Problem 23 - Calculus Early Transcendentals

Question

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. $$ x=1+2 \sqrt{t}, \quad y=t^{3}-t, z=t^{3}+t ; \quad(3,0,2) $$

Step-by-Step Solution

Answer

Parametric equations: \(x=3+t\), \(y=2t\), \(z=2+4t\).

Step 1: Identify the point on the curve

First, find the value of the parameter \( t \) at which the curve passes through the point \((3, 0, 2)\). Substitute \( x = 3 \) into the equation \( x = 1 + 2 \sqrt{t} \). Solve \( 3 = 1 + 2 \sqrt{t} \) for \( t \).

Step 2: Solve for t

From \( 3 = 1 + 2\sqrt{t} \), we get \( 2\sqrt{t} = 2 \). Thus, dividing both sides by 2, we have \( \sqrt{t} = 1 \). Squaring both sides, we obtain \( t = 1 \).

Step 3: Verify the parameter value

Substitute \( t = 1 \) into the other two parametric equations: \( y = t^3 - t = 1^3 - 1 = 0 \) and \( z = t^3 + t = 1^3 + 1 = 2 \). Since the calculated coordinates match the given point \( (3, 0, 2) \), \( t = 1 \) is correct.

Step 4: Find derivative vectors

To find the tangent line, compute the derivatives \( \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \) at \( t = 1 \).

Step 5: Compute derivatives at t=1

The derivative \( \frac{dx}{dt} = \frac{d}{dt}(1 + 2\sqrt{t}) = \frac{d}{dt}(2t^{1/2}) = t^{-1/2} \). At \( t = 1 \), \( \frac{dx}{dt} = 1 \). \( \frac{dy}{dt} = \frac{d}{dt}(t^3 - t) = 3t^2 - 1 \). At \( t = 1 \), \( \frac{dy}{dt} = 2 \). \( \frac{dz}{dt} = \frac{d}{dt}(t^3 + t) = 3t^2 + 1 \). At \( t = 1 \), \( \frac{dz}{dt} = 4 \).

Step 6: Write the parametric equations of the tangent line

The tangent line at \( t = 1 \) is parallel to the vector \( (1, 2, 4) \). Given a point \((3, 0, 2)\) and the direction vector \((1, 2, 4)\), the parametric equations are: \(x = 3 + 1t\), \(y = 0 + 2t\), \(z = 2 + 4t\).

Key Concepts

Tangent Line

A tangent line to a curve is like a straight path that just kisses the curve at a specific point. It's the line that best represents the instantaneous direction of the curve at that point. In our exercise, we're looking for a tangent line to a curve described by parametric equations.
Think of the tangent line as your best guess at where the curve is heading next. It gives us important information about the behavior of the curve near the point of tangency. To find this line, we need to know exactly where on the curve we're interested in, which is why we solve for the parameter that gives us the correct coordinates of the point.

• Key point: A tangent line touches a curve at exactly one point and has the same slope at that point.
• Why it matters: Finding the tangent helps in understanding the curve's direction and allows for approximations in the local area around the point.

Calculus

Calculus is the branch of mathematics that deals with change and motion. It's like the mathematics of movement, allowing us to find how things change over time or space. In this exercise, we're using calculus to find derivatives, which tell us about the rate of change along the curve.
Calculus splits into two major ideas:

• Differential Calculus: This involves finding derivatives, which help us understand curves and slopes.
• Integral Calculus: Focuses on the accumulation of quantities and areas under curves, but isn't our focus here.

In the context of our problem, differential calculus is used to derive the tangent line's direction by finding the slope of our curve at a particular point.
Understanding calculus is crucial because it forms the foundation of how we model and predict changes in countless aspects of the world, from the trajectories of planets to the flow of electricity.

Derivatives

Derivatives are essential tools in calculus that measure how a function changes as its input changes. In simpler terms, they are like speedometers for curves, telling us how fast something is changing at any given point.
For our parametric equations, we find the derivative of each coordinate separately with respect to the parameter, here noted as \( t \). Derivatives help us get the direction or slope of the tangent line at the specified point:

• \( \frac{dx}{dt} \): Measures the rate of change of \( x \) with respect to \( t \).
• \( \frac{dy}{dt} \): Measures the rate of change of \( y \) with respect to \( t \).
• \( \frac{dz}{dt} \): Measures the rate of change of \( z \) with respect to \( t \).

Computing these at \( t = 1 \) gives us the components of the direction vector for our tangent line. Essentially, derivatives provide us with the ingredients to "cook up" the tangent line.

Vector Calculus

Vector calculus extends regular calculus into the realm of vectors—quantities with both magnitude and direction. This is especially useful when working with curves in three-dimensional space, like in our exercise.
In this problem, we're using vector calculus to manage the parametric equations:

• The curve is defined in terms of vectors: \( \langle x, y, z \rangle = \langle 1 + 2 \sqrt{t}, t^3 - t, t^3 + t \rangle \).
• The tangent line's direction is found using a vector derivative, which combines each derivative \( \langle \frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \rangle \).
• The final tangent line is also a vector expression: \( x = 3 + 1t, \) \( y = 0 + 2t, \) \( z = 2 + 4t \).

This approach simplifies the process by allowing us to work in a coordinate-free setting where we can easily manage the directions and lengths of these vector components. Vector calculus is a powerful tool, enabling us to solve complex problems in physics and engineering efficiently.