Question

Slopes of tangent lines Find all the points at which the following curves have the given slope. $$x=4 \cos t, y=4 \sin t ; \text { slope }=\frac{1}{2}$$

Short answer

Question: Determine all the points on the parametric curve defined by \(x=4\cos{t}\) and \(y=4\sin{t}\), where the slope of the tangent line is equal to \(\frac{1}{2}\). Answer: \((x, y) = \left(4\cos{\left(\arccot{-\frac{1}{2}}+k\pi\right)}, 4\sin{\left(\arccot{-\frac{1}{2}}+k\pi\right)}\right)\), where \(k\) is any integer.

Step-by-step solution

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

Start by differentiating \(x\) and \(y\) with respect to \(t\). This will give: $$\frac{dx}{dt} = -4\sin{t}$$ $$\frac{dy}{dt} = 4\cos{t}$$ Now, use the chain rule to express \(\frac{dy}{dx}\) in terms of \(t\). The chain rule states that \(\frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx}\). We have the derivatives with respect to \(t\). To find \(\frac{dt}{dx}\), simply take the reciprocal of \(\frac{dx}{dt}\). $$\frac{dt}{dx} = -\frac{1}{4\sin{t}}$$ Now, substitute \(\frac{dy}{dt}\) and \(\frac{dt}{dx}\) into the chain rule formula to get: $$\frac{dy}{dx} = 4\cos{t}\cdot\left(-\frac{1}{4\sin{t}}\right)$$ Simplify this: $$\frac{dy}{dx} = -\cot{t}$$

Set \(\frac{1}{2}\) as the slope and solve for \(t\)

To find the points where the slope of the tangent line is \(\frac{1}{2}\), set \(\frac{dy}{dx}\) to \(\frac{1}{2}\) and solve for \(t\). We have: $$-\cot{t} = \frac{1}{2}$$ $$\cot{t} = -\frac{1}{2}$$ Now, we want to find the values of \(t\) for which this equation holds. To do that, take the inverse cotangent of both sides: $$t = \arccot{-\frac{1}{2}}+k\pi, \quad \text{where } k \in \mathbb{Z}$$

Find the points on the curve

Finally, substitute the values of \(t\) back into the parametric equations to get the points \((x, y)\) on the curve: $$x=4\cos{\left(\arccot{-\frac{1}{2}}+k\pi\right)}$$ $$y=4\sin{\left(\arccot{-\frac{1}{2}}+k\pi\right)}$$ For each integer value of \(k\), we get a point on the curve where the slope of the tangent line is \(\frac{1}{2}\).

Key concepts

Slopes

The slope of a line is a measure of how steep it is. In the context of curves, the slope of a tangent line at a particular point tells us the rate at which the function is changing at that point. For parametric curves, like the one given in this exercise, finding the slope involves a bit more work.

We use the derivative of the curve's equations to determine the slope of its tangent line. When dealing with parametric equations, the slope is found by dividing how much y changes with a tiny change in t (\(\frac{dy}{dt}\)), by how much x changes with the same change in t (\(\frac{dx}{dt}\)). This ratio \(\frac{dy}{dx}\) gives us the slope of the curve at any point defined by t.

• Positive slope: The curve goes upwards to the right.
• Negative slope: The curve goes downwards to the right.
• Zero slope: The curve is flat at that point.

In this example, we need to find where the slope is \(\frac{1}{2}\), meaning it rises gently as it moves from left to right.

Parametric Equations

Parametric equations describe a set of related quantities as functions of one or more independent variables, called parameters. For example, in our exercise:

• \(x = 4 \cos{t}\)
• \(y = 4 \sin{t}\)

Here, both x and y are defined in terms of a third variable, \(t\), known as the parameter.

These parameterized forms enable us to elegantly describe curves, especially those that are not simple circles or lines. They are highly useful for representing paths of objects not expressible by single y = f(x) functions. An advantage of parametric equations is the ability to efficiently calculate positions and movements along paths by just adjusting the parameter variable.

Our task involves finding specific values of \(t\) that ensure a given slope, utilizing these parametric forms.

Chain Rule

The chain rule is a fundamental calculus concept utilized to differentiate composite functions. In simpler terms, it helps us break down complex derivatives where one function is inside another. For parametric curves, we use the chain rule to switch from differentiating with respect to t to differentiating with respect to x.

The chain rule states:\[ \frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}\]This captures how y changes as x changes by letting us use the intermediary variable t, which simplifies calculations. We first differentiate y and x separately with respect to t. Then, we take what's essentially the inverse of that rate of x change with respect to t, multiplying it to get \(\frac{dy}{dx}\).

• First: Find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).
• Second: Calculate \(\frac{dt}{dx}\) as the reciprocal of \(\frac{dx}{dt}\).
• Finally: Use these to find \(\frac{dy}{dx}\) by multiplying \(\frac{dy}{dt}\) and \(\frac{dt}{dx}\).

Being proficient with the chain rule is crucial for solving problems involving variable changes and multivariable functions, commonly seen in physics and engineering applications.