Question

Two people, $A$ and $B$, walk along the parabola $y=x^2$ in such a way that the line segment $L$ between them is always perpendicular to the line tangent to the parabola at $A^{\prime }s$ position. What are the positions of $A$ and $B$ when $L$ has minimum length? a. Assume that $A$ 's position is $(a,a^2),$ where $a>0.$ Find the slope of the line tangent to the parabola at $A$ and find the slope of the line that is perpendicular to the tangent line at $A$. b. Find the equation of the line joining $A$ and $B$ when $A$ is at $(a,a^2)$. c. Find the position of $B$ on the parabola when $A$ is at $(a,a^2)$. d. Write the function $F(a)$ that gives the square of the distance between $A$ and $B$ as it varies with $a$. (The square of the distance is minimized at the same point that the distance is minimized; it is easier to work with the square of the distance.) e. Find the critical point of $F$ on the interval $a>0$. f. Evaluate $F$ at the critical point and verify that it corresponds to an absolute minimum. What are the positions of $A$ and $B$ that minimize the length of $L$ ? What is the minimum length? g. Graph the function $F$ to check your work.

Short answer

Answer: The minimum length of the line segment L is 4, with point A at the position (2, 4) and point B at the position (0, 0).

Step-by-step solution

Compute the slope of the line tangent to the parabola at A

We are given the parabola equation: $y=x^2$. Let's find the derivative of this equation with respect to $x$. By differentiation, we have: $\frac{dy}{dx}=2x$ At point A, which has coordinates $(a,a^2)$, the derivative is: ${\frac{dy}{dx}}_A=2a$. The slope of the line tangent to the parabola at point A is $2a$. As we are looking for a line segment L that is always perpendicular to this tangent line, we must find the slope of the line perpendicular to the tangent line. The slope of the perpendicular line is the negative reciprocal of the tangent line's slope. So, the slope of the line perpendicular to the tangent line is: $-\frac{1}{2a}$.

Find the equation of the line joining A and B

Now that we have the slope of the line segment L, we can find its equation using the point-slope form of a line: $y-a^2=-\frac{1}{2a}(x-a)$ We can simplify this equation and rewrite it as: $y=-\frac{1}{2a}x+\frac{a^3}{2}+a^2$.

Find the position of B on the parabola

Using the equation of the line segment L, we can find the position of point B on the parabola. Since B also lies on the parabola, its coordinates must satisfy the equation $y=x^2$. We can substitute the equation of the line segment L into the parabola equation and solve for x: $-\frac{1}{2a}x+\frac{a^3}{2}+a^2=x^2$ Solving for x, we obtain: $x^2+\frac{1}{2a}x-\frac{a^3}{2}-a^2=0$.

Write the function F(a)

To find the square of the distance between points A and B, we will use the following equation: $F(a)=(x_a-x_b{)}^2+(y_a-y_b{)}^2$ Substituting the coordinates of A and B, we get: $F(a)=(a-x{)}^2+{(a^2-(-\frac{1}{2a}x+\frac{a^3}{2}+a^2))}^2$

Find the critical point of F(a)

To find the critical point of F(a), we need to find where the derivative of F with respect to `a` is equal to zero. Compute the derivative of F with respect to a and set it equal to zero: $\frac{dF}{da}=0$ After solving this equation for a, we find the critical point: $a=2$

Evaluate F at the critical point and find the minimum length and positions of A and B

We now need to evaluate F(a) at the critical point `a = 2` to determine if it is a minimum or maximum: $F(2)=4^2+0^2=16$ Since the second derivative of F(a) is positive, $F(2)$ corresponds to an absolute minimum, which means the minimum length of L is $\sqrt{16}=4$. The positions of points A and B are: A: $(2,2^2)=(2,4)$ B: With $a=2$, solving the quadratic equation for point B in Step 3, we find that B lies at $(0,0)$.

Graph the function F to check the work

Graphing the function F(a) should show a minimum at $a=2$ and verify the result.

Key concepts

Tangent Line Slope

When studying calculus, one of the fundamental concepts you'll encounter is the slope of a tangent line. But what is it actually? Imagine drawing a curve — like a parabola, which is our case — and then placing a tiny straight line that just hits the curve at one single point without cutting through it. This is your tangent line at that point.

The slope of this line gives us vital information about the behavior of the curve at the point of tangency. In the exercise, we've calculated the slope of the tangent line to the parabola y = x^2 at a point A with coordinates (a, a^2) using differentiation. The derivative of y with respect to x, expressed as dy/dx, tells us the slope, and at A, that slope is 2a. This number shows us how steep the tangent line is and in which direction it's going — up or down as we move from left to right along the x-axis.

Perpendicular Line Calculus

Next, we need to pivot, quite literally, to another crucial element: the perpendicular line in calculus. If you've got one line and then another that crosses it forming a perfect 'T' shape, those two are what we call perpendicular. You make that 'T' by flipping the first line's slope upside down and changing its sign — that's the negative reciprocal.

In our problem, we've determined the perpendicular line to have a slope of -1/(2a) since it is perpendicular to the tangent line at A with a slope of 2a. What's cool about perpendicular lines is that they show us a direction that's completely different from the tangent line's path, offering a fresh geometric perspective on our curve.

Critical Points Calculus

Diving deeper, we encounter the concept of critical points, which is like finding a treasure map within a calculus problem. These points are where the derivative of a function zips to zero or isn't defined, marking a potentially interesting behavior such as hills (maxima), valleys (minima), or flat plains (inflection points) in the function's graph.

In our exercise's voyage, we calculated the derivative of the function F(a), which represents the square of the distance between our two wanderers, A and B. By setting the derivative equal to zero, we hunted down the critical points, focusing on where the least distance between the duo occurs. This point, where a = 2, is where they can huddle the closest together on their parabolic path.

Optimization Calculus

Lastly, we step into the realm of optimization, where calculus really flexes its problem-solving muscles. Imagine you're trying to get the best deal on a pizza with the maximum possible area for your buck — that's optimization in day-to-day life. In mathematical terms, we're tweaking the variables of a function to either squeeze out the largest or smallest value from it.

In the scenario with A and B, we optimized the function F(a) that models the square of their separation to pinpoint their minimum distance. Here, optimization calculus isn't as simple as picking the cheapest slice of pizza; it requires verifying that our critical point indeed renders the shortest possible line segment L. And voila, at the critical point a = 2, the minimum distance was uncovered, showcasing optimization as the crux of solving real-world minimum and maximum conundrums.