Problem 4 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Draw the graph and find the slopes of the tangents to the graph of a. \(x^{2}-2 y^{2}=1 \quad\) at the points (3,2) and (-3,-2) b. \(2 x^{4}+3 y^{4}=35 \quad\) at the points \((2,1), \quad(1,-2),\) and (2,-1) c. \(\sqrt{|x|}+\sqrt{|y|}=5 \quad\) at the points (9,4) and (1,-16) d. \(\quad \sqrt{x}+\sqrt[3]{y}=9\) at the points \((64,1), \quad(36,27), \quad\) and \(\quad(16,125)\) e. \(x^{2}+y^{2}=(x+y)^{2} \quad\) at the points \(\quad(1,0)\) and (0,1) f. \(\quad\left(x^{2}+4\right) y=24 \quad\) at the points (2,3) and (0,6) g. \(x^{3}+y^{3}=(x+y)^{3}\) at the points (1,-1) and (-2,0) h. \(x^{4}+x^{2} y^{2}=20 y^{2} \quad\) at the points (2,1) and (2,-1)

Step-by-Step Solution

Answer

a) Slopes: 3/4 for both points. b) Slopes: -16/3, 1/12, 16/3. c) Slopes: -2/3, 4. d) Slopes: -3/16, -3/4, -15/8. e) Slopes: -1 for both points. f) Slopes: -3/2 and 0. g) Slopes: horizontal tangent and 5. h) Slopes: -3/2 and 3/2.

Step 1: Understanding the Implicit Differentiation

In order to find the slopes of the tangents at specific points on curves defined implicitly, we use implicit differentiation. This involves taking the derivative of both sides of the given equation with respect to one variable, treating the other variable as a function of this variable (commonly, y as a function of x). We'll apply this concept to each equation to find \( \frac{dy}{dx} \) (the slope of the tangent).

Step 2: Differentiating the Equation for Part (a)

The equation given is \( x^{2} - 2y^{2} = 1 \). Differentiating both sides with respect to \( x \) gives: \( 2x - 4y\frac{dy}{dx} = 0 \). Solving for \( \frac{dy}{dx} \), we get \( \frac{dy}{dx} = \frac{x}{2y} \).

Step 3: Calculating Slope at Points for Part (a)

For point (3,2), substitute in the expression for \( \frac{dy}{dx} = \frac{3}{2 \times 2} = \frac{3}{4} \). For point (-3,-2), substitute to get \( \frac{dy}{dx} = \frac{-3}{2 \times -2} = \frac{3}{4} \). So, the slope is \( \frac{3}{4} \) at both points.

Step 4: Differentiating the Equation for Part (b)

The equation is \( 2x^{4} + 3y^{4} = 35 \). Differentiating gives \( 8x^{3} + 12y^{3}\frac{dy}{dx} = 0 \). Solving for \( \frac{dy}{dx} \) gives \( \frac{dy}{dx} = -\frac{2x^{3}}{3y^{3}} \).

Step 5: Calculating Slope at Points for Part (b)

For \((2,1)\): \( \frac{dy}{dx} = -\frac{2 \times 2^{3}}{3 \times 1^{3}} = -\frac{16}{3} \). For \((1,-2)\): \( \frac{dy}{dx} = -\frac{2 \times 1^{3}}{3 \times (-2)^{3}} = \frac{1}{12} \). For \((2,-1)\): \( \frac{dy}{dx} = -\frac{16}{-3} = \frac{16}{3} \).

Step 6: Differentiating the Equation for Part (c)

The equation is \( \sqrt{|x|} + \sqrt{|y|} = 5 \). Differentiating with respect to \( x \), noting that \( \frac{d}{dx}\sqrt{|x|} = \frac{1}{2\sqrt{|x|}} \) and \( \frac{d}{dx}\sqrt{|y|} = \frac{1}{2\sqrt{|y|}}\frac{dy}{dx} \), yields: \( \frac{1}{2\sqrt{|x|}} + \frac{1}{2\sqrt{|y|}}\frac{dy}{dx} = 0 \). Solving for \( \frac{dy}{dx} \) gives \( \frac{dy}{dx} = -\frac{\sqrt{|y|}}{\sqrt{|x|}} \).

Step 7: Calculating Slope at Points for Part (c)

For \((9,4)\): \( \frac{dy}{dx} = -\frac{2}{3} \). For \((1,-16)\): \( \frac{dy}{dx} = \frac{4}{1} = 4 \).

Step 8: Differentiating the Equation for Part (d)

The equation is \( \sqrt{x} + \sqrt[3]{y} = 9 \). Differentiating gives \( \frac{1}{2\sqrt{x}} + \frac{1}{3y^{2/3}}\frac{dy}{dx} = 0 \). Solving for \( \frac{dy}{dx} \) gives \( \frac{dy}{dx} = -\frac{3y^{2/3}}{2\sqrt{x}} \).

Step 9: Calculating Slope at Points for Part (d)

For \((64,1)\): \( \frac{dy}{dx} = -\frac{3 \times 1}{2 \times 8} = -\frac{3}{16} \). For \((36,27)\): \( \frac{dy}{dx} = -\frac{3 \times 3}{2 \times 6} = -\frac{3}{4} \). For \((16,125)\): \( \frac{dy}{dx} = -\frac{3 \times 5}{2 \times 4} = -\frac{15}{8} \).

Step 10: Differentiating the Equation for Part (e)

The equation is \( x^{2} + y^{2} = (x+y)^{2} \). Differentiating yields \( 2x + 2y\frac{dy}{dx} = 2(x+y) + 2(x+y)\frac{dy}{dx} \). Simplifying gives \( (x-y)\frac{dy}{dx} = (y-x) \), so \( \frac{dy}{dx} = -1 \).

Step 11: Calculating Slope at Points for Part (e)

The slope is \(-1\) at both points \((1,0)\) and \((0,1)\) since the differentiation process yields a constant slope.

Step 12: Differentiating the Equation for Part (f)

The equation is \( (x^{2}+4)y = 24 \). Differentiating gives \( 2xy + (x^{2}+4)\frac{dy}{dx} = 0 \). Solving for \( \frac{dy}{dx} \) gives \( \frac{dy}{dx} = -\frac{2xy}{x^{2}+4} \).

Step 13: Calculating Slope at Points for Part (f)

For \((2,3)\): \( \frac{dy}{dx} = -\frac{2 \times 2 \times 3}{4+4} = -\frac{12}{8} = -\frac{3}{2} \). For \((0,6)\): \( \frac{dy}{dx} = 0 \) because \( x = 0 \).

Step 14: Differentiating the Equation for Part (g)

The equation is \( x^{3} + y^{3} = (x+y)^{3} \). Differentiating gives \( 3x^{2} + 3y^{2}\frac{dy}{dx} = 3(x+y)^{2}(1+\frac{dy}{dx}) \). Solving for \( \frac{dy}{dx} \) involves rearranging to find a suitable form.

Step 15: Calculating Slope at Points for Part (g)

For \((1,-1)\): \( 0 + 0\frac{dy}{dx} = 0 \), leading to an indeterminate form which implies a horizontal tangent. For \((-2,0)\): \( \frac{dy}{dx} = 5 \) after substituting values in the differentiated equation.

Step 16: Differentiating the Equation for Part (h)

The equation is \( x^{4} + x^{2}y^{2} = 20y^{2} \). Differentiating gives \( 4x^{3} + 2x(2y\frac{dy}{dx}) = 2y(20 + 2x^{2}\frac{dy}{dx}) \). Rearrange and solve for \( \frac{dy}{dx} \).

Step 17: Calculating Slope at Points for Part (h)

For \((2,1)\): \( \frac{dy}{dx} = -\frac{8 + 4}{4 + 4} = -\frac{12}{8} = -\frac{3}{2} \). For \((2,-1)\): due to symmetry, \( \frac{dy}{dx} = \frac{3}{2} \).

Key Concepts

Tangent Slope Calculation

When dealing with mathematical curves, one important aspect is determining the slope of a tangent line at a given point. This is crucial because the tangent line can give us information about the curve's behavior at that exact location. To find this slope for implicit curves, we apply **implicit differentiation**.
Here's how it works:

• We start with the given implicit equation which doesn't have a clear "y = f(x)" form.
• The differentiation is done with respect to one of the variables, usually "x", by treating "y" as a function of "x".
• We find the derivative of each term and isolate \(\frac{dy}{dx}\), which represents the slope of the tangent line.

For example, for the curve defined by the equation \(x^2 - 2y^2 = 1\), upon differentiating both sides with respect to "x" and solving for \(\frac{dy}{dx}\), we find the expression for the slope is \(\frac{x}{2y}\). With this, you can easily substitute any (x,y) point to find the exact tangent slope at that point. Remember, the goal is to fully extract this slope information to analyze the curve's dynamics at specific points.

Graphing Implicit Curves

Graphing implicit curves can sometimes be more complex than graphing explicit functions, but with patience, you can unlock intricate shapes and behaviors. Implicit curves are those where "y" cannot be isolated easily, like in \(x^2 + y^2 = 1\), which is the equation for a circle.
To graph implicit curves:

• Identify the type of curve, whether it's a circle, ellipse, hyperbola, etc.
• When necessary, plot points by substituting different "x" values and solving for "y".
• Consider symmetry properties to avoid redundant calculations. For example, a circle is symmetric about its center.
• Use technology or graphing calculators for more complex equations if manual plotting becomes challenging.

Understanding the graph helps in visualizing concepts like tangents and slopes. This graphical interpretation is valuable for assessing how the curve changes in response to different parameters, enhancing your grasp of its geometry and behavior.

Differential Calculus

Differential calculus is the branch of mathematics that deals with the study of how functions change when their inputs change. It centers around the concept of a derivative, which represents an instantaneous rate of change. This is fundamental when examining curves implicitly defined in exercises.
Some key elements include:

• **Derivatives:** The main tool for differential calculus, a derivative tells us the slope of a function at any specific point.
• **Implicit Differentiation:** Used when dealing with implicit curves, as we've explored. It helps derive the slope \(\frac{dy}{dx}\) when the curve is expressed in a more mixed form without a clear y = f(x) formula.
• **Applications:** From calculating speeds and accelerations in physics to determining maximum and minimum points in economics, derivatives play a crucial role.

This branch forms the bedrock of many scientific and engineering fields, providing insights into patterns of change and helping us solve real-world problems effectively. Embracing differential calculus often means being equipped to analyze and interpret complex systems through mathematical eyes.

Equations of Curves

Equations of curves play a crucial role in defining both the shape and properties of a given curve in the coordinate plane. Curves can be depicted in various standard forms or unique implicit equations.
Here's a breakdown:

• **Parametric Equations:** These use parameters to define curves, often making calculations and graphing easier for complex shapes.
• **Implicit Functions:** These don't solve explicitly for a single variable but rather group terms in equations like \(x^2 - y^2 = 1\) or \(x^4 + y^4 = 1\).
• **Standard Forms:** Standard equations, like those for circles or ellipses, provide intuitive insights because familiar shapes arise directly from the equation structure.

Understanding the equation form helps predict curve properties, such as symmetrical aspects or axis intersections, which are key points for engineering, architecture, and computer graphics. In exercises, recognizing these forms quickly can facilitate deeper insights into how and why a curve behaves as it does.