Question

Consider the parabola \(\mathbf{r}(t)=\left\langle a t^{2}+1, t\right\rangle,\) for \(-\infty

Short answer

Question: For the parametrized curve, a parabola defined by \(\mathbf{r}(t)=\left\langle a t^{2}+1, t\right\rangle\), find all points on this curve at which the tangent vector is orthogonal to the curve itself. Answer: The only point on the parabola at which the tangent vector is orthogonal to the curve itself is (1, 0).

Step-by-step solution

Compute first derivative of \(\mathbf{r}(t)\)

To find the first derivative of the vector-valued function \(\mathbf{r}(t)=\left\langle a t^{2}+1, t\right\rangle\), we will compute the derivative of each component with respect to \(t\). So, \(\mathbf{r}^{\prime}(t) = \left\langle \frac{d}{dt}(a t^{2}+1), \frac{d}{dt}(t) \right\rangle = \left\langle 2at, 1\right\rangle\).

Find the dot product of \(\mathbf{r}(t)\) and \(\mathbf{r}^{\prime}(t)\)

Now we need to find the dot product of \(\mathbf{r}(t)\) and \(\mathbf{r}^{\prime}(t)\): \(\mathbf{r}(t)\cdot\mathbf{r}^{\prime}(t) = \langle a t^{2}+1, t\rangle \cdot \langle 2at, 1\rangle = (a t^{2}+1)(2at) + t(1)\).

Set the dot product equal to zero

To find the \(t\) values at which the curve and its tangent are orthogonal, we will set their dot product equal to zero, and solve for \(t\): \((a t^{2}+1)(2at) + t = 0\).

Solve for \(t\)

To find the value of \(t\) that solves the equation above, we can factor it: \(t(2a^2 t^3+2at)=0\). Now, we can see that there are two solutions for \(t\): 1. \(t=0\) 2. \(2a^2 t^3+2at=0\) For the second solution, we can factor out \(2at\) to get \(2at( a t^2 + 1) = 0\). Since \(a\) is a positive real number, this equation is only true if \(t^2 + 1 = 0\), which has no real solutions. Therefore, the only solution for \(t\) is \(t=0\).

Find the points on the parabola corresponding to \(t=0\)

Finally, we substitute our solution for \(t\) back into the original parametric equation: \(\mathbf{r}(0)=\left\langle a (0)^{2}+1, 0\right\rangle = \left\langle 1, 0\right\rangle\). Thus, the only point on the parabola at which \(\mathbf{r}\) and \(\mathbf{r}^{\prime}\) are orthogonal is \(\left(1, 0\right)\).

Key concepts

Parametric Equations

A parametric equation provides a way to define a curve by using one or more parameters. Rather than writing a relation in terms of just variables like x and y, a curve can be expressed through equations involving a third variable, often noted as \(t\). In this context, \(t\) typically represents a time or a different variable that traces a path along the curve as it changes.

In our study on the given parabola, \(\mathbf{r}(t) = \langle at^2 + 1, t \rangle\), the equation is parameterizing a path on a two-dimensional plane, as \(t\) varies across all real numbers. Here:\(at^2 + 1\) represents the x-component of the curve, while \(t\) is the y-component. This form allows us to examine and compute specific characteristics like direction, intersection, and orthogonality.

Dot Product

The dot product is a crucial tool in understanding the relationship between two vectors. It provides a measure of how much one vector goes in the direction of another. Mathematically, if we have two vectors \(\mathbf{a} = \langle a_1, a_2 \rangle\) and \(\mathbf{b} = \langle b_1, b_2 \rangle\), the dot product is calculated as \(\mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2\).

In our exercise, the dot product was used to find when the vector \(\mathbf{r}(t)\) from the curve and its derivative \(\mathbf{r}^{\prime}(t)\) are orthogonal. The solution involved computing \(\mathbf{r}(t) \cdot \mathbf{r}^{\prime}(t)\). When this product equals zero, the two vectors are perpendicular. This is because a dot product of zero indicates exact orthogonality between the vectors.

Orthogonality

Orthogonality is the concept of vectors being perpendicular to each other. This is essential in various areas, including computer graphics, physics, and engineering. When two vectors are orthogonal, they have a dot product of zero, indicating they do not influence each other in the direction they span.

In the context of the problem, finding orthogonal vectors means determining when the direction of the tangent to the curve represented by the derivative \(\mathbf{r}^{\prime}(t)\) is perpendicular to the curve's position \(\mathbf{r}(t)\) itself. This reveals points where the curve's path and its instantaneous change of direction do not align in any shared direction. In our example, the parabola has such a point at \((1, 0)\), where the vectors are exactly perpendicular.

Vector-Valued Functions

Vector-valued functions are functions that output vectors rather than single real numbers. They are particularly useful for describing curves and surfaces in two or three dimensions. In the function \(\mathbf{r}(t) = \langle at^2 + 1, t \rangle\), vector components describe a curve by mapping \(t\) to a point in space.

Each vector returned by the function represents a point on the curve for a given \(t\). The derivative of a vector-valued function like \(\mathbf{r}^{\prime}(t)\) provides insight into the curve's behavior and direction at any position \(t\). Using the derivative, we can find tangent vectors, enabling us to explore gradients, slopes or - as we did - the point of orthogonality to gain further insights into the curve's geometry.