dot r(t) = c^2. We differentiate both sides with respect to t, using the analogue of the product rule for dot products: [r'(t) dot r(t)] + [r(t) dot r'(t)] = 0. Since dot product is commutative, it immediately follows that r'(t) dot r(t) is zero, so the velocity vector is perpendicular to the position vector assuming that the position vector's magnitude is constant.

Why did we have to square and then take the radical sqrt(6^2 + 27^2) = 27.6? Couldn't we just sum the output 27 + 6 = 33?

https://en.m.wikipedia.org/wiki/Pythagorean_theorem

On the second prompted challenge, the speed of the particle at t = 3 must be root(9^2 + 27^2) and not 6^2 as the first term. Just wanted to point out for an edit!

The derivative is 2*t and 2*3=6 so the article appears to be correct. Best wishes.

Main content

Multivariable calculus

Course: Multivariable calculus > Unit 2

Lesson 8: Differentiating vector-valued functions (articles)

Derivatives of vector-valued functions

Google Classroom

How to compute, and more importantly how to interpret, the derivative of a function with a vector output.

Background

What we're building to

To take the derivative of a vector-valued function, take the derivative of each component:

\begin{array}{r} \frac{d}{d t} [\begin{array}{c} x (t) \\ y (t) \end{array}] = [\begin{array}{c} x^{'} (t) \\ y^{'} (t) \end{array}] \end{array}

If you interpret the initial function as giving the position of a particle as a function of time, the derivative gives the velocity vector of that particle as a function of time.

The derivative of a vector-valued function

Good news! Computing the derivative of a vector-valued function is nothing really new. As such, I'll keep this article pretty short. The main new takeaway is interpreting the vector derivative.

Dive in with an example

Let's start with a relatively simple vector-valued function

\vec{s} (t)

, with only two components,

\begin{array}{r} \vec{s} (t) = [\begin{array}{c} 2 \sin (t) \\ 2 \cos (t / 3) t \end{array}] \end{array}

To take the derivative of

\vec{s}

, just take the derivative of each component:

\begin{aligned} \frac{d \vec{s}}{d t} (t) & = [\begin{array}{c} \frac{d}{d t} (2 \sin (t)) \\ \frac{d}{d t} (2 \cos (t / 3)) t \end{array}] \\ = [\begin{array}{c} 2 \cos (t) \\ 2 \cos (t / 3) - \frac{2}{3} \sin (t / 3) t \end{array}] \end{aligned}

You might also write this derivative as

{\vec{s}}^{'} (t)

. This derivative is a new vector-valued function, with the same input

t

that

\vec{s}

has, and whose output has the same number of dimensions.

More generally, if we write the components of

\vec{s}

x (t)

and

y (t)

, we write its derivative like this:

\begin{array}{r} {\vec{s}}^{'} (t) = [\begin{array}{c} x^{'} (t) \\ y^{'} (t) \end{array}] \end{array}

Derivative gives a velocity vector.

For the example above, how can we visualize what the derivative means? First, to visualize

\begin{array}{r} \vec{s} (t) = [\begin{array}{c} 2 \sin (t) \\ 2 \cos (t / 3) t \end{array}] \end{array}

we note that the output has more dimensions than the input, so it is well-suited to be viewed as a parametric function.

Each point on the curve represents the tip of a vector

[\begin{matrix} 2 \sin (t_{0}) \\ 2 \cos (t_{0} / 3) t_{0} \end{matrix}]

for some specific number

t_{0}

. For instance, when

t_{0} = 2

we draw a vector to the point

\begin{array}{r} \vec{s} (2) = [\begin{array}{c} 2 \sin (2) \\ 2 \cos (2 / 3) \cdot 2 \end{array}] \approx [\begin{array}{c} 1.819 \\ 3.144 \end{array}] \end{array}

When we do this for all possible inputs

t

, the tips of the vectors

\vec{s} (t)

will trace out a certain curve:

What do we get when we plug in some value of

t

, perhaps

2

again, to the derivative?

\begin{aligned} \frac{d \vec{s}}{d t} (2) & = [\begin{array}{c} 2 \cos (2) \\ 2 \cos (2 / 3) - \frac{2}{3} \sin (2 / 3) \cdot 2 \end{array}] \\ \approx [\begin{array}{c} - 0.832 \\ 0.747 \end{array}] \end{aligned}

This is also some two-dimensional vector.

It's hard to see what this derivative vector represents when it just sits at the origin, but if we shift it so that its tail sits on the tip of the vector

\vec{s} (2)

, it has a wonderful interpretation:

If $\vec{s} (t)$ ‍ represents the position of a traveling particle as a function of time, $\frac{d \vec{s}}{d t} (t_{0})$ ‍ is the velocity vector of that particle at time $t_{0}$ ‍.
Derivative is a velocity vector tangent to the curve.

In particular, this means the direction of the vector is tangent to the curve, and its magnitude indicates the speed at which one travels along this curve as

t

increases at a constant rate (as time tends to do).

Concept Check: Suppose the position in two-dimensional space of a particle, as a function of time, is given by the function

\begin{array}{r} \vec{s} (t) = [\begin{array}{c} t^{2} \\ t^{3} \end{array}] \end{array}

Problem 1

What is $\frac{d \vec{s}}{d t}$ ‍?

Problem 2

What is the speed of the particle at time $t = 3$ ‍?

Summary

To take the derivative of a vector-valued function, take the derivative of each component.
If you interpret the initial function as giving the position of a particle as a function of time, the derivative gives the velocity vector of that particle as a function of time.

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herosaitama756
Posted 4 years ago. Direct link to herosaitama756's post “how do we prove that the ...”
how do we prove that the velocity vector is perpendicular to the position vector using the dot product?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Ian Pulizzotto
  Posted 4 years ago. Direct link to Ian Pulizzotto's post “In general, it is not nec...”
  In general, it is not necessarily true that the velocity vector is perpendicular to the position vector (for example, the particle could be traveling in a straight line through the origin). An additional assumption is needed, such as assuming that the position vector's magnitude is constant.
  
  Assume that |r(t)| = c for some constant c. We need to show that r'(t) and r(t) are perpendicular, or equivalently r'(t) dot r(t) is zero.
  
  Since the square of the magnitude of any vector is the dot product of the vector and itself, we have
  r(t) dot r(t) = c^2.
  
  We differentiate both sides with respect to t, using the analogue of the product rule for dot products:
  [r'(t) dot r(t)] + [r(t) dot r'(t)] = 0.
  
  Since dot product is commutative, it immediately follows that r'(t) dot r(t) is zero, so the velocity vector is perpendicular to the position vector assuming that the position vector's magnitude is constant.
  Comment on Ian Pulizzotto's post “In general, it is not nec...”
  (11 votes)
eugene
Posted 5 years ago. Direct link to eugene's post “Why did we have to square...”
Why did we have to square and then take the radical sqrt(6^2 + 27^2) = 27.6? Couldn't we just sum the output 27 + 6 = 33?
Button navigates to signup pageComment on eugene's post “Why did we have to square...”
(0 votes)
Answer
- amadeusz.sitnicki1
  Posted 5 years ago. Direct link to amadeusz.sitnicki1's post “https://en.m.wikipedia.or...”
  https://en.m.wikipedia.org/wiki/Pythagorean_theorem
  Button navigates to signup page
  (7 votes)
aminyahyaabadi74
Posted 4 years ago. Direct link to aminyahyaabadi74's post “Just to clarify: Here it ...”
Just to clarify: Here it only talks about taking derivative relative to the same frame. If the derivative is being calculated relative to another moving frame, things are changed. ( https://en.wikipedia.org/wiki/Rotating_reference_frame )
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(1 vote)
Answer
Rafael Dolfe
Posted 6 years ago. Direct link to Rafael Dolfe's post “On the second prompted ch...”
On the second prompted challenge, the speed of the particle at t = 3 must be root(9^2 + 27^2) and not 6^2 as the first term. Just wanted to point out for an edit!
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(0 votes)
Answer
- Richard
  Posted 5 years ago. Direct link to Richard's post “The derivative is 2*t and...”
  The derivative is 2*t and 2*3=6 so the article appears to be correct. Best wishes.
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  (11 votes)