17Calculus Vector Functions - Acceleration Vector
On this page, we take the acceleration vector that you learned to find on the projectile motion page and express it in terms of the unit tangent and the principal unit normal vectors.
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Explanation of the Acceleration Vector Components
On the projectile motion page, we showed you how to find the acceleration vector from the position or velocity vector. The acceleration vector may be written in the form
\(\vec{a}(t)=a_x(t)\vhat{i}+a_y(t)\vhat{j}+a_z(t)\vhat{k}\) |
As written above, the acceleration vector \(\vec{a}(t)\) is expressed using the three standard unit vectors, \(\vhat{i}\), \(\vhat{j}\) and \(\vhat{k}\). This is only one of an infinite number of ways to write a vector. As long as the unit vectors cover (or span) all of 3-dim space (a concept from linear algebra), we can use them to write any vector in terms of them. We will not worry about the linear algebra details here other than to say the unit tangent vector and the principal unit normal vector cover all of the space required to describe the acceleration vector. Another way of saying this is that the acceleration vector lies in the plane determined by \(\vhat{T}\) and \(\vhat{N}\). So, we can write the acceleration vector in terms of these two unit vectors in the following form.
\(\vec{a}(t) = a_{\vhat{T}}(t)\vhat{T} + a_{\vhat{N}}(t)\vhat{N}\) |
Notes - -
1. On this page, we write the unit tangent vector and principal unit normal vector with a 'hat' sign (\(\vhat{T}\)) instead of an arrow (\(\vec{T}\)). The 'hat' sign emphasizes that we are talking about vectors of length one (unit vectors). Your textbook and instructor may or may not follow this convention.
2. You may have learned in previous math classes or in physics that the acceleration is always perpendicular to the velocity. This is true only for constant speed, i.e. \(\|\vec{v}\| = C\) where C is a constant.
How To Calculate Acceleration Vector Components
Let's think about this for a moment. We have a unit vector (let's work with the unit tangent vector) and we want to find the projection of the acceleration vector in the direction of the unit tangent vector. You already know how to do this, right? What operation does this sound like? The dot product, right? Basically, the dot product calculates the projection of one vector onto another. Knowing this, our basic equations are easy.
\(a_{\vhat{T}} = \vec{a} \cdot \vhat{T}\) |
\(a_{\vhat{N}} = \vec{a} \cdot \vhat{N}\) |
We call \(a_{\vhat{T}}\) the tangential component of acceleration and \(a_{\vhat{N}}\) the normal component of acceleration.
Other Equations To Calculate The Acceleration Vector Components
There are other equations used to calculate the acceleration vector components in terms of the unit tangent vector and the principal unit normal vector. Which equation you use depends on the problem statement and what data you have. Let's develop a few here starting with the unit tangent vector.
Our basic equation for the tangential component of acceleration is \(a_{\vhat{T}} = \vec{a} \cdot \vhat{T}\). Since \(\vhat{T} = \vec{v}/\|\vec{v}\|\), we can write \(\displaystyle{a_{\vhat{T}} = \frac{\vec{a} \cdot \vec{v}}{\|\vec{v}\|} }\), which we can also use to calculate \(a_{\vhat{T}}\).
For the next equation, we are going to derive the tangential and normal components of the acceleration by what will seem like 'going in the back door,' i.e. indirectly but the equations are equally valid and usable. (Note - Most instructors will not expect you to come up with these equations this way.)
We start with \(\vhat{T}=\vec{v}/\|\vec{v}\|\), which is the basic definition of the unit tangent vector. Solving for \(\vec{v}\), we have \(\vec{v}=\|\vec{v}\|\vhat{T}\). Now we will take the derivative of both sides. On the left, we have the acceleration vector. On the right, we need to use the product rule, which gives us
\(\vec{a} = \vec{v}' = \|\vec{v}\|\vhat{T}' + \|\vec{v}\|'\vhat{T}\)
Next, we are going to do something that is really common when working with equations. It is a technique that you need to take note of, if you haven't already. We are going to multiply one of the factors by a special form of the constant 1, namely \(\|\vhat{T}'\|/\|\vhat{T}'\|\) to get
\(\displaystyle{\vec{a} = \|\vec{v}\|\vhat{T}' \frac{ \|\vhat{T}'\| }{ \|\vhat{T}'\| } + \|\vec{v}\|'\vhat{T} }\).
Now, we know that the principal unit normal vector is \(\displaystyle{ \vhat{N} = \frac{\vhat{T}'}{\|\vhat{T}'\|}}\), so we use that in the first term above to get
\(\displaystyle{\vec{a} = \|\vec{v}\| \|\vhat{T}'\| \vhat{N} + \|\vec{v}\|'\vhat{T} }\).
Okay, so if we look carefully at this last equation, we can pick off the tangential component of acceleration and the normal component of acceleration. The tangential component of acceleration is the coefficient of \(\vhat{T}\), namely \(\|\vec{v}\|'\). Similarly, the normal component of acceleration is the coefficient of \(\vhat{N}\), namely \(\|\vec{v}\| \|\vhat{T}'\|\). Now we have two more equations for the components of acceleration.
\(a_{\vhat{T}} = \| \vec{v} \|' \) |
\(a_{\vhat{N}} = \|\vec{v}\| \|\vhat{T}'\|\) |
Careful - - Take extra care to note where the prime mark is each of these equations. For \(a_{\vhat{T}}\), you need to find the magnitude of the velocity before taking the derivative. With \(a_{\vhat{N}}\), you need to take the derivative of the principal unit normal vector first, then find the magnitude.
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Figure 1 |
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Okay, so those are all the equations we are going to develop for the unit tangent vector. Let's get two more equations for the principal unit normal vector. We will use Figure 1 to help visualize the vector equations.
If we designate θ as the angle between the unit tangent vector and the acceleration vector, as we show in Figure 1, then the normal component of acceleration can be written as \(a_{\vhat{N}} = \|\vec{a}\| \sin\theta\). Multiplying and dividing by \(\|\vec{v}\|\), we have \(\displaystyle{a_{\vhat{N}} = \frac{\|\vec{v}\| \|\vec{a}\| \sin\theta}{\|\vec{v}\|} }\). Now, remember from the cross product that \(\|\vec{v}\| \|\vec{a}\| \sin\theta = \|\vec{v} \times \vec{a}\|\). So now we have another equation for the normal component of acceleration, \(\displaystyle{a_{\vhat{N}} = \frac{\|\vec{v} \times \vec{a}\|}{\|\vec{v}\|} }\).
Our last equation for the normal component of acceleration is pretty easy to get. We just use the Pythagorean Theorem on Figure 1, i.e. \( \|\vec{a}\|^2 = a_{\vhat{T}}^2 + a_{\vhat{N}}^2\) and solve for \(a_{\vhat{N}}\) to get \(a_{\vhat{N}} = \sqrt{\|\vec{a}\|^2 - a_{\vhat{T}}^2}\).
If you think about it, there are other equations we could have derived. However, these are the ones used the most and, now that you understand some of the techniques we use, you can derive other ones that you might need in the future. Here is quick video clip that discusses the equations on this page but does not derive them.
MIP4U - Determining the Tangential and Normal Components of Acceleration [2mins-9secs]
video by MIP4U |
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Okay, time for some practice problems.
Practice
Unless otherwise instructed, calculate the tangential and normal components of the acceleration of vector for the given position vector. If a value for \(t\) is given, also find the values at that time. Give your answers in exact form.
\( \vec{r}(t) = \langle t, t^2/4\rangle \), \( t=2 \)
Problem Statement |
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Calculate the tangential and normal components of the acceleration of vector \( \vec{r}(t) = \langle t, t^2/4\rangle \) at the point \( t=2 \).
Final Answer |
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\(\displaystyle{ a_{\vec{T}}(t) = \frac{t/4}{\sqrt{1+t^2/4}}}\), \(a_{\vec{T}}(2) = 1/(2\sqrt{2})\)
\(\displaystyle{a_{\vec{N}}(t) = \frac{1/2}{\sqrt{1+t^2/4}}}\), \(a_{\vec{N}}(2) = 1/(2\sqrt{2})\)
Problem Statement |
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Calculate the tangential and normal components of the acceleration of vector \( \vec{r}(t) = \langle t, t^2/4\rangle \) at the point \( t=2 \).
Solution |
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2069 video
video by MIP4U |
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Final Answer |
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\(\displaystyle{ a_{\vec{T}}(t) = \frac{t/4}{\sqrt{1+t^2/4}}}\), \(a_{\vec{T}}(2) = 1/(2\sqrt{2})\)
\(\displaystyle{a_{\vec{N}}(t) = \frac{1/2}{\sqrt{1+t^2/4}}}\), \(a_{\vec{N}}(2) = 1/(2\sqrt{2})\)
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\( \vec{r}(t) = \langle t,t^3 \rangle \), \( t=1 \)
Problem Statement |
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Calculate the tangential and normal components of the acceleration of vector \( \vec{r}(t) = \langle t,t^3 \rangle \) at the point \( t=1 \).
Final Answer |
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\(\displaystyle{ a_{\vec{T}}(t) = \frac{18t^3}{\sqrt{1+9t^4}} }\), \( a_{\vec{T}}(1) = 18/\sqrt{10} \)
\(\displaystyle{ a_{\vec{N}}(t) = \frac{6t}{\sqrt{1+9t^4}} }\), \( a_{\vec{N}}(1) = 6/\sqrt{10} \)
Problem Statement |
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Calculate the tangential and normal components of the acceleration of vector \( \vec{r}(t) = \langle t,t^3 \rangle \) at the point \( t=1 \).
Solution |
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2070 video
video by MIP4U |
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Final Answer |
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\(\displaystyle{ a_{\vec{T}}(t) = \frac{18t^3}{\sqrt{1+9t^4}} }\), \( a_{\vec{T}}(1) = 18/\sqrt{10} \)
\(\displaystyle{ a_{\vec{N}}(t) = \frac{6t}{\sqrt{1+9t^4}} }\), \( a_{\vec{N}}(1) = 6/\sqrt{10} \)
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\( \vec{r}(t) = \cos t\vhat{i} + \sin t\vhat{j} + t\vhat{k} \)
Problem Statement |
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Calculate the tangential and normal components of the acceleration of vector \( \vec{r}(t) = \cos t\vhat{i} + \sin t\vhat{j} + t\vhat{k} \).
Final Answer |
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\( a_{\vec{T}}(t) = 0 \), \( a_{\vec{N}}(t) = 1 \)
Problem Statement |
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Calculate the tangential and normal components of the acceleration of vector \( \vec{r}(t) = \cos t\vhat{i} + \sin t\vhat{j} + t\vhat{k} \).
Solution |
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2071 video
video by Krista King Math |
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Final Answer |
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\( a_{\vec{T}}(t) = 0 \), \( a_{\vec{N}}(t) = 1 \)
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Calculate \( a_{\vec{T}}(1) \) and \( a_{\vec{N}}(1) \) for \( \vec{r}(t) = \langle \cos(\pi t), \sin(\pi t), t^3\rangle \).
Problem Statement |
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Calculate \( a_{\vec{T}}(1) \) and \( a_{\vec{N}}(1) \) for \( \vec{r}(t) = \langle \cos(\pi t), \sin(\pi t), t^3\rangle \).
Final Answer |
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\(\displaystyle{ a_{\vec{T}}(1) = \frac{18}{\sqrt{\pi^2+9}} }\), \(\displaystyle{ a_{\vec{N}}(1) = \frac{\pi\sqrt{\pi^4+9\pi^2+36}}{\sqrt{\pi^2+9}} }\)
Problem Statement |
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Calculate \( a_{\vec{T}}(1) \) and \( a_{\vec{N}}(1) \) for \( \vec{r}(t) = \langle \cos(\pi t), \sin(\pi t), t^3\rangle \).
Solution |
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2072 video
video by David Lippman |
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Final Answer |
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\(\displaystyle{ a_{\vec{T}}(1) = \frac{18}{\sqrt{\pi^2+9}} }\), \(\displaystyle{ a_{\vec{N}}(1) = \frac{\pi\sqrt{\pi^4+9\pi^2+36}}{\sqrt{\pi^2+9}} }\)
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Topics You Need To Understand For This Page
Equations On This Page
This is a partial list of equations for calculating the tangential and normal components of acceleration. For a full list, see the vector functions equations page.
\(\vec{a}\) |
acceleration vector |
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\(\vhat{T} = \vec{T}\) |
unit tangent vector |
\(\vhat{N} = \vec{N}\) |
principal unit normal vector |
\( \vec{a} = a_{\vhat{T}}\vhat{T} + a_{\vhat{N}}\vhat{N}\) | |
tangential component of acceleration | |
\(a_{\vhat{T}} = \vec{a} \cdot \vhat{T} \) | |
\(\displaystyle{a_{\vhat{T}} = \frac{\vec{a} \cdot \vec{v}}{\|\vec{v}\|} }\) | |
\(a_{\vhat{T}} = \| \vec{v} \|' \) | |
normal component of acceleration | |
\(a_{\vhat{N}} = \vec{a} \cdot \vhat{N} \) | |
\(a_{\vhat{N}} = \|\vec{v}\| \|\vhat{T}'\|\) | |
\(\displaystyle{a_{\vhat{N}} = \frac{\|\vec{v} \times \vec{a}\|}{\|\vec{v}\|} }\) | |
\(a_{\vhat{N}} = \sqrt{\|\vec{a}\|^2 - a_{\vhat{T}}^2}\) | |
Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
|---|---|
\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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Practice Instructions
Unless otherwise instructed, calculate the tangential and normal components of the acceleration of vector for the given position vector. If a value for \(t\) is given, also find the values at that time. Give your answers in exact form.
