Problem 19 - Calculus - AP Edition

Question

Find the slope of the tangent line to the curve at the given points in two ways: first by solving for \(y\) in terms of \(x\) and differentiating and then by implicit differentiation. $$x^{2}+y^{2}=1 ;(1 / 2, \sqrt{3} / 2),(1 / 2,-\sqrt{3} / 2)$$

Step-by-Step Solution

Answer

Slopes: \(-\frac{1}{\sqrt{3}}\) and \(\frac{1}{\sqrt{3}}\).

Step 1: Solve for y in terms of x

Starting with the equation of the circle, \(x^2 + y^2 = 1\), solve for \(y\) to express it explicitly in terms of \(x\). This gives us \(y = \pm \sqrt{1 - x^2}\).

Step 2: Differentiate explicitly

Differentiate \(y = \sqrt{1 - x^2}\) and \(y = -\sqrt{1 - x^2}\) with respect to \(x\). Applying the chain rule, for \(y = \sqrt{1 - x^2}\), we get \(\frac{dy}{dx} = \frac{-x}{\sqrt{1 - x^2}}\). Similarly, for \(y = -\sqrt{1 - x^2}\), we obtain \(\frac{dy}{dx} = \frac{x}{\sqrt{1 - x^2}}\).

Step 3: Evaluate the derivative at the given points

Substitute \(x = \frac{1}{2}\) into the derivative expression for each branch. For \(y = \sqrt{1 - x^2}\), \(\frac{dy}{dx} = \frac{-\frac{1}{2}}{\sqrt{\frac{3}{4}}} = -\frac{1}{\sqrt{3}}\). For \(y = -\sqrt{1 - x^2}\), \(\frac{dy}{dx} = \frac{\frac{1}{2}}{\sqrt{\frac{3}{4}}} = \frac{1}{\sqrt{3}}\).

Step 4: Apply implicit differentiation

Differentiate the original equation implicitly with respect to \(x\): \(2x + 2y\frac{dy}{dx} = 0\). Simplify to find \(\frac{dy}{dx} = -\frac{x}{y}\).

Step 5: Evaluate the implicit derivative at the given points

Plug in the points \(\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)\) and \(\left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)\) into \(\frac{dy}{dx} = -\frac{x}{y}\). For the first point, the slope is \(-\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}}\). For the second point, the slope is \(-\frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}\).

Key Concepts

Slope of a Curve

In mathematics, the slope of a curve at a given point represents how steep the curve is at that specific point. It's a crucial concept linked to the derivative of a function. Think of it like a snapshot of the curve's direction at any point. For the problem given, we were asked to find the slope of a curve, specifically a circle described by the equation \(x^{2} + y^{2} = 1\). This equation depicts a circle. At different points along this circle, its slope varies.

In this scenario, the slope of the tangent line, which is a line that touches the circle at exactly one point, was determined using both explicit and implicit differentiation methods. The slope helps us to understand how the function behaves at that particular point and provides insight into the curve's structure.

With circles, this results in interesting behavior, as we see slopes changing direction (positive or negative) depending on the exact point you choose to examine. In short, knowing how to find the slope of a curve empowers you to better model and predict the behavior of different functions.

Implicit Differentiation

Implicit Differentiation is a powerful calculus tool used when you have functions defined by equations that are difficult to solve for one variable in terms of another. Instead of isolating the variable, like in explicit differentiation, you differentiate both sides of the equation simultaneously, keeping both variables intact.

In the case of the circle equation \(x^{2} + y^{2} = 1\), implicit differentiation allows us to differentiate directly without needing to rearrange the equation explicitly for \(y\). Using this technique, the derivative \(\frac{dy}{dx}\) was found by differentiating each term:

• \(2x\) is differentiated to get \(2\).
• \(2y\frac{dy}{dx}\) where we use the chain rule, remains \(2y\frac{dy}{dx}\).

Setting the derivative of the whole equation to zero gives: \(2x + 2y\frac{dy}{dx} = 0\). From here, solve for \(\frac{dy}{dx}\), resulting in \(-\frac{x}{y}\).

This method is particularly useful when working with curves and shapes where solving for \(y\) explicitly is tricky or not feasible.

Explicit Differentiation

Unlike implicit differentiation, explicit differentiation requires one to first solve the equation for one variable in terms of the other. It's often the starting point in calculus courses to introduce the concept of derivatives.

In this exercise, we explicitly solved the equation \(x^{2} + y^{2} = 1\) to express \(y\) in terms of \(x\), resulting in \(y = \pm \sqrt{1 - x^{2}}\). Once expressed in this form, finding the derivative \(\frac{dy}{dx}\) becomes straightforward using the chain rule.

The chain rule was applied to both the positive and negative scenarios of \(y\), giving:

• For \(y = \sqrt{1 - x^2}\), \(\frac{dy}{dx} = \frac{-x}{\sqrt{1 - x^2}}\).
• For \(y = -\sqrt{1 - x^2}\), \(\frac{dy}{dx} = \frac{x}{\sqrt{1 - x^2}}\).

This approach, although generally more cumbersome due to the need to isolate one variable, clearly shows how derivatives describe changes by comparing small deviations around a point on the curve.