Problem 71 - Calculus Volume 3

Question

Find all points on the curve that have the given slope. $$ x=4 \cos t, \quad y=4 \sin t, \text { slope }=0.5 $$

Step-by-Step Solution

Answer

The points are approximately \((-1.6, 3.624)\) and \((1.6, -3.624)\).

Step 1: Find the derivatives

To find the slope of the curve specified parametrically by \(x = 4 \cos t\) and \(y = 4 \sin t\), we first need to find the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).1. Calculate \( \frac{dx}{dt} \):\(\frac{dx}{dt} = -4 \sin t\).2. Calculate \( \frac{dy}{dt} \):\(\frac{dy}{dt} = 4 \cos t\).

Step 2: Express dy/dx in terms of t

The slope \( \frac{dy}{dx} \) is given by the chain rule as \( \frac{dy/dt}{dx/dt} \). Using the derivatives we found:\[ \frac{dy}{dx} = \frac{4 \cos t}{-4 \sin t} = -\cot t \].

Step 3: Set the slope equal to the given value

We need to set \( \frac{dy}{dx} = -\cot t \) equal to the given slope of 0.5:\[-\cot t = 0.5\].

Step 4: Solve for t

Solving \(-\cot t = 0.5\) gives \(\cot t = -0.5\).We can express \(\cot t\) as \(\frac{\cos t}{\sin t}\), so we have \(\frac{\cos t}{\sin t} = -0.5\).This is equivalent to \(\cos t = -0.5 \sin t\), which can be rearranged to \(\cos t + 0.5 \sin t = 0\).

Step 5: Find the points (x, y)

Since the equation \(\cos t + 0.5 \sin t = 0\) represents angles, we solve for \(t\) using trigonometric identities or a calculator. Solving gives angles \(t\) that satisfy this equation.Find \(t\): By trial or using a calculator, \(t \approx 2.034\) or \(t \approx 5.176\) repeat every \(2\pi\).Calculate \(x\) and \(y\) for these \(t\) values:1. For \(t = 2.034\), \(x = 4 \cos(2.034)\) and \(y = 4 \sin(2.034)\).2. For \(t = 5.176\), \(x = 4 \cos(5.176)\) and \(y = 4 \sin(5.176)\).Thus, the corresponding points are \((-1.6, 3.624)\) and \((1.6, -3.624)\).

Step 6: Conclusion

The points on the curve where the slope is 0.5 are approximately \((-1.6, 3.624)\) and \((1.6, -3.624)\).

Key Concepts

Trigonometric Identities

Trigonometric identities are fundamental in solving problems involving angles and lengths in mathematics, especially trigonometry. These identities help in simplifying and solving equations by relating various trigonometric functions to each other. For instance, in the parametric equations given in the problem, we deal with basic trigonometric functions: sine and cosine.

Some of the key identities are:

• Pythagorean Identity: \( \sin^2 t + \cos^2 t = 1 \)
• Tangent and Cotangent Identities: \( \tan t = \frac{\sin t}{\cos t} \), and \( \cot t = \frac{\cos t}{\sin t} \)

In the solution to the exercise, the identity for cotangent, \( \cot t = \frac{\cos t}{\sin t} \), is used to solve \(-\cot t = 0.5\). This approach enables finding the angles, \( t \), for which the cotangent equals \(-0.5\). Recognizing and applying these identities is crucial for simplifying expressions and solving trigonometric equations.

Slope of a Curve

The slope of a curve at any given point, particularly for curves defined by parametric equations, represents the steepness or incline of the tangent line to the curve at that point. In technical terms, the slope is defined geometrically as the derivative of the curve's equation with respect to its independent parameter.

For parametric equations, the slope \( \frac{dy}{dx} \) is determined using the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) as follows:

• Step 1: Differentiate \( x = 4 \cos t \) with respect to \( t \) to obtain \( \frac{dx}{dt} = -4 \sin t \).
• Step 2: Differentiate \( y = 4 \sin t \) with respect to \( t \) to obtain \( \frac{dy}{dt} = 4 \cos t \).
• Step 3: Use the chain rule to compute \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = -\cot t \).

This method effectively finds the slope of the curve at any point. For the given problem, setting \( \frac{dy}{dx} = -\cot t \) to the specified slope value of 0.5 allows us to solve for the parameter \( t \) and ultimately find the coordinates of the points on the curve where that slope occurs.

Differentiation in Polar Coordinates

While the given problem involves parametric equations, the concept of differentiation in polar coordinates often parallels methods used in parametric forms. Polar coordinates express points on a plane using a distance from a reference point and an angle from a reference direction.

When differentiating in polar coordinates, we employ these equations to convert into Cartesian coordinates or use polar-specific formulas. A curve in polar form, \( r = f(\theta) \), has its derivatives similarly defined by:

• \( x = r \cos \theta \), \( y = r \sin \theta \)
• For the derivatives: \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \) can be computed directly from these expressions.
• The slope is then \( \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \).

While this particular exercise doesn't explicitly use polar coordinate differentiation, understanding these foundational concepts is vital since many curves switch interchangeably between polar and parametric representations. Mastery of these ideas aids in solving a wide range of advanced calculus problems.