Problem 34 - Calculus Early Transcendentals: Pearson New International Edition

Question

, find the equation of the tangent line to the given curve at the given value of \(t\) without eliminating the parameter. Make a sketch. $$ x=2 e^{t}, y=\frac{1}{3} e^{-t} ; t=0 $$

Step-by-Step Solution

Answer

The tangent line is \( y = -\frac{1}{6}x + \frac{2}{3} \) at \( (2, \frac{1}{3}) \).

Step 1: Parameterized Equation Evaluation

First, evaluate the curve's position when \( t=0 \). Substitute \( t=0 \) into the parameterized equations: \( x = 2e^t \) and \( y = \frac{1}{3}e^{-t} \). Thus, \( x = 2e^0 = 2 \) and \( y = \frac{1}{3}e^0 = \frac{1}{3} \). The point on the curve is \((2, \frac{1}{3})\).

Step 2: Differentiate the Parameterized Equations

Find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). For \( x = 2e^t \), \( \frac{dx}{dt} = 2e^t \). For \( y = \frac{1}{3}e^{-t} \), \( \frac{dy}{dt} = -\frac{1}{3}e^{-t} \).

Step 3: Find the Slope of the Tangent Line

To find the slope of the tangent line, use \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \). Substitute \( t=0 \) into the derivatives: \( \frac{dx}{dt} = 2e^0 = 2 \) and \( \frac{dy}{dt} = -\frac{1}{3}e^0 = -\frac{1}{3} \). Therefore, the slope is \( \frac{-\frac{1}{3}}{2} = -\frac{1}{6} \).

Step 4: Write the Equation of the Tangent Line

Use the point-slope form of a line equation: \( y - y_1 = m(x - x_1) \). The point \((x_1, y_1)\) is \((2, \frac{1}{3})\), and the slope \( m \) is \(-\frac{1}{6} \). Substitute these values into the equation: \( y - \frac{1}{3} = -\frac{1}{6}(x - 2) \). Simplify to get the equation of the tangent line: \( y = -\frac{1}{6}x + \frac{1}{3} + \frac{1}{3} \), or \( y = -\frac{1}{6}x + \frac{2}{3} \).

Step 5: Sketch the Curve and the Tangent Line

Sketch the parametric curve derived from \( x = 2e^t \) and \( y = \frac{1}{3}e^{-t} \), noting that it plots a curve that tries to replicate an exponential growth for \( x \) and exponential decay for \( y \). Mark the point \( (2, \frac{1}{3}) \) on the graph. Draw the tangent line with the slope \(-\frac{1}{6}\) that passes through this point. The tangent line touches the curve at this single point.

Key Concepts

Parametric Equations

Parametric equations are a fascinating way to describe a curve using parameters, usually represented as variables. In our exercise, we have parameterized equations for both x and y, dependent on a parameter, t. This means instead of one equation involving x and y, we have two: one for x in terms of t and one for y in terms of t.
For the given problem, our equations are: \( x = 2e^t \) and \( y = \frac{1}{3}e^{-t} \).
These equations mark the curve's position at any given time t. By using parametric equations, you can represent more complex shapes and movements that might be cumbersome with conventional equations. This approach is fairly common, especially in calculus and physics, as it allows us to consider motion as a function of time.

Tangent Line

The tangent line is a straight line that "just touches" a curve at a point, matching the curve's direction at that point. It's a powerful concept because it gives us the best linear approximation of the curve at that point. For the curve given by the parametric equations, the tangent line can be found by calculating the derivative to get the slope.
At \( t = 0 \), the point on the curve where we need the tangent line is \( (2, \frac{1}{3}) \).
Once we acquire the slope at this point using derivatives of the parametric equations, we can use the point-slope form of a line: \ y - y_1 = m(x - x_1) \ to write the equation of the tangent line. This process grounds abstract mathematics in a visual and concrete form, useful in understanding changes on a curve at a specific point.

Differentiation

Differentiation is key in calculus. It's the process of finding the derivative, which signifies how a function changes at any point. For parametric equations, differentiation lets us understand how x and y change with respect to t.
For the given problem:

• To find \( \frac{dx}{dt} \) where \( x = 2e^t \), we differentiate to get \( 2e^t \).
• For \( y = \frac{1}{3}e^{-t} \), \( \frac{dy}{dt} \) comes out as \(-\frac{1}{3}e^{-t} \).

To find the slope of the tangent line \( \frac{dy}{dx} \), we divide these derivatives: \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \).
That's how differentiation helps in understanding and calculating the tangent line's slope, offering insight into the instantaneous rate of change at any given point on the curve.

Exponential Functions

Exponential functions are crucial, often appearing in numerous applications like growth models, decay processes, and more. They're characterized by having a constant base raised to a variable exponent. The two parameterized equations \( x = 2e^t \) and \( y = \frac{1}{3}e^{-t} \) involve exponential functions with base \( e \), which is an important mathematical constant approximately equal to 2.718.
In the exercise, the exponential function describes exponential growth for \( x \) and exponential decay for \( y \).
Growth is modeled by \( 2e^t \), which signifies that x increases rapidly as t increases. On the other hand, \( \frac{1}{3}e^{-t} \) models decay, meaning y decreases as t increases. These equations showcase how exponential functions can model different types of change, explaining not just static properties but dynamic behaviors in mathematical phenomena.