Problem 70 - Calculus Volume 3

Question

Determine the slope of the tangent line, then find the equation of the tangent line at the given value of the parameter. $$ x=\sqrt{t}, \quad y=2 t, \quad t=4 $$

Step-by-Step Solution

Answer

The equation of the tangent line is \( y = 8x - 8 \).

Step 1: Express Derivatives

We are given the parametric equations \( x = \sqrt{t} \) and \( y = 2t \). To find the slope of the tangent, we need to calculate \( \frac{dy}{dx} \). This is the derivative of \( y \) with respect to \( x \), which can be found using \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \). First, find these derivatives: \( \frac{dx}{dt} = \frac{d}{dt}(\sqrt{t}) = \frac{1}{2\sqrt{t}} \) and \( \frac{dy}{dt} = \frac{d}{dt}(2t) = 2 \).

Step 2: Calculate the Slope of the Tangent Line

Use the chain rule to find the slope of the tangent line: \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2}{\frac{1}{2\sqrt{t}}} = 4 \sqrt{t} \). Plug in \( t = 4 \) to find the slope at this point: \( 4 \sqrt{4} = 8 \). The slope of the tangent line at \( t = 4 \) is 8.

Step 3: Evaluate the Parametric Equations at \( t=4 \)

Substitute \( t = 4 \) into the parametric equations to find the point of tangency. For \( x \), \( x = \sqrt{4} = 2 \). For \( y \), \( y = 2 \times 4 = 8 \). The point is (2, 8).

Step 4: Write the Equation of the Tangent Line

Use the point-slope form of the equation of a line, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \((x_1, y_1)\) is the point. Substitute \( m = 8 \), \( x_1 = 2 \), \( y_1 = 8 \): \( y - 8 = 8(x - 2) \). Simplify to get \( y = 8x - 8 \).

Key Concepts

Parametric Equations

Parametric equations are a powerful way to describe curves using parameters. In simple words, rather than directly relating two variables, like in the function \( y = f(x) \), we use a third variable, usually \( t \), to express both \( x \) and \( y \). This makes it handy to describe more complex curves you cannot define using standard equations.
For the given exercise, the parametric equations are \( x = \sqrt{t} \) and \( y = 2t \). This means both \( x \) and \( y \) depend on the parameter \( t \), forming a pair for every value of \( t \). Parametric equations can easily describe paths and motions, which are vital in calculus, physics, and engineering. Some common uses include:

• Modeling trajectories of objects and paths of moving points.
• Representing circles, ellipses, and other curves not expressible as functions.
• Defining curves that loop or change direction.

Understanding how to work with parametric equations is key to solving this exercise and has many applications in real-world problems.

Tangent Line

A tangent line is a straight line that touches a curve at a single distinct point but does not cross it at that point. This line represents the instantaneous rate of change or the slope at that particular spot on the curve.
Consider drawing a line to touch a circle exactly at one point. That touch point is the concept behind a tangent. For the parametrically defined path described by \( x = \sqrt{t} \) and \( y = 2t \), we're interested in determining the tangent line at when \( t = 4 \).
Here, the tangent line we calculate will help us understand the behavior of the curve at \( t = 4 \), giving insights into how the path is changing right at that location. Tangent lines are extremely useful because they allow for approximations and predictions around points on a curve.

Derivatives

Derivatives are foundational in understanding how functions change. They provide the slope of a function at any given point, defining an instantaneous rate of change. In this exercise, we are asked to find \( \frac{dy}{dx} \), which is the derivative of \( y \) with respect to \( x \).
To derive parametric equations like \( x = \sqrt{t} \) and \( y = 2t \), use intermediate derivatives \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \). These calculations are crucial:

• \( \frac{dx}{dt} = \frac{1}{2\sqrt{t}} \) indicates how \( x \) changes as \( t \) increases.
• \( \frac{dy}{dt} = 2 \) shows the rate at which \( y \) changes with \( t \).

The derivative \( \frac{dy}{dx} \) represents the slope of the tangent at any point \( t \). Derivatives are wide-ranging in application, calculating speed, acceleration, and other changes over time.

Chain Rule

The chain rule is a fundamental concept that lets us compute the derivative of composite functions. In the context of parametric equations, it helps in finding how one variable changes concerning another, when both are dependent on a third variable.
When determining \( \frac{dy}{dx} \) given \( x = \sqrt{t} \) and \( y = 2t \), the chain rule shines. It states:

\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \]

This is crucial as it breaks the problem into manageable pieces. Compute \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \) first. Then, use their ratio to find the overall derivative \( \frac{dy}{dx} \).
This method simplifies complex derivatives often encountered in situations where direct calculations are tricky. The chain rule is immensely useful not only in calculus but also in areas involving multivariable functions.