Finding curvature of the vector function

Steps for finding curvature

To find the curvature $\kappa(t)$ of a vector function $r(t)=r(t)_1\bold i+r(t)_2\bold j+r(t)_3\bold k$, we’ll use the equation

$\kappa(t)=\frac{\left|T'(t)\right|}{\left|r'(t)\right|}$

where $\left|T'(t)\right|$ is the magnitude of the derivative of the unit tangent vector $T(t)$, which we can find using

$\left|T'(t)\right|=\sqrt{\left[T'(t)_1\right]^2+\left[T'(t)_2\right]^2+\left[T'(t)_3\right]^2}$

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where $T(t)$ is the unit tangent vector, which we can find using

$T(t)=\frac{r'(t)}{\left|r'(t)\right|}$

where $r'(t)$ is the derivative of the vector function and where $\left|r'(t)\right|$ is the magnitude of the derivative of the vector function, which we can find using

$\left|r'(t)\right|=\sqrt{\left[r'(t)_1\right]^2+\left[r'(t)_2\right]^2+\left[r'(t)_3\right]^2}$

In other words, in order to find $\kappa(t)$, we’ll

1. Find $r'(t)$, and use it to

2. Find $\left|r'(t)\right|$, and then use $r'(t)$ and $\left|r'(t)\right|$ to

3. Find $T(t)$, and then use it to

4. Find $T'(t)$, and then use it to

5. Find $\left|T'(t)\right|$, and then use $\left|r'(t)\right|$ and $\left|T'(t)\right|$ to

6. Find $\kappa(t)$

How to find curvature of a vector function

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How to find curvature step-by-step for a vector function

Example

Find the curvature of the vector function.

$r(t)=4t\bold i+t^2\bold j+2t\bold k$

We’ll start by calculating the derivative of the vector function.

$r'(t)=4\bold i+2t\bold j+2\bold k$

Then we’ll find $\left|r'(t)\right|$.

$\left|r'(t)\right|=\sqrt{\left[r'(t)_1\right]^2+\left[r'(t)_2\right]^2+\left[r'(t)_3\right]^2}$

$\left|r'(t)\right|=\sqrt{(4)^2+(2t)^2+(2)^2}$

$\left|r'(t)\right|=\sqrt{16+4t^2+4}$

$\left|r'(t)\right|=\sqrt{4t^2+20}$

$\left|r'(t)\right|=\sqrt{4\left(t^2+5\right)}$

$\left|r'(t)\right|=2\sqrt{t^2+5}$

Now we’ll use the derivative and its magnitude to find an equation for the unit tangent vector $T(t)$.

$T(t)=\frac{r'(t)}{\left|r'(t)\right|}$

$T(t)=\frac{4\bold i+2t\bold j+2\bold k}{2\sqrt{t^2+5}}$

$T(t)=\frac{4}{2\sqrt{t^2+5}}\bold i+\frac{2t}{2\sqrt{t^2+5}}\bold j+\frac{2}{2\sqrt{t^2+5}}\bold k$

$T(t)=\frac{2}{\sqrt{t^2+5}}\bold i+\frac{t}{\sqrt{t^2+5}}\bold j+\frac{1}{\sqrt{t^2+5}}\bold k$

To find curvature, we need the derivative of the vector function, the magnitude of the derivative, the unit tangent vector, its derivative, and the magnitude of its derivative.

Now we can find the derivative of the unit tangent vector $T'(t)$. We’ll need to use quotient rule to find the derivatives of the coefficients on $\bold i$, $\bold j$, and $\bold k$.

$T'(t)=\frac{(0)\sqrt{t^2+5}-(2)\left[\frac12\left(t^2+5\right)^{-\frac12}(2t)\right]}{\left(\sqrt{t^2+5}\right)^2}\bold i+\frac{(1)\sqrt{t^2+5}-(t)\left[\frac12\left(t^2+5\right)^{-\frac12}(2t)\right]}{\left(\sqrt{t^2+5}\right)^2}\bold j$

$+\frac{(0)\sqrt{t^2+5}-(1)\left[\frac12\left(t^2+5\right)^{-\frac12}(2t)\right]}{\left(\sqrt{t^2+5}\right)^2}\bold k$

$T'(t)=\frac{-2t\left(t^2+5\right)^{-\frac12}}{t^2+5}\bold i+\frac{\sqrt{t^2+5}-t^2\left(t^2+5\right)^{-\frac12}}{t^2+5}\bold j+\frac{-t\left(t^2+5\right)^{-\frac12}}{t^2+5}\bold k$

$T'(t)=-\frac{2t}{\left(t^2+5\right)^\frac32}\bold i+\frac{\sqrt{t^2+5}-\frac{t^2}{\sqrt{t^2+5}}}{t^2+5}\bold j-\frac{t}{\left(t^2+5\right)^\frac32}\bold k$

$T'(t)=-\frac{2t}{\left(t^2+5\right)^\frac32}\bold i+\frac{\frac{t^2+5}{\sqrt{t^2+5}}-\frac{t^2}{\sqrt{t^2+5}}}{t^2+5}\bold j-\frac{t}{\left(t^2+5\right)^\frac32}\bold k$

$T'(t)=-\frac{2t}{\left(t^2+5\right)^\frac32}\bold i+\frac{\frac{t^2+5-t^2}{\sqrt{t^2+5}}}{t^2+5}\bold j-\frac{t}{\left(t^2+5\right)^\frac32}\bold k$

$T'(t)=-\frac{2t}{\left(t^2+5\right)^\frac32}\bold i+\frac{\frac{5}{\sqrt{t^2+5}}}{t^2+5}\bold j-\frac{t}{\left(t^2+5\right)^\frac32}\bold k$

$T'(t)=-\frac{2t}{\left(t^2+5\right)^\frac32}\bold i+\frac{5}{\left(t^2+5\right)^\frac32}\bold j-\frac{t}{\left(t^2+5\right)^\frac32}\bold k$

Then we’ll find the magnitude of the derivative of the unit tangent vector $\left|T'(t)\right|$.

$\left|T'(t)\right|=\sqrt{\left[T'(t)_1\right]^2+\left[T'(t)_2\right]^2+\left[T'(t)_3\right]^2}$

$\left|T'(t)\right|=\sqrt{\left[-\frac{2t}{\left(t^2+5\right)^\frac32}\right]^2+\left[\frac{5}{\left(t^2+5\right)^\frac32}\right]^2+\left[-\frac{t}{\left(t^2+5\right)^\frac32}\right]^2}$

$\left|T'(t)\right|=\sqrt{\frac{4t^2}{\left(t^2+5\right)^3}+\frac{25}{\left(t^2+5\right)^3}+\frac{t^2}{\left(t^2+5\right)^3}}$

$\left|T'(t)\right|=\sqrt{\frac{5t^2+25}{\left(t^2+5\right)^3}}$

$\left|T'(t)\right|=\sqrt{\frac{5\left(t^2+5\right)}{\left(t^2+5\right)^3}}$

$\left|T'(t)\right|=\sqrt{\frac{5}{\left(t^2+5\right)^2}}$

$\left|T'(t)\right|=\frac{\sqrt{5}}{t^2+5}$

Finally we can solve for the curvature $\kappa(t)$ of the vector function

$\kappa(t)=\frac{\left|T'(t)\right|}{\left|r'(t)\right|}$

$\kappa(t)=\frac{\frac{\sqrt{5}}{t^2+5}}{2\sqrt{t^2+5}}$

$\kappa(t)=\frac{\sqrt{5}}{2\left(t^2+5\right)^\frac32}$

This is the curvature of the vector function.

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