17Calculus Vector Functions - Curvature
For the following discussion, we will consider a parameterized curve defined by the vector function \(\displaystyle{ \vec{r}(t) = \langle x(t), y(t), z(t) \rangle }\) which is traversed once on the continuous interval \( a \leq t \leq b \).
Some books use the Greek letter \(\kappa\) (kappa) for curvature. We use a capital K.
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The curvature of a smooth curve is a measure of how 'tight' or 'sharp' the curve is. If we have a smooth curve \(\vec{r}\) and we have a function s which is the arc length function, the curvature is defined to be
\(\displaystyle{ K(s) = \left\| \frac{d\vec{T}}{ds} \right\|}\).
This equation for the curvature is not particularly useful for calculations. So we have several other ways to write the equation of the curvature. But first notice, that the curvature is a scalar function, not a vector function. And since it is the norm of a vector, the curvature will always be positive.
Curvature Formula #1
For our first equation to use when calculating the curvature, we will use the chain rule to write \(\displaystyle{ \frac{d\vec{T}}{dt} = \frac{d\vec{T}}{ds} \cdot \frac{ds}{dt} }\). We can solve for \(d\vec{T}/ds\) to get
\(\displaystyle{ K = \left\| \frac{d\vec{T}}{ds} \right\| = \frac{\| d\vec{T}/dt \|}{ \| ds/dt \| } = \frac{1}{\| \vec{v} \|} \left\| \frac{d\vec{T}}{dt} \right\| }\)
Notice in the previous equation, we used \(ds/dt = \|\vec{v}\|\) to simplify the equation somewhat.
Now we can write the first curvature formula in a form that we can use for calculations.
\[\displaystyle{ K(t) = \frac{1}{\|\vec{v}\|} \left\| \frac{d\vec{T}}{dt} \right\| = \frac{\| \vec{T}'(t) \|}{\|\vec{r}'(t)\|} }\]
Curvature Formula #2
If we define a vector \(\vec{a} = d\vec{v}/dt\) as an acceleration vector, a second curvature formula is \[\displaystyle{ K=\frac{\|\vec{v} \times \vec{a} \|}{\|\vec{v}\|^3} }\] Before working some practice problems, here is a quick video clip for you that should help you understand curvature a bit better.
MIP4U - Determining Curvature of a Curve Defined by a Vector Valued Function [3mins-9secs]
video by MIP4U |
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Practice
Unless otherwise instructed, calculate the curvature of these vector functions. If a point is given, calculate the curvature at that point also. Give your answers in exact, completely factored form.
\( \vec{r}(t) = 3t\vhat{i} + 4\sin t\vhat{j} + \) \( 4\cos t\vhat{k} \)
Problem Statement |
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Calculate the curvature of the vector function \( \vec{r}(t) = 3t\vhat{i} + 4\sin t\vhat{j} + \) \( 4\cos t\vhat{k} \). Give your answer in exact, completely factored form.
Final Answer |
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\( K(t) = 4/25 \)
Problem Statement |
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Calculate the curvature of the vector function \( \vec{r}(t) = 3t\vhat{i} + 4\sin t\vhat{j} + \) \( 4\cos t\vhat{k} \). Give your answer in exact, completely factored form.
Solution |
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2073 video
video by Krista King Math |
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Final Answer |
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\( K(t) = 4/25 \) |
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\( \vec{r}(t) = \langle 2\cos(t), 2\sin(t) \rangle \)
Problem Statement |
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Calculate the curvature of the vector function \( \vec{r}(t) = \langle 2\cos(t), 2\sin(t) \rangle \). Give your answer in exact, completely factored form.
Final Answer |
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\( K(t) = 1/2 \)
Problem Statement |
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Calculate the curvature of the vector function \( \vec{r}(t) = \langle 2\cos(t), 2\sin(t) \rangle \). Give your answer in exact, completely factored form.
Solution |
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2074 video
video by MIP4U |
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Final Answer |
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\( K(t) = 1/2 \) |
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\( \vec{r}(t) = \langle t, t^2/2, t^3/3 \rangle \), \(t=1\)
Problem Statement |
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Find the curvature at \(t=1\) for \( \vec{r}(t) = \langle t, t^2/2, t^3/3 \rangle \).
Final Answer |
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\( K(1) = \sqrt{2}/3 \)
Problem Statement |
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Find the curvature at \(t=1\) for \( \vec{r}(t) = \langle t, t^2/2, t^3/3 \rangle \).
Solution |
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2075 video
video by MIP4U |
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Final Answer |
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\( K(1) = \sqrt{2}/3 \) |
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\( \vec{r}(t) = \langle -5t, 2t^3, 3t^4 \rangle \), \( t = 2 \).
Problem Statement |
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Find the curvature of \( \vec{r}(t) = \langle -5t, 2t^3, 3t^4 \rangle \) at \( t = 2 \).
Final Answer |
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\( K(t) = 24\sqrt{3229}/(9817)^{3/2} \)
Problem Statement |
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Find the curvature of \( \vec{r}(t) = \langle -5t, 2t^3, 3t^4 \rangle \) at \( t = 2 \).
Solution |
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2076 video
video by MIP4U |
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Final Answer |
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\( K(t) = 24\sqrt{3229}/(9817)^{3/2} \) |
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\( \vec{r}(t) = \langle 2\cos(3t),2\sin(3t), \) \(4t \rangle \), \( t = 0 \).
Problem Statement |
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Find the curvature of \( \vec{r}(t) = \langle 2\cos(3t),2\sin(3t), \) \(4t \rangle \) at \( t = 0 \).
Final Answer |
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9/26
Problem Statement |
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Find the curvature of \( \vec{r}(t) = \langle 2\cos(3t),2\sin(3t), \) \(4t \rangle \) at \( t = 0 \).
Solution |
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This problem is solved in each of these videos. The first video uses the cross product version of the curvature formula, the second video does not use the cross product.
2077 video
video by MIP4U |
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2077 video
video by MIP4U |
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Final Answer |
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9/26 |
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You CAN Ace Calculus
Topics You Need To Understand For This Page
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 | ||
|---|---|---|
\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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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 | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
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 curvature of these vector functions. If a point is given, calculate the curvature at that point also. Give your answers in exact, completely factored form.
