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Horizontal and Vertical Tangency In Exercises \(29-38\) , find all points (if any) of horizontal and vertical tangency to the curve. Use a graphing utility to confirm your results. $$ x=t+4, \quad y=t^{3}-3 t $$

Short Answer

Expert verified
The points of horizontal tangency to the curve are (5, -2) and (3, 2), and there are no points of vertical tangency.

Step by step solution

01

Find the derivative

Firstly, to find the derivative of the parametric equation, we use the formula \(y'(t) = dy/dx = (dy/dt) / (dx/dt)\). With the given equations we can obtain \(dx/dt = 1\) and \(dy/dt = 3t^{2}-3\). Therefore, \(y'(t) = (3t^{2} - 3) / 1 = 3t^{2} - 3\).
02

Find the points of horizontal tangency

For horizontal tangency, the derivative of the function is equal to zero. Hence, we set \(3t^{2}-3 = 0\). By solving this equation, we find \(t = ±1\). Substituting these values back into the original parametric equations gives us the points of horizontal tangency: (5, -2) when t=-1, and (3, 2) when t=1.
03

Find the points of vertical tangency

A point of vertical tangency would occur when the denominator of the derivative (\(dx/dt = 1\)) is equal to 0. However, in this case, since the denominator (the derivative of x with respect to t) is always 1, there is no value of t that can make the slope of the tangent line vertical. Therefore, there are no points of vertical tangency for this parametric curve.
04

Confirm results with a graphing utility

By plotting the parametric equation, we should be able to visually confirm that the curve indeed has horizontal tangents at the points (5, -2) and (3, 2), and no vertical tangents.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Horizontal Tangency
Understanding horizontal tangency in the context of parametric equations requires looking at the slope of the tangent line to the curve. In this case, a horizontal tangent occurs where the slope is zero. When we have a parametric equation, such as the one in our exercise \( x=t+4, y=t^{3}-3t \), we can find the horizontal tangents by setting the derivative \( y'(t) \) equal to zero:

For our equation, \( y'(t) = 3t^{2}-3 \), solving for \( y'(t) = 0 \) provides us with specific values of \( t \) that correspond to horizontal tangency. In our example, we find two points of horizontal tangency, occurring at \( t = -1 \) and \( t = 1 \) which correspond to the points (5, -2) and (3, 2) on the graph.
Vertical Tangency
Vertical tangency is characterized by a tangent line that is vertical, or in other words, the slope of the tangent is undefined or infinite. To locate points of vertical tangency in parametric equations, we investigate where the denominator of the derivative expression, \( dx/dt \) approaches zero. With \( x'(t) = 1 \) for our provided exercise, we note that there are no values of \( t \) where \( x'(t) \) is zero since it remains constant. Consequently, this implies that no vertical tangency will occur for the curve described by the given set of parametric equations.
Parametric Derivative Calculation
To find points of tangency, we must first understand how to compute the derivative of a parametric equation. The derivative \( y'(t) \) represents the slope of the tangent line and is calculated by dividing the derivative of \( y \) with respect to \( t \) by the derivative of \( x \) with respect to \( t \) i.e., \( y'(t) = \frac{dy/dt}{dx/dt} \). In this exercise, the derivatives \( \frac{dx}{dt} = 1 \) and \( \frac{dy}{dt} = 3t^{2}-3 \) are used to compute \( y'(t) \) which is then analyzed to determine the points of tangency. This step is crucial for understanding the behavior of the graph at different points.
Graphing Utility Confirmation
Once we have theoretically determined the points of horizontal and vertical tangency, it's essential to visually confirm the findings with a graphing utility. By plotting the curve defined by the parametric equations \( x=t+4 \) and \( y=t^{3}-3t \), we can inspect the presence of horizontal tangency at the points found in our calculations. Using a graphing utility serves as a visual concretization of the abstract derivatives and algebraic manipulations we perform, providing an excellent tool to affirm that the computed points (5, -2) and (3, 2) align with the graph's apparent horizontal tangent lines. Notably, for this specific curve, there will be no vertical tangent lines to confirm, as illustrated when the derivative of \( x \) with respect to \( t \) never equals zero.

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Most popular questions from this chapter

Sketching and Identifying a Conic In Exercises \(13-22\) , find the eccentricity and the distance from the pole to the directrix of the conic. Then sketch and identify the graph. Use a graphing utility to confirm your results. $$ r=\frac{4}{1+\cos \theta} $$

Area of a Region For each polar equation, sketch its graph, determine the interval that traces the graph only once, and find the area of the region bounded by the graph using a geometric formula and integration. (a) \(r=10 \cos \theta \quad\) (b) \(r=5 \sin \theta\)

Folium of Descartes A curve called the folium of Descartes can be represented by the parametric equations $$x=\frac{3 t}{1+t^{3}}\( and \)y=\frac{3 t^{2}}{1+t^{3}}$$ (a) Convert the parametric equations to polar form. (b) Sketch the graph of the polar equation from part (a). (c) Use a graphing utility to approximate the area enclosed by the loop of the curve.

Sketching and Identifying a Conic In Exercises \(13-22\) , find the eccentricity and the distance from the pole to the directrix of the conic. Then sketch and identify the graph. Use a graphing utility to confirm your results. $$ r=\frac{300}{-12+6 \sin \theta} $$

Writing In Exercises 33 and \(34,\) use a graphing utility to graph the polar equations and approximate the points of intersection of the graphs. Watch the graphs as they are traced in the viewing window. Explain why the pole is not a point of intersection obtained by solving the equations simultaneously. $$ \begin{array}{l}{r=4 \sin \theta} \\ {r=2(1+\sin \theta)}\end{array} $$

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