Problem 91 - Thomas Calculus

Question

In Exercises \(87-94\) , 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. $$ x=2 t^{2}+3, \quad y=t^{4}, \quad t=-1 $$

Step-by-Step Solution

Answer

The tangent line equation is \( y = x - 4 \) and \( \frac{d^2 y}{dx^2} \) at \( t = -1 \) is \( \frac{1}{2} \).

Step 1: Find the Derivative dx/dt

Given that \( x = 2t^2 + 3 \), we differentiate with respect to \( t \): \[ \frac{dx}{dt} = \frac{d}{dt}(2t^2 + 3) = 4t. \]

Step 2: Find the Derivative dy/dt

Given that \( y = t^4 \), we differentiate with respect to \( t \): \[ \frac{dy}{dt} = \frac{d}{dt}(t^4) = 4t^3. \]

Step 3: Find dy/dx

The derivative \( \frac{dy}{dx} \) is found by dividing \( \frac{dy}{dt} \) by \( \frac{dx}{dt} \): \[ \frac{dy}{dx} = \frac{4t^3}{4t} = t^2. \]

Step 4: Evaluate the Gradient at t = -1

Substitute \( t = -1 \) into \( \frac{dy}{dx} \): \[ \frac{dy}{dx} \bigg|_{t=-1} = (-1)^2 = 1. \]

Step 5: Find the Point on the Curve

Substitute \( t = -1 \) into the expressions for \( x \) and \( y \):\[ x = 2(-1)^2 + 3 = 5, \quad y = (-1)^4 = 1. \]The point on the curve is \((5, 1)\).

Step 6: Write the Equation of the Tangent Line

Using the point-slope form for a line \( y - y_1 = m(x - x_1) \), where \( m \) is the gradient:\[ y - 1 = 1(x - 5). \]This simplifies to \( y = x - 4 \).

Step 7: Compute dx/dt for the Second Derivative (d²y/dx²)

To find \( \frac{d^2 y}{dx^2} \), use the formula:\[ \frac{d^2 y}{dx^2} = \frac{d}{dt}\left( \frac{dy}{dx} \right) \div \frac{dx}{dt}. \]Recalculate derivatives: \( \frac{dy}{dx} = t^2 \), so \( \frac{d}{dt}(t^2) = 2t \).

Step 8: Evaluate d²y/dx² at t = -1

Substitute derivatives into the formula:\[ \frac{d^2 y}{dx^2} = \frac{2t}{4t} = \frac{1}{2}. \]Evaluate at \( t = -1 \):\[ \frac{d^2 y}{dx^2} \bigg|_{t=-1} = \frac{1}{2}. \]

Key Concepts

Parametric Equations

Parametric equations are a powerful way to express mathematical relationships. Unlike conventional equations which use one variable (like x) to define another (like y), parametric equations use a third variable, typically t, to define both x and y independently. This approach is especially helpful for describing curves that aren't just simple lines or circles.

For example, we have the parametric equations given in our original problem:

• \( x = 2t^2 + 3 \)
• \( y = t^4 \)

Both x and y are expressed in terms of t, which allows for more complex shapes when the full set of points on the curve is plotted. By varying t, you can plot a curve by getting values for x and y, which shows a different way of understanding functions and how variables relate to one another.

It’s important to understand how these equations can be manipulated to determine properties of the curve – like finding tangent lines or determining curvature.

Implicit Differentiation

Implicit differentiation is a technique used to find the derivative of an equation that defines y implicitly as a function of x. In simpler terms, it allows us to find derivatives when y and x are jumbled together in an equation, rather than neatly separated.

In the context of parametric equations, we often encounter the need to find \( \frac{dy}{dx} \) without having y expressed explicitly in terms of x. Instead, we can differentiate both x and y with respect to t, a shared variable, and then create a ratio to find \( \frac{dy}{dx} \).
For instance, we found:

• \( \frac{dx}{dt} = 4t \)
• \( \frac{dy}{dt} = 4t^3 \)
• \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4t^3}{4t} = t^2 \)

Thus, implicit differentiation gives us a way to determine slopes (represented by \( \frac{dy}{dx} \)) for curves described by parametric equations, leading to the information needed for tangent lines.

Second Derivative

The second derivative, denoted \( \frac{d^2 y}{dx^2} \), is a measure of the curvature of a function. It provides insight into how the slope of a curve is changing at any given point.

After finding \( \frac{dy}{dx} \) from our parametric equations, we used another layer of differentiation to obtain \( \frac{d^2 y}{dx^2} \). This involved differentiating \( \frac{dy}{dx} \) with respect to t and dividing the result by \( \frac{dx}{dt} \).
From our example:

• Differentiate \( t^2 \) to get \( 2t \)
• Divide by \( \frac{dx}{dt} = 4t \) to find \( \frac{d^2 y}{dx^2} = \frac{2t}{4t} = \frac{1}{2} \)

Evaluating at \( t = -1 \) showed \( \frac{d^2 y}{dx^2} = \frac{1}{2} \). Understanding the second derivative is crucial for analyzing concavity and points of inflection on the curve, showing how the curve bends and twists.