Problem 79 - Calculus Volume 3

Question

For \(x=\sin (2 t), y=2 \sin t\) where \(0 \leq t<2 \pi\). Find all values of \(t\) at which a vertical tangent line exists.

Step-by-Step Solution

Answer

Values of t: \(\frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}\).

Step 1: Understand what a vertical tangent means

A vertical tangent line to a curve means that at that point, the slope of the tangent is undefined. For parametric equations, this occurs when the derivative of y with respect to t is non-zero and the derivative of x with respect to t is zero.

Step 2: Find derivatives dx/dt and dy/dt

First, find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). \(x = \sin(2t)\), thus \(\frac{dx}{dt} = 2\cos(2t)\). Similarly, \(y = 2\sin(t)\), so \(\frac{dy}{dt} = 2\cos(t)\).

Step 3: Set dx/dt to zero

For a vertical tangent, set \(\frac{dx}{dt} = 0\): \[2\cos(2t) = 0\] This simplifies to \(\cos(2t) = 0\).

Step 4: Solve for values of t

The values of t that satisfy \(\cos(2t) = 0\) are those where \(2t = \frac{\pi}{2} + n\pi\) for integer n. Solving for t gives \(t = \frac{\pi}{4} + \frac{n\pi}{2}\).

Step 5: Consider the given range for t

Since \(0 \leq t < 2\pi\), find the values of t within this range:- When \(n=0\), \(t = \frac{\pi}{4}\).- When \(n=1\), \(t = \frac{3\pi}{4}\).- When \(n=2\), \(t = \frac{5\pi}{4}\).- When \(n=3\), \(t = \frac{7\pi}{4}\).Therefore, the possible values of t are \(\frac{\pi}{4}\), \(\frac{3\pi}{4}\), \(\frac{5\pi}{4}\), and \(\frac{7\pi}{4}\).

Step 6: Verify dy/dt is not zero at these t-values

Ensure that \(\frac{dy}{dt}\) is not also zero at these points, as both being zero implies indeterminate rather than vertical tangent.\(\frac{dy}{dt} = 2\cos(t)\) should not be zero:- For \(t = \frac{\pi}{4}, \frac{dy}{dt} = 2\times\frac{1}{\sqrt{2}} eq 0\).- For \(t = \frac{3\pi}{4}, \frac{dy}{dt} = 2\times\frac{-1}{\sqrt{2}} eq 0\).- For \(t = \frac{5\pi}{4}, \frac{dy}{dt} = 2\times\frac{-1}{\sqrt{2}} eq 0\).- For \(t = \frac{7\pi}{4}, \frac{dy}{dt} = 2\times\frac{1}{\sqrt{2}} eq 0\).Thus, all these values of t are valid.

Key Concepts

Vertical Tangent

A vertical tangent occurs on a curve when the slope of the tangent line is undefined. In parametric equations, like in this exercise, we have two functions: one for the x-coordinates and one for the y-coordinates. The slope of a tangent line is given by the derivative of y with respect to x. However, in parametric form, this is expressed as: \[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\]A vertical tangent implies that

• \( \frac{dx}{dt} = 0 \) (since division by zero leads to undefined slope)
• \( \frac{dy}{dt} eq 0 \) (to ensure the tangent isn't indeterminate)

Finding these conditions helps identify the specific points where the curve becomes vertical.

Derivatives

The derivative is a fundamental tool in calculus, representing the rate at which a function changes at any given point. For parametric equations, we take the derivative of both x and y with respect to the parameter t. This provides the rates at which x and y change as t changes. For the given equations:

• \(x = \sin(2t)\) yields \(\frac{dx}{dt} = 2\cos(2t)\)
• \(y = 2\sin(t)\) gives \(\frac{dy}{dt} = 2\cos(t)\)

These derivatives help us analyze the curve's behavior, such as finding tangent lines. Setting \(\frac{dx}{dt} = 0\) leads to potential vertical tangents.

Trigonometry

Trigonometry plays a significant role in this particular problem as it deals with the sine and cosine functions. Understanding these functions is crucial for solving equations involving them. For instance, knowing that \[\cos(2t) = 0\]leads to finding specific values of t. This holds true when the angle

• \(2t = \frac{\pi}{2} + n\pi \)

where 'n' is an integer, as cosine reaches zero at those points. These trigonometric identities bring about solutions that fit within a given range like \(0 \leq t < 2\pi\). These allow us to explore valid solutions for vertical tangents in detail.

Calculus

Calculus encompasses both the study of change (differentiation) and the study of accumulation. In this exercise, it deeply involves differentiation. As we work with parametric equations, we need to understand how the components (x and y) change with respect to time or parameter 't'. This requires using derivatives to ensure we correctly define the behavior of the parametric curve. Finding vertical tangents involves setting the derivative of one parameter to zero to find critical points where the slope becomes undefined. Then verifying the second derivative isn't zero. These concepts of calculus are essential to finding key points on curves. These points help us understand movements and shapes within the mathematical and physical world.