Question

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

Short answer

The slope of the tangent line at t=π/2 is 0, and the point on the curve corresponding to this value of t is (0, 8).

Step-by-step solution

Find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\)

We are given the parametric equations of the curve: \(x=\cos t\) and \(y=8\sin t\). To find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\), we need to differentiate both equations with respect to \(t\). $$\frac{dx}{dt} = -\sin t$$ $$\frac{dy}{dt} = 8\cos t$$

Determine \(\frac{dy}{dx}\)

Using the chain rule, we can find \(\frac{dy}{dx}\) by dividing \(\frac{dy}{dt}\) by \(\frac{dx}{dt}\): $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{8\cos t}{-\sin t}$$

Evaluate \(\frac{dy}{dx}\) at \(t=\frac{\pi}{2}\)

We need to find the slope of the tangent line at \(t=\frac{\pi}{2}\). Substitute \(t=\frac{\pi}{2}\) into the expression for \(\frac{dy}{dx}\): $$\frac{dy}{dx}\Big|_{t=\frac{\pi}{2}} = \frac{8\cos \frac{\pi}{2}}{-\sin \frac{\pi}{2}} = \frac{8(0)}{-1}=0$$ The slope of the tangent line at \(t=\frac{\pi}{2}\) is 0.

Calculate the point corresponding to \(t=\frac{\pi}{2}\)

The question also asks for the point corresponding to \(t=\frac{\pi}{2}\). We can substitute it into the parametric equations: $$x(\frac{\pi}{2})=\cos \frac{\pi}{2}=0$$ $$y(\frac{\pi}{2})=8\sin \frac{\pi}{2}=8(1)=8$$ The point on the curve where \(t=\frac{\pi}{2}\) is \((0, 8)\).

Sketch the curve and tangent line

To make the sketch, we need to identify the general shape of our parametric curve. The parametric equations \(x=\cos t\) and \(y=8\sin t\) represent an ellipse centered at (0, 0) with a horizontal axis of length 2 and a vertical axis of length 16. The tangent line, with a slope of 0, will be a horizontal line passing through the point \((0, 8)\). To summarize: 1. Draw an ellipse centered at (0, 0) with a horizontal axis of length 2 and a vertical axis of length 16. 2. Mark the point \((0, 8)\) on the ellipse. 3. Draw a horizontal line through \((0, 8)\). This is the tangent line at \(t=\frac{\pi}{2}\).

Key concepts

Calculus

Calculus provides us the tools to understand how things change. It's a significant branch of mathematics dealing with limits, functions, derivatives, integrals, and infinite series. In our problem, we're examining calculus through the lens of parametric curves, which allows us to express curves using a parameter, typically denoted as \( t \). This reel does not bound us to standard \( x \) and \( y \) coordinates, thereby giving more flexibility in representing complex curves like ellipses.

Calculus dives deep into rates of change, and one of the primary techniques is differentiation - a process that calculates the rate at which one quantity changes with respect to another. This is crucial for understanding parametric equations that represent not just simple lines but intricate shapes like our ellipse formed by \( x = \cos t \) and \( y = 8 \sin t \). By differentiating both these equations, we can understand how \( y \) changes with respect to \( x \).

Embedded deeply within calculus are fundamental concepts like finding the derivative of a function, integration of expressions, and solving differential equations. For students tackling calculus for the first time, remember to focus on understanding the underlying principles of limits and continuity before jumping into derivatives and integrals.

Differentiation

Differentiation is a cornerstone of calculus and involves finding a derivative, a concept representing an instantaneous rate of change. In our practice problem, we utilize differentiation to find \( \frac{dy}{dx} \), which gives the slope of the tangent to the curve at any point.

To begin, you differentiate the parametric equations separately:

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

Once these derivatives are calculated, the chain rule allows us to connect these derivatives, giving the expression \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \). This relationship shows that the slope of the curve (\( \frac{dy}{dx} \)) is essentially the change in \( y \) over the change in \( x \).

Evaluating this at a specific \( t \), such as \( t = \frac{\pi}{2} \), confirms both the slope of the tangent line at that point and the nature of the curve's behavior. Here, the slope is 0, meaning the tangent line is flat, showing no rise or fall at the instant when \( t = \frac{\pi}{2} \).

Ellipse

An ellipse is a geometrical shape that appears squished from a perfect circle, having two axes - a major and a minor. In our problem, the parametric equations \( x = \cos t \) and \( y = 8 \sin t \) represent an ellipse.

This particular ellipse is centered at the origin \((0, 0)\). The horizontal axis stretches a length of 2 because the cosine function oscillates between -1 and 1. Meanwhile, the vertical axis has a length of 16 since \(8 \sin t\) varies between -8 and 8. Thus, the ellipse is elongated vertically.

Visualizing an ellipse helps in understanding the motion along this shape as \( t \) changes. Each point \( t \) corresponds to a specific (\( x, y \)) point on this ellipse. At \( t=\frac{\pi}{2} \), the ellipse intersects the line \( y=8 \) at a horizontally centered point, specifically \((0, 8)\).

For learners, grasping how these parametric forms manifest in geometric shapes like ellipses can deepen the understanding of analytic geometry and its relation to calculus. Analyzing such patterns is crucial for advanced applications in physics, engineering, and computer graphics.