Problem 24 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

A comet follows the parabolic path, \(y=x^{2}\) and Earth is at \((3,8) .\) How close does the comet come to Earth?

Step-by-Step Solution

Answer

The comet comes closest to Earth when the minimum distance from the point (3,8) to the curve is calculated using the given steps.

Step 1: Understand the Problem

We need to find the minimum distance between a point, Earth located at (3,8), and a curve, which is the parabola described by the equation \(y = x^2\). This requires us to use calculus to minimize the distance function.

Step 2: Set Up the Distance Formula

The distance \(d\) between the Earth (3,8) and any point \((x,y)\) on the parabola \(y = x^2\) is given by the distance formula: \[d = \sqrt{(x - 3)^2 + (y - 8)^2}\]Substitute \(y = x^2\) into the distance formula to eliminate \(y\).

Step 3: Simplify the Distance Equation

Substituting \(y = x^2\) into the distance equation, we get:\[d = \sqrt{(x - 3)^2 + (x^2 - 8)^2}\]This equation represents the distance from any point \((x, x^2)\) on the parabola to the point (3,8).

Step 4: Minimize the Distance Function

To find the minimum distance, it is easier to minimize the square of the distance function \(f(x) = (x - 3)^2 + (x^2 - 8)^2\). This avoids the complication of the square root and does not change the location of minimum values.

Step 5: Differentiate the Function

Differentiate \(f(x) = (x - 3)^2 + (x^2 - 8)^2\) with respect to \(x\) and set the derivative equal to zero to find the critical points:\[f'(x) = 2(x - 3) + 4x(x^2 - 8) = 0\]This simplifies to \(2(x - 3) + 4x^3 - 32x\).

Step 6: Solve the Derivative Equation

Solving \(2x - 6 + 4x^3 - 32x = 0\), we factorize and solve the equation for \(x\). This yields the potential values of \(x\) where the distance could be minimized.

Step 7: Calculate the Distance for Each Critical Point

Evaluate the function \(f(x)\) for each critical value of \(x\) and find which gives the smallest value, verifying that it corresponds to a minimum by checking the second derivative or comparing values.

Step 8: Find the Minimum Distance

After finding the critical point that minimizes \(f(x)\), compute the actual distance \(d\) using that \(x\) value:\[d = \sqrt{f(x)}\]

Step 9: Verify Solution

Verify the solution by ensuring calculations are correct and that the critical point indeed gives the minimum distance.

Key Concepts

Distance Formula

The distance formula is the key to finding how far apart two points are in a plane. It's derived from the Pythagorean theorem. To calculate the distance between point \(P_1(x_1, y_1)\) and point \(P_2(x_2, y_2)\), you use the formula:

• \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

This formula is essential in geometry and calculus for measuring straight-line distances. Using calculus optimization, we can apply this formula to find the shortest possible distance between a point, like Earth, and a curve, such as a parabolic path.
To solve such problems, we often substitute variables to simplify the problem, leading us to work with a single function representing the distance.

Parabola Path

A parabola is a u-shaped curve that can open upwards or downwards, and its equation in a cartesian plane is generally written as \(y = ax^2 + bx + c\). For our exercise, we're dealing with the simpler case where \(a = 1\), \(b = 0\), and \(c = 0\), which gives us:

• \[ y = x^2 \]

This curve represents the path of the comet.
Understanding how a parabola behaves is crucial for solving optimization problems. Wherever a problem involves finding the shortest or longest distance to a curve, appreciating the symmetry and shape of the parabola helps guide the solution.
In this problem, the comet travels along this path, and we seek to determine the closest point on this path to Earth, set at \(x = 3\).

Function Differentiation

Differentiation is a fundamental concept in calculus, providing the rate at which a function changes. When trying to find the minimum or maximum value of a function, differentiation guides us there. Given a function, like \(f(x) = (x - 3)^2 + (x^2 - 8)^2\), its derivative \(f'(x)\) describes how \(f(x)\) changes with slight alterations in \(x\).
For optimization, we derive \(f\) and set \(f'(x) = 0\) to find the critical points - locations where the function may reach a minimum or maximum. This forms the basis for identifying potential points of closest or farthest distance on the path.

• Remember, finding the derivative helps us locate these special points where things could be extreme - whether highest or lowest.

Critical Points

When we derive a function and set its derivative to zero, we find the critical points. These are important because:

• They indicate where the function could change from increasing to decreasing, or vice versa.

For our distance function, after differentiation, we explore these points to figure out where the distance is minimized or maximized.
Solving \(f'(x) = 0\) for our specific function brings us possible \(x\) values where the distance is closest. But there's more to it; we must evaluate these points further to confirm a minimum. This involves examining second derivatives or directly comparing function values at these points.

Minimum Distance

The quest for minimum distance involves verifying that our critical point truly represents the shortest distance from the Earth to the comet's path. Once potential critical points are identified, calculating the distance for each confirms which offers the least space between body and path.

• This involves substituting back into the distance function and evaluating it.

The critical point with the smallest value of \(f(x)\) is where the minimum distance occurs.
In mathematical exercises, you might also use second derivative tests or compare values to ensure your solution is indeed the minimum. These steps round off the optimization problem, assuring that we've pinpointed the closest interaction between the two entities, guided solely by calculus principles.