Problem 7 - Calculus Early Transcendentals

Question

Carry out the following steps. a. Use implicit differentiation to find \(\frac{d y}{d x}\) b. Find the slope of the curve at the given point. $$y^{2}=4 x ;(1,2)$$

Step-by-Step Solution

Answer

Answer: The slope of the curve at the point (1, 2) is 1.

Step 1: Apply implicit differentiation to the given equation

We are given the equation \(y^2 = 4x\). To find \(\frac{dy}{dx}\), we need to differentiate both sides of the equation with respect to x, treating y as a function of x. Differentiating \(y^2\) with respect to x gives: $$\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}$$ Differentiating \(4x\) with respect to x gives: $$\frac{d}{dx}(4x) = 4$$ Now we write the resulting equation: $$2y \frac{dy}{dx} = 4$$

Step 2: Solve for \(\frac{dy}{dx}\)

To find \(\frac{dy}{dx}\), we need to solve for it in the equation we derived in Step 1. $$2y \frac{dy}{dx} = 4$$ Divide both sides of the equation by \(2y\): $$\frac{dy}{dx} = \frac{4}{2y}$$ Simplify the expression: $$\frac{dy}{dx} = \frac{2}{y}$$

Step 3: Evaluate the slope at the given point

Now we will find the slope of the curve at the given point (1, 2) by evaluating the expression for \(\frac{dy}{dx}\) at this point. Substitute the given point (1, 2) into the expression: $$\frac{dy}{dx} = \frac{2}{y} = \frac{2}{2}$$ Simplify the expression: $$\frac{dy}{dx} = 1$$ The slope of the curve at the point (1, 2) is 1.

Key Concepts

Slope of a Curve

Imagining a curve on a graph, the slope at any given point represents how steep the curve is at that specific location. It's critical for understanding how a function behaves as we move along different points on its path.

To find the slope using calculus, we focus on determining the derivative of the function, which is the main tool for understanding the rate of change at any specific point on the curve. Using implicit differentiation, as done in the exercise, we can calculate the slope even when we have an equation that describes y implicitly as a function of x. In our example, the expression achieved after differentiating gives us a recipe to calculate the slope at any point of the curve just by knowing the y-coordinate. Evaluating this expression at the given point (1,2), we determined the slope to be 1, which indicates a linear and constant increase of y with respect to x at that point.

Derivatives in Calculus

The concept of derivatives in calculus is all about understanding how a function changes at an infinitesimally close point. It is the cornerstone of calculus, widely used to find the rates at which one quantity changes with respect to another.

When we talk about deriving functions implicitly as in the given exercise, it means we are dealing with cases where y is not isolated on one side of the equation. A special technique called implicit differentiation is employed, which allows for the derivative of y with respect to x, \( \frac{dy}{dx} \), to be found without explicitly solving for y first. Implicit differentiation applies the same rules as normal differentiation, but it takes into account the fact that y is a function of x, so whenever we differentiate an expression with y in it, we multiply by \( \frac{dy}{dx} \) to account for y's own rate of change.

Evaluating Expressions

When we evaluate expressions, especially in the context of calculus, we're substituting specific values for variables to calculate a numerical result. This is often the case after finding a general formula or derivative for a function, where we use known values to get an exact answer.

In our exercise, after finding the derivative of y with respect to x, we evaluated it at a particular point on the curve, (1, 2), to find the slope of the curve at that point. Evaluating expressions is a crucial step in calculus problems because it allows us to apply general formulas to specific situations, giving us meaningful results that can describe phenomena such as velocity, acceleration, or, as in this case, the slope of a curve at a given point.