Problem 3 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Shown Figure Ex. 4.4 .3 is the ellipse, $$ \frac{2 x^{2}}{35}+\frac{3 y^{2}}{35}=1 $$ and tangents to the ellipse at (2,3) and at (4,-1) . a. Find the slopes of the tangents. b. Find equations of the tangents. c. Find the point of intersection of the tangents. Figure for Exercise 4.4 .3 Graph of the ellipse \(2 x^{2} / 35+3 y^{2} / 35=1\) and tangents to the ellipse at the points (2,3) and (4,-1) .

Step-by-Step Solution

Answer

Tangent slopes: -4/9 and 8/3. Tangent equations: y = -4/9x + 38/9 and y = 8/3x - 35/3. Intersection: (71/14, 1/18).

Step 1: Differentiate the Ellipse Equation

The equation of the ellipse is \( \frac{2x^2}{35} + \frac{3y^2}{35} = 1 \). To find the slopes of the tangents, we need the derivative with respect to \( x \). Simplify it to \( \frac{2}{35}x^2 + \frac{3}{35}y^2 = 1 \) to derive implicitly: \( \frac{4x}{35} + \frac{6y}{35} \frac{dy}{dx} = 0 \). Solve for \( \frac{dy}{dx} \): \( \frac{dy}{dx} = -\frac{4x}{6y} = -\frac{2x}{3y} \).

Step 2: Evaluate Slopes at Given Points

Now plug the given points into the derivative to find tangent slopes. For point (2, 3), \( \frac{dy}{dx} = -\frac{2(2)}{3(3)} = -\frac{4}{9} \). For point (4, -1), \( \frac{dy}{dx} = -\frac{2(4)}{3(-1)} = \frac{8}{3} \). Hence, the slopes are \(-\frac{4}{9}\) and \(\frac{8}{3}\).

Step 3: Find Equations of Tangents

Use the point-slope form \( y - y_1 = m(x - x_1) \) to find the tangent lines. For slope \(-\frac{4}{9}\) at (2, 3), the equation is \( y - 3 = -\frac{4}{9}(x - 2) \), or \( y = -\frac{4}{9}x + \frac{38}{9} \). For slope \(\frac{8}{3}\) at (4, -1), the tangent is \( y + 1 = \frac{8}{3}(x - 4) \), simplifying to \( y = \frac{8}{3}x - \frac{35}{3} \).

Step 4: Find Point of Intersection

Set the two tangent equations equal to find their intersection: \( -\frac{4}{9}x + \frac{38}{9} = \frac{8}{3}x - \frac{35}{3} \). Multiply to clear fractions: \( -4x + 38 = 24x - 105 \). Solve for \( x \): \( 142 = 28x \), hence \( x = \frac{71}{14} \). To find \( y \), substitute \( x \) back into either equation; using \( y = -\frac{4}{9}x + \frac{38}{9} \), we find \( y = \frac{1}{18} \). Thus, the intersection point is \( \left(\frac{71}{14}, \frac{1}{18}\right) \).

Key Concepts

Ellipse

An ellipse is a smooth, rounded shape that looks like a stretched-out circle. It is defined by its equation, which typically takes the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \( a \) and \( b \) are the semi-major and semi-minor axes. In our specific problem, the ellipse is given by the equation \( \frac{2x^2}{35} + \frac{3y^2}{35} = 1 \). This means it has been scaled differently along its x and y axes, leading to its elongated shape. Ellipses have unique properties, such as having two focal points, and are often found in physics, especially in orbital motion.

Understanding the equation of an ellipse is key in calculus when solving for tangent lines. Knowing how to manipulate and differentiate an ellipse equation empowers you to solve complex calculus problems related to curves and shapes.

Implicit Differentiation

Implicit differentiation is a crucial technique in calculus used to find the derivative of an equation not solved in terms of a single variable, such as our ellipse equation. Unlike explicit functions where \( y \) is isolated, implicit equations blend \( x \) and \( y \) intricately. To tackle this, differentiate both sides of the equation with respect to \( x \).

For our ellipse, starting from \( \frac{2x^2}{35} + \frac{3y^2}{35} = 1 \), you derive each term separately. The term \( \frac{2x^2}{35} \) differentiates to \( \frac{4x}{35} \), and \( \frac{3y^2}{35} \) differentiates to \( \frac{6y}{35} \frac{dy}{dx} \), thanks to the chain rule.

This method helps find \( \frac{dy}{dx} \) by algebraically rearranging the terms, revealing the slope of the tangent line at any specific point on the ellipse, illustrating how implicit differentiation can address complex curves.

Tangent Line

The tangent line to a curve at a point is a straight line that touches the curve at that single point without crossing it. It provides the slope that represents how steep the curve is at that point. For an ellipse, determining the tangent line involves both the point of tangency and the slope of the curve at that point.

Using the point-slope form of a line, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the point of tangency, you can derive the equation for the tangent line. In the provided solution, we calculated the slopes at points (2, 3) and (4, -1), then formed linear equations for the tangents. For the slope \(-\frac{4}{9}\) at (2, 3), the tangent line equation becomes \( y = -\frac{4}{9}x + \frac{38}{9} \).

Tangent lines are significant in calculus as they approximate curves locally and provide essential insights into the behavior of functions, especially for optimization and integration.

Intersection Point

The intersection point is where two lines meet or cross each other. In analytical geometry, you find this point by solving the equations of the lines simultaneously. For our problem with tangent lines, this requires setting the equations of the two tangent lines equal and solving for the variables \( x \) and \( y \).

Once the two equations, \( -\frac{4}{9}x + \frac{38}{9} = \frac{8}{3}x - \frac{35}{3} \), are solved, we identify the \( x \)-coordinate of the intersection by equating and then isolating \( x \). After finding \( x = \frac{71}{14} \), substitute it into one of the tangent line equations to get the \( y \)-coordinate, resulting in \( y = \frac{1}{18} \).

The intersection point \( \left(\frac{71}{14}, \frac{1}{18}\right) \) is crucial for understanding the spatial configuration of these lines, especially in tasks involving curve analysis and optimization problems in calculus.