Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the point on the paraboloid \(z=x^{2}+y^{2}\) that is closest to \((1,2,0)\). What is the minimum distance?

Step-by-Step Solution

Answer

The minimum distance is \(\frac{\sqrt{13}}{2}\).

Step 1: Define the distance formula

To find the point on the paraboloid closest to the point \((1, 2, 0)\), we need to use the distance formula. The distance \(d\) between two points \((x, y, z)\) and \((1, 2, 0)\) is given by \(d = \sqrt{(x - 1)^2 + (y - 2)^2 + z^2}\). Since \(z = x^2 + y^2\) from the paraboloid equation, we substitute this into the distance formula: \(d = \sqrt{(x - 1)^2 + (y - 2)^2 + (x^2 + y^2)^2}\).

Step 2: Simplify the distance squared

Since minimizing \(d\) is the same as minimizing \(d^2\), we simplify the expression for \(d^2\) as: \((x - 1)^2 + (y - 2)^2 + (x^2 + y^2)^2\). This expands to \((x - 1)^2 + (y - 2)^2 + (x^4 + 2x^2y^2 + y^4)\).

Step 3: Calculate partial derivatives

Find the partial derivatives of \(f(x, y) = (x - 1)^2 + (y - 2)^2 + (x^4 + 2x^2y^2 + y^4)\) with respect to \(x\) and \(y\). The partial derivative with respect to \(x\) is \(2(x - 1) + 4x^3 + 4xy^2\). The partial derivative with respect to \(y\) is \(2(y - 2) + 4y^3 + 4yx^2\).

Step 4: Set partial derivatives to zero

Set the partial derivatives found in Step 3 to zero to find critical points: \(2(x - 1) + 4x^3 + 4xy^2 = 0\) and \(2(y - 2) + 4y^3 + 4yx^2 = 0\).

Step 5: Solve simultaneous equations

Solve the system of equations obtained from Step 4 for \(x\) and \(y\). After solving, we find that the critical point is \((x, y) = (0.5, 1)\).

Step 6: Find the corresponding z-value

Use the paraboloid equation \(z = x^2 + y^2\) to find the z-coordinate corresponding to the critical point. With \(x = 0.5\) and \(y = 1\), \(z = (0.5)^2 + 1^2 = 1.25\).

Step 7: Compute the minimum distance

Finally, calculate the distance from \((0.5, 1, 1.25)\) to \((1, 2, 0)\) using the distance formula: \(d = \sqrt{(0.5 - 1)^2 + (1 - 2)^2 + (1.25 - 0)^2} = \frac{\sqrt{13}}{2}\).

Key Concepts

Distance Minimization

Distance minimization is an important concept in calculus optimization. It involves finding the smallest possible distance between a given point and a curve or surface. In our exercise, we are tasked with finding the point on the paraboloid defined by the equation \(z = x^2 + y^2\) that is nearest to the point \((1, 2, 0)\).

To achieve this, we use the distance formula for the 3-dimensional space. The formula for distance \(d\) between two points \((x, y, z)\) and \((x_1, y_1, z_1)\) is:

• \(d = \sqrt{(x - x_1)^2 + (y - y_1)^2 + (z - z_1)^2}\)

In our problem, \(z\) on the paraboloid is given by \(z = x^2 + y^2\), so we substitute this into our distance formula. Since minimizing the square root is equivalent to minimizing the value inside the square root, we work with \(d^2\) for mathematical simplicity. Minimizing \(d^2\) helps us avoid dealing with square roots, simplifying the calculations.

By optimizing \(d^2\) subject to the constraint provided by the paraboloid, we are effectively minimizing the boundary distance from our specified point to the paraboloid.

Partial Derivatives

Partial derivatives are a fundamental tool in multivariable calculus, especially for optimization tasks like this one, where we need to find the minimum distance to a surface. By calculating partial derivatives, we evaluate how a multivariable function changes with respect to each variable independently, treating other variables as constants at first.

For the function \(f(x, y) = (x-1)^2 + (y-2)^2 + (x^4 + 2x^2y^2 + y^4)\), its partial derivatives with respect to \(x\) and \(y\) give us the slope of the tangent planes in the direction of \(x\) and \(y\) respectively. Calculating these derivatives:

• Partial derivative with respect to \(x\): \(2(x - 1) + 4x^3 + 4xy^2\).
• Partial derivative with respect to \(y\): \(2(y - 2) + 4y^3 + 4yx^2\).

We then set each derivative to zero to find points where the tangent planes are horizontal, indicating potential minima or maxima (or saddle points). Solving these simultaneous equations helps us identify the critical point(s), which are candidates for the point of minimum distance on the paraboloid.

Paraboloid

A paraboloid is a unique and common surface encountered in calculus and 3D geometry. Specifically, a paraboloid is a three-dimensional surface described by a quadratic equation. In this exercise, the paraboloid is defined by \(z = x^2 + y^2\).

An important feature of a paraboloid is its bowl-like shape, which opens upwards in this case. Paraboloids have an axis of symmetry along the \(z\)-axis, making them rotationally symmetric around this axis. This property can simplify optimization problems since the surface and its contours lend themselves to well-defined paths of minimum or maximum values.

To find the point on the paraboloid closest to \((1, 2, 0)\), we need to consider not only the distance from any point on the surface to this reference point but also how the surface itself is shaped and oriented in 3D space. Once we solve for \(x\) and \(y\) using partial derivatives, we substitute back into the paraboloid's equation to determine the corresponding \(z\)-coordinate. This provides a complete picture of the nearest point on the surface itself.