Question

Consider the following parametric curves. a. Determine dy/dx in terms of t and evaluate it at the given value oft. b. Make a sketch of the curve showing the tangent line at the point corresponding to the given value of \(t\) $$x=2+4 t, y=4-8 t ; t=2$$

Short answer

Answer: At t=2, the value of dy/dx is -2. The equation of the tangent line at this point is y = -2x + 8.

Step-by-step solution

Finding dx/dt and dy/dt

Given: $$x=2+4 t$$ $$y=4-8 t$$ To find dx/dt, differentiate x with respect to t: $$\frac{dx}{dt} = \frac{d}{dt}(2+4t)$$ $$\frac{dx}{dt} = 4$$ Now, differentiate y with respect to t to find dy/dt: $$\frac{dy}{dt} = \frac{d}{dt}(4-8t)$$ $$\frac{dy}{dt} = -8$$

Finding dy/dx

Now that we have dx/dt and dy/dt, we can find dy/dx by dividing dy/dt by dx/dt: $$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$$ $$\frac{dy}{dx} = \frac{-8}{4}$$ $$\frac{dy}{dx} = -2$$

Evaluate dy/dx at the given value of t

We are given the value of t as 2. Since we already found that dy/dx is -2, which is a constant and does not depend on t, evaluating it at t=2 will give us the same result: -2.

Finding the point corresponding to the given value of t

Now, let's find the point on the curve corresponding to t=2: $$x = 2 + 4(2) = 10$$ $$y = 4 - 8(2) = -12$$ So the point is (10, -12).

Finding the equation of the tangent line

We have the slope(dy/dx) as -2, and the point (10, -12). To find the equation of the tangent line, we can use the point-slope form of equation: $$y - y_1 = m(x - x_1)$$ where m is the slope, and (x_1, y_1) is the given point. Plug in the values: $$y - (-12) = -2(x - 10)$$ $$y + 12 = -2x + 20$$ $$y = -2x + 8$$ This is the equation of the tangent line.

Making the sketch

Now that we have the curve and the equation of the tangent line, we can make a sketch: 1. Sketch the curve $$x=2+4t, y=4-8t$$ 2. Plot the point (10, -12) on the curve 3. Draw the tangent line with the equation $$y = -2x + 8$$ passing through the point (10, -12) The sketch will show the curve, the point (10, -12), and the tangent line passing through that point with a slope of -2.

Key concepts

Tangent Line

A tangent line is a straight line that touches a curve at a single point without crossing it. At this specific point, the tangent line's slope matches the slope of the curve. This means it's a perfect snapshot of how steep the curve is right at that moment. Think of it like leaning a straight stick against a bendy road at just the right angle.

For the parametric curve given by the equations \(x = 2 + 4t\) and \(y = 4 - 8t\), we found the point (10, -12) to be where the tangent line is drawn. At this point, the slope of the tangent line was calculated to be -2. This slope tells us the direction and steepness of the tangent line at that specific location.

• The tangent line gives us insight into the curve's behavior locally.
• It can also be used to make approximate predictions about the curve near the tangent point.

Understanding tangent lines helps in various applications, from engineering to physics, where analyzing the rate of change is crucial.

Derivative

The derivative is a powerful tool used in calculus to measure how a function changes as its input changes. In simpler terms, it's like taking the pulse of a function's rate of change. For parametric curves, we often find the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\), which describe the rate of change in the x and y directions, respectively.

In our exercise, we determined

\[\frac{dx}{dt} = 4\]
\[\frac{dy}{dt} = -8\]

Dividing \(\frac{dy}{dt}\) by \(\frac{dx}{dt}\) gives us the derivative \(\frac{dy}{dx}\), which is -2 in this case. This derivative represents the slope of the tangent line at any point \(t\).

• Derivatives help us understand how functions behave, showing us where they're increasing or decreasing.
• They are essential for finding maximum and minimum values, and for solving many real-world problems.

The concept of derivatives is foundational, aiding in boosting our comprehension of motion and change.

Parametric Equations

Parametric equations express curves by defining their x and y coordinates with respect to a third variable, often denoted as \(t\). This parameter \(t\) often represents time, but it can be any variable that moves you along the curve.

In our exercise, the parametric equations are:
\[x = 2 + 4t\]
\[y = 4 - 8t\]

These equations map out a straight line in the xy-plane as \(t\) changes. Each pair (x, y) is a result of replacing \(t\) with a specific value, providing a pathway along the curve as \(t\) varies.

• Parametric equations are crucial when dealing with motions and trajectories, such as the path of a projectile.
• They allow more flexible descriptions of curves that other formulas might not easily capture.

Through parametric equations, we explore complex shapes and movements beyond regular functions.

Calculus

Calculus is the branch of mathematics dedicated to studying change. It provides tools, like derivatives and integrals, to analyze how quantities vary. In this context, using calculus allows us to explore how curves behave through their slopes and areas.

When working with parametric curves, calculus lets us:

• Find the slope of tangent lines through derivatives.
• Determine lengths and areas under curves using integration, although not directly involved in this exercise.

Applying calculus to parametric equations, as we did, helps us break down complex motions and relationships into manageable pieces.

By analyzing the given parametric equations \(x = 2 + 4t\) and \(y = 4 - 8t\) through calculus, we gained insight into the slope and behavior of the tangent line. This analysis is invaluable for predicting patterns and understanding dynamics in various scientific fields. Calculus transforms our understanding of the world, revealing the math behind continuous change and growth.