Question

Cusps and noncusps a. Graph the curve \(\mathbf{r}(t)=\left\langle t^{3}, t^{3}\right\rangle .\) Show that \(\mathbf{r}^{\prime}(0)=\mathbf{0}\) and the curve does not have a cusp at \(t=0 .\) Explain. b. Graph the curve \(\mathbf{r}(t)=\left\langle t^{3}, t^{2}\right\rangle .\) Show that \(\mathbf{r}^{\prime}(0)=\mathbf{0}\) and the curve has a cusp at \(t=0 .\) Explain. c. The functions \(\mathbf{r}(t)=\left\langle t, t^{2}\right\rangle\) and \(\mathbf{p}(t)=\left\langle t^{2}, t^{4}\right\rangle\) both satisfy \(y=x^{2} .\) Explain how the curves they parameterize are different. d. Consider the curve \(\mathbf{r}(t)=\left\langle t^{m}, t^{n}\right\rangle,\) where \(m>1\) and \(n>1\) are integers with no common factors. Is it true that the curve has a cusp at \(t=0\) if one (not both) of \(m\) and \(n\) is even? Explain.

Short answer

Explain your reasoning. a) \(\mathbf{r}(t) = \langle t^3, t^3 \rangle\) b) \(\mathbf{r}(t) = \langle t^3, t^2 \rangle\) c) Compare \(\mathbf{r}(t) = \langle t, t^2 \rangle\) and \(\mathbf{p}(t) = \langle t^2, t^4 \rangle\) d) If one of \(m\) and \(n\) is even, is there always a cusp at \(t = 0\) for the curve \(\mathbf{r}(t) = \langle t^m, t^n \rangle\) where \(m > 1\) and \(n > 1\) are integers with no common factors? Answer: There is a cusp at \(t = 0\) in curve b) \(\mathbf{r}(t) = \langle t^3, t^2 \rangle\). In curve a, the graph is a straight line which does not have a cusp, and in curve c, both \(\mathbf{r}(t) = \langle t, t^2 \rangle\) and \(\mathbf{p}(t) = \langle t^2, t^4 \rangle\) represent the same parabolic curve but with different domains. For part d, it is true that if one of \(m\) and \(n\) is even, there will always be a cusp at \(t = 0\) for the curve \(\mathbf{r}(t) = \langle t^m, t^n \rangle\).

Step-by-step solution

Graph the curve and find the first derivative for part a

Consider the curve \(\mathbf{r}(t) = \langle t^3, t^3 \rangle\). To graph this curve, we can rewrite it in terms of \(x\) and \(y\) as \(x = t^3\) and \(y = t^3\). This implies \(y = x\). The graph of this curve is a straight line passing through the origin with a slope of \(1\). Now, let's find the first derivative of \(\mathbf{r}(t)\). It is given by \(\mathbf{r}'(t) = \langle 3t^2, 3t^2 \rangle\).

Check the first derivative at \(t = 0\) for part a

To check if the curve has a cusp or not at \(t = 0\), we need to evaluate the first derivative at the given point: $$\mathbf{r}'(0) = \langle 3(0)^2, 3(0)^2 \rangle = \langle 0, 0 \rangle = \mathbf{0}.$$ However, the graph of the curve is a simple straight line, and it does not have a cusp at \(t=0\). This is because the tangent of the curve is defined everywhere, including at \(t=0\).

Graph the curve and find the first derivative for part b

Now, let's consider the curve \(\mathbf{r}(t) = \langle t^3, t^2 \rangle\). To graph this curve, we can rewrite it in terms of \(x\) and \(y\) as \(x = t^3\) and \(y = t^2\). This is the graph of a polynomial curve with a cusp at the origin. The first derivative of \(\mathbf{r}(t)\) is given by \(\mathbf{r}'(t) = \langle 3t^2, 2t \rangle\).

Check the first derivative at \(t = 0\) for part b

To check if the curve has a cusp or not at \(t = 0\), we need to evaluate the first derivative at the given point: $$\mathbf{r}'(0) = \langle 3(0)^2, 2(0) \rangle = \langle 0, 0 \rangle = \mathbf{0}.$$ This time, the curve actually has a cusp at \(t=0\). The tangent at this point is vertical, which means the first derivative is undefined at \(t=0\).

Compare the parameterized functions in part c

To compare the parameterized functions \(\mathbf{r}(t) = \langle t, t^2 \rangle\) and \(\mathbf{p}(t) = \langle t^2, t^4 \rangle\), we need to look at their graphs. Both functions satisfy the equation \(y = x^2\). The first function, \(\mathbf{r}(t)\), represents the graph of a parabola opening upwards with its vertex at the origin. The second function, \(\mathbf{p}(t)\), represents the same parabolic graph but with a restricted domain, only considering the values of \(x \ge 0\). In other words, it is the right half of the same parabola. Therefore, the difference between the two parameterized functions is the domain they cover.

Consider the general case for part d

Given the curve \(\mathbf{r}(t) = \langle t^m, t^n \rangle\), where \(m > 1\) and \(n > 1\) are integers with no common factors, we are to determine if the curve has a cusp at \(t = 0\) if one (but not both) of \(m\) and \(n\) is even. First, let's find the first derivative of the curve: $$\mathbf{r}'(t) = \langle mt^{m-1}, nt^{n-1} \rangle.$$ Now, let's consider the cases when either \(m\) or \(n\) is even: Case 1: \(m\) is even, and \(n\) is odd. The first derivative at \(t=0\) becomes: $$\mathbf{r}'(0) = \langle 0, nt^b \rangle$$ where \(b = n - 1 > 0\). Case 2: \(m\) is odd, and \(n\) is even. The first derivative at \(t=0\) is: $$\mathbf{r}'(0) = \langle mt^a, 0 \rangle$$ where \(a = m - 1 > 0\). We see that in both cases the first derivative at \(t=0\) is either vertical or horizontal (neither of both components is non-zero), meaning that there will be a cusp at that point. So, the statement is true: if one and only one of \(m\) and \(n\) is even, then the curve will have a cusp at \(t=0\).

Key concepts

Derivative Evaluation

When dealing with parametric curves, the derivative evaluation is key to understanding the behavior of the curve. In a regular function, the derivative tells us the slope of the tangent. For parametric equations, derivatives are needed for both components of the vector function. Consider a parametric curve defined as \( \mathbf{r}(t) = \langle f(t), g(t) \rangle \). The first derivative, \( \mathbf{r}'(t) = \langle f'(t), g'(t) \rangle \), represents the rate of change for each component.

When we evaluate this derivative at a specific point, such as \( t = 0 \), we determine the behavior of the curve at that point. Checking if \( \mathbf{r}'(0) = \langle 0, 0 \rangle \) can indicate the presence of a unique feature, like a cusp, especially if further analysis suggests the tangent slope does not maintain a regular finite or infinite value.

Graphing Parametric Curves

Graphing parametric curves may initially seem challenging, but it provides valuable insights into curve behaviors such as cusps and loops.

To graph a curve parametrically expressed as \( \mathbf{r}(t) = \langle x(t), y(t) \rangle \), you can eliminate the parameter \( t \) by expressing one variable in terms of the other or by plotting points for selected values of \( t \).

• For example, if \( x(t) = t^3 \) and \( y(t) = t^2 \), plotting for values around \( t = 0 \) illustrates critical characteristics and transitions of the curve.
• Another example, \( \mathbf{r}(t) = \langle t^3, t^3 \rangle \), translates to \( y = x \), presenting as a straight line.
  

This graphing technique reveals unique features like symmetry, inflection points, and, notably, cusps, when the derivative evaluation results in \( \mathbf{r}'(0) = \mathbf{0} \).

Tangent Vector at a Point

The concept of a tangent vector is integral to understanding the geometric nature of parametric curves. In simple terms, a tangent vector at a point of a curve indicates the direction and rate at which the curve "moves" through that point.

For parametric curves \( \mathbf{r}(t) = \langle x(t), y(t) \rangle \), the tangent vector is given by \( \mathbf{r}'(t) = \langle x'(t), y'(t) \rangle \). For instance, evaluating at \( t = 0 \), if \( \mathbf{r}'(0) = \langle 0, 0 \rangle \), it indicates the speed of change is zero, suggesting a possible cusp.

• At a typical point like \( \mathbf{r}(t) = \langle t, t^2 \rangle \), \( \mathbf{r}'(t) \) provides directional insight.

Specifically, typical behaviour and unusual features like horizontal or vertical cusps at \( t = 0 \) can be detected by examining whether \( x'(t) \) or \( y'(t) \) becomes zero.

Parametric Equations

Parametric equations differ from typical equations as they utilize a parameter, often \( t \), to define a curve in a plane. Instead of a single \( x,y \) function, they split into two separate functions \( x(t) \) and \( y(t) \).

Such equations are beneficial because:

• They allow representation of complex curves, not easily expressed in one variable.
• They are very flexible and can describe motion, generating the position as a function of time.
  
• Examples include \( \mathbf{r}(t) = \langle t^2, t^4 \rangle \), which provides insight into the shape and key points such as cusps.

Utilizing parametric equations, the variety and pathways available extend well beyond typical Cartesian graphing, illustrating features like loops or cusps that are critical in advanced curve analysis.