Problem 28 - Calculus Early Transcendentals: Pearson New International Edition

Question

find \(d y / d x\) and \(d^{2} y / d x^{2}\) without eliminating the parameter. $$ x=\cot t-2, y=-2 \csc t+5 ; 0

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = -2 \cos t \); \( \frac{d^2y}{dx^2} = -2 \).

Step 1: Differentiate x with respect to t

Given the parametric equation for x: \( x = \cot t - 2 \). Differentiate this with respect to \( t \) to find \( \frac{dx}{dt} \).Use the derivative of \( \cot t \), which is \( -\csc^2 t \):\[ \frac{dx}{dt} = -\csc^2 t \]

Step 2: Differentiate y with respect to t

Given the parametric equation for y: \( y = -2 \csc t + 5 \). Differentiate this with respect to \( t \) to find \( \frac{dy}{dt} \).Use the derivative of \( \csc t \), which is \( -\csc t \cot t \):\[ \frac{dy}{dt} = -2(-\csc t \cot t) = 2\csc t \cot t \]

Step 3: Find \( \frac{dy}{dx} \) using \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \)

To find \( \frac{dy}{dx} \), use the formula:\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \]Substitute the values from Step 1 and Step 2:\[ \frac{dy}{dx} = \frac{2\csc t \cot t}{-\csc^2 t} = -2\frac{\cot t}{\csc t} = -2\cos t \]

Step 4: Differentiate \( \frac{dy}{dt} \) again for \( \frac{d^2y}{dt^2} \)

Differentiate the expression for \( \frac{dy}{dt} \) again with respect to \( t \) to find \( \frac{d^2y}{dt^2} \):\[ \frac{d^2y}{dt^2} = \frac{d}{dt}(2 \csc t \cot t) \ = 2 (-\csc t \cot^2 t - \csc^3 t) \ = -2 \csc t (\cot^2 t + \csc^2 t) \]

Step 5: Differentiate \( \frac{dy}{dx} \) with respect to t to find \( \frac{d^2y}{dx^2} \)

Now, differentiate \( \frac{dy}{dx} = -2 \cos t \) with respect to \( t \) and then use the chain rule to divide by \( \frac{dx}{dt} \):First, differentiate \( \frac{dy}{dx} \):\[ \frac{d}{dt}(-2\cos t) = 2\sin t \]Now, find \( \frac{d^2y}{dx^2} \) by dividing this derivative by \( \frac{dx}{dt} \):\[ \frac{d^2y}{dx^2} = \frac{2\sin t}{-\csc^2 t} = -2 \cos^2 t \]

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

Notice that \( \csc t \) is \( \frac{1}{\sin t} \), so multiplication results in:\[ \frac{d^2y}{dx^2} = -2 (\sin t)^2 = -2 \]

Key Concepts

Parametric Equations

Parametric equations help us describe a curve in the plane without restricting it to a single coordinate system like Cartesian or polar coordinates. Instead of relying on expressions like \( y = f(x) \), parametric equations use an independent parameter to express \( x \)and \( y \):

• x = \( f(t) \)
• y = \( g(t) \)

In this format, \( t \)is the independent parameter, usually representing time in physics but can serve other purposes in mathematics.
In our example, the parametric equations are:

• \( x = \cot t - 2 \)
• \( y = -2 \csc t + 5 \)

When working with parametric equations, instead of differentiating \( y \)with respect to \( x \), we differentiate each expression with respect to \( t \), and then use the chain rule. This method is particularly useful when the relationship between \( x \)and \( y \)is not straightforward and involves more complex forms, like trigonometric functions as seen here.

Second Derivative

The second derivative \( \frac{d^2y}{dx^2} \) measures how the rate of change \( \frac{dy}{dx} \)itself changes with respect to \( x \).With parametric equations, to find the second derivative, we perform the following steps:

• First, obtain \( \frac{dy}{dx} \) by dividing \( \frac{dy}{dt} \) by \( \frac{dx}{dt} \).
• Second, differentiate \( \frac{dy}{dx} \) with respect to \( t \).
• Finally, divide this result by \( \frac{dx}{dt} \)to obtain \( \frac{d^2y}{dx^2} \).

For our problem, \( \frac{dy}{dx} = -2 \cos t \).The second step gives us \( 2 \sin t \).After dividing by \( -\csc^2 t \), which simplifies to \( \sin^2 t \), we get \( \frac{d^2y}{dx^2} = -2 \).This approach effectively captures the behavior of the curve and highlights concavity or convexity, indicating whether the curve is bending upwards or downwards.

Trigonometric Differentiation

Trigonometric differentiation is pivotal when handling functions involving sine, cosine, tangent, and their respective reciprocals. Essentially, differentiation rules specific to trigonometric functions simplify the process:

• \( \frac{d}{dt}(\sin t) = \cos t \)
• \( \frac{d}{dt}(\cos t) = -\sin t \)
• \( \frac{d}{dt}(\tan t) = \sec^2 t \)
• \( \frac{d}{dt}(\csc t) = -\csc t \cot t \)
• \( \frac{d}{dt}(\sec t) = \sec t \tan t \)
• \( \frac{d}{dt}(\cot t) = -\csc^2 t \)

When working with parametrically defined curves, these rules allow quick differentiation of complex equations. For instance, in our solution, the expressions of \( x \)and \( y \) involve \( \cot t \)and \( \csc t \).Differentiating these using the above rules yields:

• \( \frac{dx}{dt} = -\csc^2 t \)
• \( \frac{dy}{dt} = 2\csc t \cot t \)

Understanding how each function's derivative emerges is crucial to progressing through calculus problems smoothly and accurately.