Question

Find all the points on the following curves that have the given slope. $$x=2 \cos t, y=8 \sin t ; \text { slope }=-1$$

Short answer

Question: For the given parametric equations, $x(t) = 2 \cos(t)$ and $y(t) = 8\sin(t)$, find all the points on the curve with a slope of -1. Answer: The points on the curve with a slope of -1 can be represented by the parametric equations $\begin{cases} x = 2\cos(\arctan(4) + k\pi) \\ y = 8\sin(\arctan(4) + k\pi) \end{cases}$, where k is an integer. To find specific points, plug in different values of k.

Step-by-step solution

Find the derivatives of x(t) and y(t) #

First, differentiate the given parametric equations with respect to t: $$\frac{dx}{dt}=-2\sin(t) \text{ and }\frac{dy}{dt}=8\cos(t)$$

Find the slope dy/dx #

Now, to find the slope of the curve, we will divide dy/dt by dx/dt: $$\frac{dy}{dx}=\frac{8\cos(t)}{-2\sin(t)}=\frac{-4\cos(t)}{\sin(t)}$$

Set the slope equal to the given slope and solve for t #

Set the slope equal to the given slope -1: $$\frac{-4\cos(t)}{\sin(t)} = -1$$ Solve for t: $$4\cos(t)=\sin(t)$$ Since we are solving for points on a trigonometric curve, the solutions for t will be in the form of angles. So, by dividing both sides by 4, we can represent the equation as $$\frac{\sin(t)}{\cos(t)} = 4$$ Or, in the tangent form: $$\tan(t) = 4$$ Now, find the values of t: $$t = \arctan(4) + k\pi$$ where k is an integer.

Substitute the values of t back into the parametric equations #

Substitute the values of t back into the parametric equations x(t) and y(t): For x-coordinate: $$x = 2\cos(\arctan(4) + k\pi)$$ For y-coordinate: $$y = 8\sin(\arctan(4) + k\pi)$$ Now, we have parametric equations for the points on the curve with a slope of -1: $$\begin{cases} x = 2\cos(\arctan(4) + k\pi) \\ y = 8\sin(\arctan(4) + k\pi) \end{cases}$$ where k is an integer. To find specific points, plug in different values of k.

Key concepts

Derivatives

Derivatives are an essential concept in calculus, often used to understand how a function changes at any given point. In the context of parametric equations, derivatives help us find the rate at which one variable changes with respect to another. In this exercise, we're given parametric equations in terms of a parameter \( t \).
To find how \( x \) and \( y \) change with respect to \( t \), we compute the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). For \( x = 2\cos(t) \), the derivative is \( \frac{dx}{dt} = -2\sin(t) \). Similarly, for \( y = 8\sin(t) \), the derivative is \( \frac{dy}{dt} = 8\cos(t) \).

• \( \frac{dx}{dt} = -2\sin(t) \)
• \( \frac{dy}{dt} = 8\cos(t) \)

Derivatives allow us to study the behavior of these parametric equations and eventually find the slope of the curve, which we need to solve for specific points.

Trigonometric Functions

Trigonometric functions such as sine and cosine are functions of an angle, commonly used to model periodic phenomena.
In the exercise, the parametric equations are expressed using trigonometric functions: \( x = 2\cos(t) \) and \( y = 8\sin(t) \). These equations describe a parametric curve, and they take advantage of the periodic properties of trigonometric functions.
In trigonometry, \( \cos(t) \) and \( \sin(t) \) cycle through their values as \( t \) varies, producing a smooth and continuous path on a graph. Here, to find the slope \(-1\), we use the relationship derived from these functions:

• \( 4\cos(t) = \sin(t) \)
• This simplifies to \( \tan(t) = 4 \), where \( \tan(t) \) is the tangent of \( t \), representing the ratio of sine to cosine.

The solution \( t = \arctan(4) + k\pi \) captures the periodic nature of the trigonometric functions, with \( k \pi \) accounting for the periodicity of \( \pi \). By substituting these into the parametric equations, we can graph points that follow this trigonometric curve.

Slope of a Curve

The slope of a curve at a point tells us how steep the curve is at that point. It's crucial for understanding the direction and shape of the curve on a graph.
In parametric equations, the slope of the curve defined by the equations \( x(t) \) and \( y(t) \) is given by \( \frac{dy}{dx} \). We calculate this by dividing the derivative of \( y \) with respect to \( t \) by the derivative of \( x \) with respect to \( t \). Here, this is \( \frac{dy}{dx} = \frac{8\cos(t)}{-2\sin(t)} = \frac{-4\cos(t)}{\sin(t)} \).
To find points where the slope is \(-1\), we set \( \frac{-4\cos(t)}{\sin(t)} = -1 \) and solve for \( t \).

• This results in \( 4\cos(t) = \sin(t) \).
• By converting this into the tangent form \( \tan(t) = 4 \), we solve for \( t = \arctan(4) + k\pi \).

By identifying these points, we can determine where the slope of the tangent to the curve equals \(-1\), leading to a deeper comprehension of the curve's geometric properties.