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 $r(t)=4t^3\bold i+6t\bold j+4t\ln(t)\bold k$.

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