Question

A curve \(C\) is described along with 2 points on \(C\). (a) Using a sketch, determine at which of these points the curvature is greater. (b) Find the curvature \(\kappa\) of \(C\), and evaluate \(\kappa\) at each of the 2 given points. \(C\) is defined by \(\vec{r}(t)=\left\langle t^{2}-1, t^{3}-t\right\rangle ;\) points given at \(t=0\) and \(t=5\).

Short answer

The curvature is greater at point \( t=0 \). \( \kappa(t=0)=2 \) and \( \kappa(t=5)\) is approximately zero.

Step-by-step solution

Understanding Curvature

The curvature, \( \kappa \), of a curve at a given point measures how quickly the curve is changing direction at that point. For a parametric curve \( \vec{r}(t) = \langle x(t), y(t) \rangle \), the curvature is given by \[\kappa = \frac{|x' y'' - y' x''|}{(x'^2 + y'^2)^{3/2}}\]where \( x' = \frac{dx}{dt}, y' = \frac{dy}{dt}, x'' = \frac{d^2x}{dt^2}, y'' = \frac{d^2y}{dt^2} \). First, we'll find these derivatives for our curve \( \vec{r}(t) = \langle t^2 - 1, t^3 - t \rangle \).

Calculate First Derivatives

Find the first derivatives:\( x(t) = t^2 - 1 \), so \( x'(t) = \frac{d}{dt}(t^2 - 1) = 2t \).\( y(t) = t^3 - t \), so \( y'(t) = \frac{d}{dt}(t^3 - t) = 3t^2 - 1 \).

Calculate Second Derivatives

Now, calculate the second derivatives:\( x''(t) = \frac{d}{dt}(2t) = 2 \).\( y''(t) = \frac{d}{dt}(3t^2 - 1) = 6t \).

Apply Curvature Formula

Substitute the derivatives into the curvature formula:\[\kappa = \frac{|(2t)(6t) - (3t^2 - 1)(2)|}{((2t)^2 + (3t^2-1)^2)^{3/2}}\] Simplify:\[\kappa = \frac{|12t^2 - (6t^2 - 2)|}{(4t^2 + (9t^4 - 6t^2 + 1))^{3/2}}\]\[= \frac{|12t^2 - 6t^2 + 2|}{(9t^4 - 2t^2 + 1)^{3/2}}\]\[= \frac{|6t^2 + 2|}{(9t^4 - 2t^2 + 1)^{3/2}}\].

Evaluate at Given Points

Evaluate \( \kappa \) at \( t=0 \) and \( t=5 \):For \( t=0 \):\[ \kappa = \frac{|6(0)^2 + 2|}{(9(0)^4 - 2(0)^2 + 1)^{3/2}} = \frac{2}{1^{3/2}} = 2 \]For \( t=5 \):\[ \kappa = \frac{|6(5)^2 + 2|}{(9(5)^4 - 2(5)^2 + 1)^{3/2}} = \frac{152}{2278^{3/2}} \].

Comparing Curvature at Points

Compare the calculated curvatures: At \( t=0 \), \( \kappa = 2 \), and at \( t=5 \), \( \kappa \) is considerably smaller, \( \frac{152}{2278^{3/2}} \). The curvature is greater at \( t=0 \) as the numerical value 2 is larger than the small fractional value at \( t=5 \).

Key concepts

Understanding Parametric Equations

Parametric equations are a powerful way to describe curves in a plane. Instead of expressing a function explicitly as \(y = f(x)\), we use an intermediary variable, often \(t\), known as the parameter. This gives us two functions: \(x(t)\) and \(y(t)\). These functions define coordinates \((x, y)\) as \(t\) varies over a certain interval.

In the given exercise, the parametric equations for the curve \(C\) are \(x(t) = t^2 - 1\) and \(y(t) = t^3 - t\). This means the position along the curve for any value of \(t\) can be determined by evaluating these two expressions for that \(t\). These equations allow us to easily describe motion along the curve, especially when curves are not just simple lines or circles.

Calculating Derivatives

The concept of derivatives plays a central role in understanding the behavior of curves. Derivatives measure how a function changes as its input changes. For parametric curves, finding the derivatives of \(x(t)\) and \(y(t)\) with respect to \(t\) tells us about the velocity of the curve at a certain point.

In the exercise, we calculated the first derivatives: \(x'(t) = 2t\) and \(y'(t) = 3t^2 - 1\). These tell us how the \(x\) and \(y\) components of the curve change. We also compute the second derivatives \(x''(t) = 2\) and \(y''(t) = 6t\), which inform us about the acceleration of the curve's \(x\) and \(y\) components. These derivatives are essential for later computing the curvature of the curve.

Using the Curvature Formula

Curvature quantifies how much a curve deviates from being a straight line. Specifically, for a parametric curve, the curvature \(\kappa\) at a point can be found using the formula:\[\kappa = \frac{|x'y'' - y'x''|}{(x'^2 + y'^2)^{3/2}}.\]
Here, \(x'\), \(y'\), \(x''\), and \(y''\) are the first and second derivatives of \(x\) and \(y\) with respect to \(t\), as found earlier. This formula takes into account how the curve is turning and is particularly useful for complex or non-linear curves described by parametric equations.

When applied to our curve \(C\), after substituting the derivatives, we get:\[\kappa = \frac{|6t^2 + 2|}{(9t^4 - 2t^2 + 1)^{3/2}}.\]This expression can be evaluated at specific points to determine how sharply the curve bends at those points.

Evaluating Curvature at Specific Points

Once we have the curvature formula, we can evaluate the curvature at any desired points on the curve. In this exercise, we are interested in \(t = 0\) and \(t = 5\). Evaluating at these points involves plugging the values into the curvature expression.- At \(t = 0\), the curvature, \(\kappa = 2\).- At \(t = 5\), calculating gives a smaller value, \(\kappa = \frac{152}{2278^{3/2}}\).
These values tell us that the curve changes direction more sharply at \(t = 0\) compared to \(t = 5\), where it is relatively less curved. Such an analysis is invaluable in various fields, from engineering to computer graphics, where understanding the properties of curves is essential.