How to find the unit tangent vector
Formula for the unit tangent vector
To find the unit tangent vector for a vector function, we use the formula
???T(t)=\frac{r'(t)}{||r'(t)||}???
where ???r'(t)??? is the derivative of the vector function ???r(t)=r(t)_1\bold i+r(t)_2\bold j+r(t)_3\bold k??? and ???t??? is given.
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Remember that ???||r'(t)||??? is the magnitude of the derivative of the vector function at time ???t???. We can find ???|r'(t)|??? using the formula
???||r'(t)||=\sqrt{[r'(t)_1]^2+[r'(t)_2]^2+[r'(t)_3]^2}???
How to find the unit tangent vector to a vector function at a particular parameter value
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Let’s do a step-by-step example of how to find the unit tangent vector for a vector function
Example
Find the unit tangent vector of the vector function at the given value of ???t=1???.
???r(t)=4t^3\bold i+6t\bold j+4t\ln(t)\bold k???
We’ll start by finding the derivative of the vector function ???r(t)=4t^3\bold i+6t\bold j+4t\ln(t)\bold k??? at time ???t=1??? so that we can plug it into the formula for the unit tangent vector. To find the derivative, we’ll just replace each of the coefficients with their derivatives. The derivative of ???4t^3??? is ???12t^2???; the derivative of ???6t??? is ???6???; the derivative of ???4t\ln{(t)}??? using product rule is ???(4)(\ln(t))+(4t)(1/t)???.
???r'(t)=12t^2\bold i+6\bold j+\left[(4)(\ln(t))+(4t)\left(\frac{1}{t}\right)\right]\bold k???
???r'(t)=12t^2\bold i+6\bold j+[4\ln(t)+4]\bold k???
Now we’ll find the value of the derivative at ???t=1???.
???r'(1)=12(1)^2\bold i+6\bold j+[4\ln(1)+4]\bold k???
???r'(1)=12\bold i+6\bold j+[4(0)+4]\bold k???
???r'(1)=12\bold i+6\bold j+4\bold k???
Remember that ||r'(t)|| is the magnitude of the derivative of the vector function at time t.
Now we’ll use the values from the derivative to find the magnitude of the vector function at ???t=1??? so that we can plug it into the formula for the unit tangent vector.
???||r'(t)||=\sqrt{[r'(t)_1]^2+[r'(t)_2]^2+[r'(t)_3]^2}???
???||r'(1)||=\sqrt{12^2+6^2+4^2}???
???||r'(1)||=\sqrt{144+36+16}???
???||r'(1)||=\sqrt{196}???
???||r'(1)||=14???
Plugging everything into the formula for the unit tangent vector, we get
???T(1)=\frac{12\bold i+6\bold j+4\bold k}{14}???
???T(1)=\frac{12}{14}\bold i+\frac{6}{14}\bold j+\frac{4}{14}\bold k???
???T(1)=\frac{6}{7}\bold i+\frac{3}{7}\bold j+\frac{2}{7}\bold k???
which is the equation of the unit tangent vector for ???r(t)=4t^3\bold i+6t\bold j+4t\ln(t)\bold k???.
