Problem 24 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

The parametric equations for the tangent line are \(x = 1 + t\), \(y = t\), \(z = 0\).

Step 1: Evaluate the curve at the given point

First, we need to find the value of the parameter \( t \) at which the point \((1, 0, 0)\) lies on the curve. The parametric equations are \(x = e^t\), \(y = te^t\), and \(z = te^{t^2}\). Set \(x = e^t = 1\). Since the exponential function is 1 when the exponent is 0, \(t = 0\). Evaluate the other coordinates: \(y = 0 \, e^0 = 0\) and \(z = 0 \, e^{0^2} = 0\). Thus, \(t=0\) is the parameter value at the given point \((1, 0, 0)\).

Step 2: Find the derivatives of the parametric equations

Next, find the derivatives \( \frac{dx}{dt} \), \( \frac{dy}{dt} \), and \( \frac{dz}{dt} \) which give the direction vector of the tangent line. For \(x=e^t\), \( \frac{dx}{dt} = e^t \). For \(y=te^t\), apply the product rule: \( \frac{dy}{dt} = e^t + te^t = e^t(1+t) \). For \(z=te^{t^2}\), use the chain rule and product rule: \( \frac{dz}{dt} = e^{t^2}(2t^2 + 1) \).

Step 3: Evaluate the derivatives at \(t=0\)

Substitute \(t=0\) into the derivatives to find the tangent direction vector at the point. \( \frac{dx}{dt} \bigg|_{t=0} = e^0 = 1 \), \( \frac{dy}{dt} \bigg|_{t=0} = e^0(1+0) = 1 \), and \( \frac{dz}{dt} \bigg|_{t=0} = 0 \). Thus, the tangent direction vector is \(\langle 1, 1, 0 \rangle\).

Step 4: Write the parametric equations for the tangent line

The general form for the parametric equations of a line through a point \((x_0, y_0, z_0)\) with direction vector \(\langle a, b, c \rangle\) is \(x = x_0 + at\), \(y = y_0 + bt\), and \(z = z_0 + ct\). Use the point \((1, 0, 0)\) and direction vector \(\langle 1, 1, 0 \rangle\): \(x = 1 + t\), \(y = 0 + t\), \(z = 0 + 0t\). Thus, the parametric equations for the tangent line are \(x = 1 + t\), \(y = t\), \(z = 0\).

Key Concepts

Tangent Line

A tangent line is a straight line that touches a curve at only one point and has the same direction as the curve at that point. This means it represents the instantaneous direction in which the curve is heading. Think of it like a snapshot of the curve’s path at a single point.
When working with parametric equations, finding a tangent line involves several steps.

• First, you identify the parameter value corresponding to the given point on the curve. This tells us where on the curve we should focus.


• Then, by calculating the derivatives of each parametric equation, you determine the rate of change or the direction in each dimension.


• Finally, the values of these derivatives at the specified parameter yield the direction vector of the tangent line.

In our problem, the tangent line at point \(1, 0, 0\) represents the curve’s direction precisely at that spot, described by the direction vector \langle 1, 1, 0 \rangle\. Make sure the direction vector calculations are accurate, as they pivotally shape the tangent line's direction.

Calculus

Calculus is a branch of mathematics that studies how things change. It's all about finding connections between changing quantities and can be divided into differential and integral calculus.
Differential calculus comes into play when we're finding tangent lines, as it deals with derivatives. A derivative represents the rate at which a quantity changes. In our exercise, the derivatives of the parametric equations give us the gradient, or the direction of the tangent line.
Using calculus to solve problems requires understanding:

• Derivatives: These provide slopes of tangent lines and can be understood as the velocity of changing quantities over time.


• Product and Chain Rules: These are essential when differentiating more complex equations, as seen with calculations like \( y = te^t \) and \( z = te^{t^2} \).

Calculus also helps predict future motion and structural understanding of functions, allowing us to find tangent lines with precision. Whether dealing with parametric or simple equations, calculus is the tool that brings clarity to how changes at specific points relate to the overall function.

Direction Vector

A direction vector indicates the direction of a line or a curve at a certain point. It acts as a guide, showing us how the line will move if extended infinitely.
In three-dimensional space, direction vectors are written as \langle a, b, c \rangle\. Each component corresponds to movement in the x, y, and z directions, respectively. In our case, direction vector \langle 1, 1, 0 \rangle\ represents a movement of:

• 1 unit in the x-direction


• 1 unit in the y-direction


• 0 units in the z-direction.

This means any point on our tangent line moves one unit along with both the x and y axes with no movement along the z-axis. The vector helps translate the mathematical derivative into a physical path for visualization and understanding which direction the line tangent to the curve at a given point will go. To visualize, imagine plotting the line starting from point \(1, 0, 0\) in the notebook, and extending it along these directions.