Problem 17 - Essential Calculus Early Transcendentals

Question

Find the points on the ellipse \(4 x^{2}+y^{2}=4\) that are farthest away from the point \((1,0)\) .

Step-by-Step Solution

Answer

The points farthest from (1,0) are \((-\frac{1}{3},\frac{4\sqrt{2}}{3})\) and \((-\frac{1}{3},-\frac{4\sqrt{2}}{3})\).

Step 1: Understand the Problem

We need to find points on the ellipse defined by the equation \(4x^2 + y^2 = 4\) that are the farthest from the point \((1,0)\). This can be done by maximizing the distance between a general point \((x,y)\) on the ellipse and the point \((1,0)\).

Step 2: Distance Formula

Use the distance formula to express the distance \(D\) between a point \((x, y)\) on the ellipse and the point \((1, 0)\) as: \[ D = \sqrt{(x - 1)^2 + y^2} \].

Step 3: Square the Distance

To simplify differentiation, maximize the square of the distance \(D^2\) instead of \(D\): \[ D^2 = (x - 1)^2 + y^2 = x^2 - 2x + 1 + y^2 \].

Step 4: Use the Ellipse Constraint

Substitute \(y^2 = 4 - 4x^2\) from the ellipse equation into the expression for \(D^2\): \[ D^2 = x^2 - 2x + 1 + (4 - 4x^2) \]. Simplify this to \[ D^2 = -3x^2 - 2x + 5 \].

Step 5: Differentiate and Find Critical Points

Differentiate \(D^2\) with respect to \(x\): \[ \frac{d(D^2)}{dx} = -6x - 2 \]. Set the derivative equal to zero to find critical points: \(-6x - 2 = 0\). Solving for \(x\), we get \(x = -\frac{1}{3}\).

Step 6: Substitute Back to Find Corresponding y-values

Substitute \(x = -\frac{1}{3}\) back into the ellipse equation \(4x^2 + y^2 = 4\) to find \(y\). Calculate \(y^2 = 4 - 4(-\frac{1}{3})^2 = 4 - \frac{4}{9}\) which gives \(y^2 = \frac{32}{9}\). Thus, \(y = \pm \frac{4\sqrt{2}}{3}\).

Step 7: Verify Extrema

To ensure these are maximums, compare to potential endpoint values given the ellipse bounds or apply the second derivative test if needed. The points calculated should be evaluated against possible minimum lengths to confirm they are indeed maximum points on the ellipse.

Key Concepts

Ellipse

An ellipse is a geometric shape resembling an elongated circle. It can be thought of as a smooth curve on a plane surrounding two focal points, where the sum of the distances to the two foci is constant for every point on the ellipse. The equation for an ellipse in its standard form is

• For horizontal major axis: \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]
• For vertical major axis: \[ \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 \]

where

• \((h, k)\) is the center
• \(a\) is the semi-major axis length
• \(b\) is the semi-minor axis length

In this particular exercise, the ellipse is defined by the equation \[ 4x^2 + y^2 = 4 \], which is an implicit form. Simplifying it to the form\[ \frac{x^2}{1} + \frac{y^2}{4} = 1 \], we identify that the ellipse has its major axis along the \(y\)-axis.

Distance Formula

The distance formula is a direct extension of the Pythagorean theorem and is used to calculate the distance between two points in a Cartesian plane. It is defined as:\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]This formula helps in determining how far apart two points are by considering the differences in their x-coordinates and y-coordinates.

• To find the distance between two points \((x_1, y_1)\) and \((x_2, y_2)\), we square the differences
• Sum them, and then take the square root of the result.

When working with problems involving ellipses or any geometric constraints, the distance formula becomes essential in finding points that maximize or minimize distances subject to such constraints.

Critical Points

Critical points are where the first derivative of a function is zero or undefined. It is essential in calculus to find these points because

• They often indicate where a function might have maximum or minimum values.
• These points help to determine the behavior of functions.

To find critical points of a function, such as the squared distance function derived from the ellipse problem, you would

• Differentiate the function concerning the variable
• Set the resulting derivative equal to zero, and solve for the variable.

For the ellipse maximization problem, differentiating the squared distance function and setting it to zero leads to finding the value \(x = -\frac{1}{3}\) as a critical point.

Differentiation

Differentiation is a fundamental concept in calculus involving finding a derivative. A derivative represents the rate of change of a function with respect to one of its variables. Derivatives are vital for solving maximization and minimization problems since they help identify points where function values change their behavior.

• The power of differentiation lies in its ability to pinpoint exact rates of change and relationships between variables in mathematical models.
• By taking the derivative of a function, one can assess whether a critical point is a maximum, minimum, or saddle point.

In the context of this exercise, differentiation is used to find the derivative of the squared distance function regarding x, to identify critical points where the distance is maximized or minimized, thereby leading to solutions that meet the problem's constraints.