Question

Derivatives Consider the following parametric curves. a. Determine \(dy/dx\) 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 \(t\). $$x=3 \sin t, y=3 \cos t ; t=\pi / 2$$

Short answer

Question: Determine the slope of the tangent line to the curve represented by the parametric equations \(x=3 \sin t\) and \(y=3 \cos t\) when \(t=\pi/2\) and provide a sketch of the curve and the tangent line at this point. Answer: The slope of the tangent line at \(t=\pi/2\) is \(-\infty\), indicating that the tangent line is vertical. The curve represents a circle with radius 3, and the tangent line is a vertical line passing through the point \((3,0)\).

Step-by-step solution

Find \(dx/dt\) and \(dy/dt\)

The given parametric equations are: $$x=3 \sin t \quad , \quad y=3 \cos t$$ Now, let's differentiate both x and y with respect to t: $$\frac{d x}{d t} = \frac{d(3\sin t)}{d t} = 3\cos t$$ $$\frac{d y}{d t} = \frac{d(3\cos t)}{d t} = -3\sin t$$

Calculate \(dy/dx\) using the chain rule

Using the chain rule: $$\frac{d y}{d x}\ =\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$$ So, \(\frac{d y}{d x}\ =\frac{-3\sin t}{3\cos t}\) which gives us: $$\frac{d y}{d x} = -\tan t$$

Evaluate \(dy/dx\) at \(t=\pi/2\)

Now, we will substitute the given value of t: $$\frac{d y}{d x} = -\tan\left(\frac{\pi}{2}\right)=-\infty$$ The slope of the tangent line is \(-\infty\) at \(t=\pi/2\), which means the tangent line is vertical.

Sketch the curve and the tangent line

The given parametric equations represent the equation of a circle with radius 3. To sketch the curve and the tangent line at the given point, follow these steps: 1. Plot the circle using the equation $$x^2 + y^2 = (3 \sin t)^2 + (3 \cos t)^2 = 3^2$$ 2. Locate the point corresponding to \(t=\pi/2\). This point is given by \(x=3\sin(\pi/2)=3\) and \(y=3\cos(\pi/2)=0\). So, the point is \((3,0)\). 3. From Step 3, we determined that the tangent line is vertical. Therefore, draw a vertical line parallel to the y-axis passing through the point \((3,0)\). Now you have a sketch of the curve and the tangent line at the point corresponding to \(t=\pi/2\).

Key concepts

Derivative of Parametric Equations

In parametric equations, both the x- and y-coordinates are expressed as functions of a parameter, commonly denoted as \( t \). Differentiating each function with respect to \( t \) gives us crucial information about how the points move along the curve. For example, if we have the parametric equations \( x=3 \sin t \) and \( y=3 \cos t \), we can find the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) to understand the changes in x and y as t changes.

When we differentiate these:

• \( \frac{dx}{dt} = 3 \cos t \)
• \( \frac{dy}{dt} = -3 \sin t \)

These derivatives reflect how the x- and y-coordinates increase or decrease relative to changes in \( t \). To find the slope \( \frac{dy}{dx} \) of the parametric curve, we make use of the chain rule. This takes us directly to our next concept.

Tangent Line to a Parametric Curve

The tangent line to a curve at a given point is a straight line that just "touches" the curve at that point. The slope of this line can be found using the derivative \( \frac{dy}{dx} \) obtained from the parametric equations. With our circle parametrization example, at each point on the circle, the tangent can change direction rapidly. So, first we must distinguish \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \), then use them to find the slope of our tangent line.

The tangent line slope is given by the expression \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \). In our solution:

• \( \frac{dy}{dx} = \frac{-3 \sin t}{3 \cos t} = -\tan t \)

This means at any point on the circle defined by \( t \), the gradient (or direction of steepness) of our tangent line relates to \( -\tan t \). Evaluating at \( t=\frac{\pi}{2} \), we find the slope is \( -\infty \), indicating a vertical tangent line.

Chain Rule in Calculus

The chain rule is a fundamental tool in calculus used to differentiate compositions of functions. In parametric equations, it aids in expressing \( \frac{dy}{dx} \) as a ratio of \( \frac{dy}{dt} \) over \( \frac{dx}{dt} \). This is because the functions \( x \) and \( y \) are both linked through the common parameter \( t \), making their derivatives dependent on \( t \) as well.

Using the earlier provided formula \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \), we apply the chain rule to solve for derivatives more efficiently when functions are nested or when dealing with parametric forms:

• Divide \( \frac{dy}{dt} = -3 \sin t \) by \( \frac{dx}{dt} = 3 \cos t \)
• The result, \( \frac{dy}{dx} = -\tan t \), gives us the relationship of the slope in terms of \( t \).

Thus, the chain rule simplifies complex differential tasks by breaking them into manageable parts, especially useful for parametric curves where each variable depends on another parameter.