Problem 92 - Calculus Early Transcendentals

Question

Find all the points at which the following curves have the given slope. $$x=2+\sqrt{t}, y=2-4 t ; \text { slope }=-8$$

Step-by-Step Solution

Answer

Answer: The point on the curve with a slope of -8 is (3, -2).

Step 1: Find the derivatives with respect to t

We'll first find the derivatives of x and y with respect to t.$$ \frac{dx}{dt}=\frac{d}{dt}(2+\sqrt{t}) $$ $$ \frac{dy}{dt}=\frac{d}{dt}(2-4t) $$

Step 2: Compute the derivatives with respect to t

Now, we'll compute the derivatives.$$ \frac{dx}{dt} = \frac{1}{2\sqrt{t}} $$ $$ \frac{dy}{dt} = -4 $$

Step 3: Calculate the slope dy/dx

Next, we'll calculate the slope by dividing dy/dt by dx/dt.$$ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-4}{\frac{1}{2\sqrt{t}}} = -8\sqrt{t} $$

Step 4: Set the slope equal to -8 and solve for t

Now, we'll set the slope equal to -8 and solve for t.$$ -8\sqrt{t} = -8 $$Divide both sides by -8.$$ \sqrt{t} = 1 $$Square both sides.$$ t = 1 $$

Step 5: Find the points where the curve has the given slope

Finally, we'll plug the value of t back into the original parametric equations to find the points where the curve has the given slope.$$ x = 2 + \sqrt{1} = 2 + 1 = 3 $$ $$ y = 2 - 4(1) = 2 - 4 = -2 $$So, the point where the curve has a slope of -8 is (3, -2).

Key Concepts

Parametric Equations

Parametric equations allow us to express the coordinates of the points that make up a curve using parameters. This is helpful for defining curves that might not easily fit into the standard form of algebraic equations. For example, using the parameter \( t \), we can specify the position along a path using two equations: one for the x-coordinate, \( x = 2 + \sqrt{t} \), and another for the y-coordinate, \( y = 2 - 4t \).

This offers flexibility because each part, the \( x \) and \( y \), can vary independently based on \( t \). Essentially, parametric equations transform complex shapes into simple frameworks that are easier to manipulate and analyze. Understanding these equations means you're breaking down a curve into its essentials, focusing on how each component contributes to the whole shape.

Derivative with respect to a parameter

When dealing with parametric equations, we often want to understand how the coordinates change as the parameter changes. This is where derivatives enter the scene, specifically, taking a derivative with respect to a parameter. In this context, it involves finding \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) which explain how \( x \) and \( y \) change as \( t \) varies.

For our given equations, the derivatives are:

• \( \frac{dx}{dt} = \frac{1}{2\sqrt{t}} \)
• \( \frac{dy}{dt} = -4 \)

This shows us how, as \( t \) changes, \( x \) and \( y \) move. The derivative with respect to \( t \) is a fundamental tool for finding slopes or any rate of change related to our parameter.

Calculating slope of a curve

The slope of a curve is essentially a measure of how steep it is at any given point. In the context of parametric equations, calculating the slope \( \frac{dy}{dx} \) requires us to utilize the derivatives with respect to the parameter \( t \). We use the formula \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \), which reconfigures our understanding from a parameter-driven perspective back into the familiar \( x \) and \( y \) relationship.

In our exercise, this was simplified to \( \frac{dy}{dx} = -8\sqrt{t} \). To find where the slope equals \(-8\), we set \(-8\sqrt{t} = -8\) and solve for \( t \), resulting in \( t = 1 \). Plug this result back into the original equations to get the actual point on the curve, \((3, -2)\).

This approach helps us understand not just where the curve is heading, but also at what rate it's doing so, reflecting the balance between independent parameter changes and their influence on overall shape.