Problem 2 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Find the slope of the graph of \(\sqrt{x}-\sqrt{5-y^{2}}=5\) at the point (36,-2) . Is there a slope to the graph at the point (46,1)\(?\) Figure 4.7: Solid curve: Graph of \(\sqrt{x}+\sqrt{5-y^{2}}=5\) and the points (9,1) and (4,2) for Example 4.6 .1 . Dashed curve: Graph of \(\sqrt{x}-\sqrt{5-y^{2}}=5\) and the point (36,-2) for Exercise 4.6 .2 .

Step-by-Step Solution

Answer

The slope of the graph at (36, -2) is \(-1/24\); there is no slope at (46, 1) as it is not on the graph.

Step 1: Identify the equation and point

The equation of the graph given is \( \sqrt{x} - \sqrt{5-y^2} = 5 \). We need to find the slope at the point (36, -2).

Step 2: Implicit differentiation

To find the slope, differentiate both sides of the equation with respect to \( x \). This gives us: \[ \frac{d}{dx}(\sqrt{x}) - \frac{d}{dx}(\sqrt{5-y^2}) = \frac{d}{dx}(5) \]The derivative of \( \sqrt{x} \) is \( \frac{1}{2\sqrt{x}} \). For \( \sqrt{5-y^2} \), use the chain rule: differentiate \( \sqrt{u} \) as \( \frac{1}{2\sqrt{u}} \), where \( u = 5-y^2 \), multiplied by the derivative of \( -y^2 \) with respect to \( x \), which is \( -2y \frac{dy}{dx} \).

Step 3: Solve for \( \frac{dy}{dx} \)

From Step 2, we have: \[ \frac{1}{2\sqrt{x}} - \frac{1}{2\sqrt{5-y^2}} (-2y) \frac{dy}{dx} = 0 \]Simplify and solve for \( \frac{dy}{dx} \):\[ \frac{1}{2\sqrt{x}} = \frac{y}{\sqrt{5-y^2}} \frac{dy}{dx} \]\[ \frac{dy}{dx} = \frac{\sqrt{5-y^2}}{2y\sqrt{x}} \].

Step 4: Substitute the point (36, -2) into the derivative

Substitute \( x = 36 \), \( y = -2 \) into \( \frac{dy}{dx} = \frac{\sqrt{5-y^2}}{2y\sqrt{x}} \): \[ \frac{dy}{dx} = \frac{\sqrt{5 - (-2)^2}}{2(-2)\sqrt{36}} \]Calculate the values: \( \sqrt{36} = 6 \) and \( \sqrt{5 - 4} = 1 \).Thus, \[ \frac{dy}{dx} = \frac{1}{-24} = -\frac{1}{24} \].

Step 5: Check the point (46, 1) for existence on the curve

For the point (46, 1), check if it satisfies the equation \( \sqrt{x} - \sqrt{5-y^2} = 5 \).Substitute \( x = 46 \) and \( y = 1 \) into: \( \sqrt{46} - \sqrt{5 - 1^2} = 5 \).Calculate: \( \sqrt{46} \approx 6.78 \), and \( \sqrt{4} = 2 \).Thus, \( 6.78 - 2 eq 5 \), so (46, 1) is not on the graph.

Key Concepts

Slope of a Curve

The slope of a curve at a given point is a measure of how steep the curve is at that particular location. In essence, it tells us the rate at which the value of one variable changes with respect to another. For a curve defined by an equation, the slope can be found by differentiating the equation with respect to one of the variables, commonly denoted as the x-axis. The concept is similar to finding the slope of a straight line but can be a bit more complex due to the curvature.

Think of riding a bicycle on a hill: the steeper the hill, the harder you have to pedal. That's what the slope of the curve describes. In our example problem, we used implicit differentiation to find the slope of the curve described by the equation \(\sqrt{x} - \sqrt{5-y^{2}} = 5\) at the point (36, -2). By finding the slope, we calculated how quickly the \(y\) value changes as \(x\) increases.

**Practical Tip:** Always check if the point lies on the curve before trying to find the slope there, just as we tested point (46,1) to ensure it's on the graph before evaluating its slope.

Chain Rule

The chain rule is a differentiation technique used when dealing with composite functions. A composite function is a function inside another one. In the context of implicit differentiation, we often encounter such functions, particularly when a variable is defined implicitly and not isolated on one side of the equation.

In our exercise, we faced this situation. The term \(\sqrt{5-y^2}\) is a composite function where \(y\) is nested inside a square root. To differentiate this with respect to \(x\), we used the chain rule. Here's how it worked:

• First, acknowledge the outer function: the square root \(\sqrt{u}\).
• Next, identify the inner function: \(u = 5-y^2\).
• Differentiate the outer function, \(\frac{1}{2\sqrt{u}}\), and multiply by the derivative of the inner function, \(-2y \frac{dy}{dx}\).

The chain rule allowed us to manage these layers of functions efficiently.

Differentiation Techniques

Differentiation is a fundamental concept in calculus, and it encompasses several techniques to handle various types of equations and functions. These techniques allow us to find derivatives—the measures of how a function changes as its input changes. They are vital in determining slopes among many other applications.

In our problem, we applied implicit differentiation, a specific differentiation technique used when it's challenging or impossible to solve an equation explicitly for one variable. Unlike explicit functions where one variable is clearly dependent on the other, implicit functions hide the relationship, such as in \(\sqrt{x} - \sqrt{5-y^{2}} = 5\).

To differentiate implicitly, recognize both variables' roles and apply the chain rule, as detailed earlier. This approach allows us to differentiate terms involving both \(x\) and \(y\) without isolating \(y\) alone. This was essential in our problem to find \(\frac{dy}{dx}\).

**Tip:** Practice different types of differentiation techniques to develop a robust toolkit. This helps in recognizing which technique suits a particular problem best.