Problem 12 - Thomas Calculus

Question

Find an equation for the line tangent to the curve at the point defined by the given value of \(t\) . Also, find the value of \(d^{2} y / d x^{2}\) at this point. $$ x=\cos t, \quad y=1+\sin t, \quad t=\pi / 2 $$

Step-by-Step Solution

Answer

The tangent line equation is \(y = 2\); \(\frac{d^2y}{dx^2} = -1\).

Step 1: Parameterize the Curve

To begin, note that the curve is parameterized by the functions \(x = \cos t\) and \(y = 1 + \sin t\). We are interested in the point where \(t = \frac{\pi}{2}\).

Step 2: Calculate the Point on the Curve

Substitute \(t = \frac{\pi}{2}\) into the parameterizations:\[ x = \cos\left(\frac{\pi}{2}\right) = 0, \quad y = 1 + \sin\left(\frac{\pi}{2}\right) = 2. \]Thus, the point on the curve is \((0, 2)\).

Step 3: Find Derivative \(\frac{dy}{dx}\)

Use the chain rule to find \(\frac{dy}{dx}\):\[ \frac{dx}{dt} = -\sin t, \quad \frac{dy}{dt} = \cos t. \]Thus, \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\cos t}{-\sin t} = -\cot t\).

Step 4: Evaluate \(\frac{dy}{dx}\) at \(t = \frac{\pi}{2}\)

Substitute \(t = \frac{\pi}{2}\) into \(\frac{dy}{dx} = -\cot t\):\[ \frac{dy}{dx}\bigg|_{t = \frac{\pi}{2}} = -\cot\left(\frac{\pi}{2}\right) = 0. \]The slope of the tangent line is 0.

Step 5: Write the Equation of the Tangent Line

The equation of the tangent line is of the form \(y - y_1 = m(x - x_1)\), where \(m\) is the slope. Here, \(m = 0\), \((x_1, y_1) = (0, 2)\), so:\[ y - 2 = 0(x - 0) \Rightarrow y = 2. \]

Step 6: Compute \(\frac{d^2y}{dx^2}\)

To find \(\frac{d^2y}{dx^2}\), use the derivatives:\[ \frac{d}{dt}\left( -\cot t \right) = \frac{d}{dt}\left(\frac{\cos t}{-\sin t}\right) = \frac{\sin^2 t + \cos^2 t}{\sin^2 t} = \frac{1}{\sin^2 t} = \csc^2 t. \]Now, differentiate \(\frac{dy}{dx} = -\cot t\) with respect to \(t\) and multiply by \(\frac{dt}{dx}\):\[ \frac{d^2y}{dx^2} = \csc^2 t \cdot \frac{1}{\frac{dx}{dt}} = \csc^2 t \cdot \frac{1}{-\sin t}. \]

Step 7: Evaluate \(\frac{d^2y}{dx^2}\) at \(t = \frac{\pi}{2}\)

Substitute \(t = \frac{\pi}{2}\) into \(\frac{d^2y}{dx^2}\):\[ \frac{d^2y}{dx^2} \bigg|_{t = \frac{\pi}{2}} = \csc^2\left(\frac{\pi}{2}\right) \cdot \frac{1}{-\sin\left(\frac{\pi}{2}\right)} = 1 \cdot \frac{1}{-1} = -1. \]Thus, \(\frac{d^2y}{dx^2} = -1\) at this point.

Key Concepts

Parametric Equations

Parametric equations allow us to define a curve in the plane using two separate equations for each coordinate of the points on the curve, typically in terms of a third variable, called the parameter. In this exercise, the parameter is denoted as \( t \). Instead of expressing \( y \) as a direct function of \( x \), we use:

• \( x = \cos t \)
• \( y = 1 + \sin t \)

These equations specify a curve through the trajectory of points \((x, y)\) as \( t \) varies. When \( t = \frac{\pi}{2} \), substituting \( t \) into these equations gives the specific point \((0, 2)\) on the curve. This method of parameterizing via \( t \) is useful in tracing more complex paths and is especially helpful in dealing with objects moving over time.

Chain Rule

The chain rule is a fundamental principle in calculus used to differentiate composite functions. When dealing with parametric equations, we often need to find \( \frac{dy}{dx} \), the derivative of \( y \) with respect to \( x \), while we begin with \( y \) and \( x \) defined in terms of \( t \). To achieve this, we differentiate each equation with respect to \( t \):

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

We can then relate these to find \( \frac{dy}{dx} \) using the formula:\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\]This gives us \( \frac{dy}{dx} = -\cot t \). At \( t = \frac{\pi}{2} \), substituting this value results in \( \frac{dy}{dx} = 0 \). Hence, the slope of the tangent line at this point is 0, indicating a horizontal tangent. This chain rule insight is crucial when dealing with parametric curves and their slopes.

Second Derivative

The second derivative \( \frac{d^2y}{dx^2} \) of a parametric curve provides information about the curve's concavity and inflection points. To find this, we initially differentiate \( \frac{dy}{dx} \) with respect to \( t \), given:\[\frac{dy}{dx} = -\cot t\]Using the derivative \( \frac{d}{dt} \left(-\cot t\right) = \csc^2 t \), we then apply the chain rule again, multiplying by \( \frac{dt}{dx} \):\[\frac{d^2y}{dx^2} = \csc^2 t \cdot \frac{1}{\frac{dx}{dt}}\]Since \( \frac{dx}{dt} = -\sin t \), we substitute it to yield:\[\frac{d^2y}{dx^2} = \csc^2 t \cdot \frac{1}{-\sin t} = \frac{1}{-\sin t} \times \frac{1}{\sin t} = \frac{1}{-\sin^2 t}\]At \( t = \frac{\pi}{2} \), the value simplifies to -1, indicating that at this point, the curve is concave down. This demonstrates how parametric calculus can offer insights into a curve’s behavior beyond merely finding tangent lines.