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Vector Functions - Unit Tangent Vector |
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In order to discuss curvature and a few other topics, we need to define a special vector called the unit tangent vector. As the name indicates, the unit tangent vector is a vector that is tangent to the curve and it's length is one. |
What may not be obvious is that there is only one unit tangent vector and it points in the direction of motion. Given the vector function \(\vec{r}(t)\), the most basic equation we use to find the unit tangent vector is
\(\displaystyle{ \vec{T}(t) = \frac{\vec{r}'(t)}{ \| \vec{r}'(t) \| } }\)
The vector function \( \vec{r}(t) \) is often a position vector. As you know from basic calculus, the derivative of the position is velocity. So you will often see \(\vec{v}(t)=\vec{r}'(t)\) where \(\vec{v}(t)\) is referred to as the velocity vector. This allows us to write the unit tangent vector as \(\displaystyle{ \vec{T}(t) = \frac{\vec{v}(t)}{ \| \vec{v}(t) \| } }\).
This unit tangent vector is used a lot when calculating the principal unit normal vector, acceleration vector components and curvature. So take a few minutes to work some practice problems before going on to the next topic.
next: principal unit normal vector → |
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Practice Problems |
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Instructions - - Unless otherwise instructed, find the unit tangent vector for the given vector function at the given point. If no point is given, find the general unit tangent vector \(\vec{T}(t)\).
Level A - Basic |
Practice A03 | |
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\(\vec{r}(t)=\langle 2\sin(t),4\cos(t),4\sin^2(t)\rangle\), \(t=\pi/6\) | |
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Practice A05 | |
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Find \(\vec{T}(t)\) and \(\vec{T}(0)\) for \(\vec{r}(t)=\langle 5t^2+1,-e^{-3t},2\sin(-3t)\rangle\). | |
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Practice A06 | |
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\(\vec{r}(t)=(-t^3+t)\vhat{i}+(\ln(t^2))\vhat{j}+(\cos(\pi t))\vhat{k}\), \(t=1\) | |
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