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# Answer to Question #122825 in Vector Calculus for Ojugbele Daniel

Question #122825
What do you understand by unit tangent vector
1
Expert's answer
2020-06-23T14:34:52-0400

Given a smooth vector-valued function  "\\overrightarrow{r}(t)". Any vector parallel to "\\overrightarrow{r}'(t_0)" is tangent to the graph of "\\overrightarrow{r}(t)" at "t=t_0." It is often useful to consider just the direction of "\\overrightarrow{r}'(t)" and not its magnitude.

Therefore we are interested in the unit vector in the direction of "\\overrightarrow{r}'(t)"

This leads to a definition.

Let "\\overrightarrow{r}(t)" be a smooth function on an open interval "I." The unit tangent vector "\\overrightarrow{T}(t)" is

"\\overrightarrow{T}(t)={1\\over ||\\overrightarrow{r}'(t)||}\\overrightarrow{r}'(t), ||\\overrightarrow{r}'(t)||\\not=0"

Let "\\overrightarrow{v}(t)=\\overrightarrow{r}'(t)" denote the velocity vector. Then we define the unit tangent vector by as the unit vector in the direction of the velocity vector.

"\\overrightarrow{T}(t)={1\\over ||\\overrightarrow{v}(t)||}\\overrightarrow{v}(t), ||\\overrightarrow{v}(t)||\\not=0"

The tangential component of acceleration is in the direction of the unit tangent vector

"\\overrightarrow{a}_\\tau(t)=a\\cdot\\overrightarrow{T}(t)=a\\cdot{1\\over ||\\overrightarrow{v}(t)||}\\overrightarrow{v}(t), ||\\overrightarrow{v}(t)||\\not=0"

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