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Chapter 14: Problem 28

Find a vector equation of the line tangent to the graph of \(\mathbf{r}(t)\) at the point \(P_{0}\) on the curve. $$\mathbf{r}(t)=4 \cos t \mathbf{i}-3 t \mathbf{j}: P_{0}(2,-\pi)$$

Short Answer

Expert verified
The tangent line is \( \mathbf{l}(t) = (2 - 2\sqrt{3}t)\mathbf{i} + (-\pi - 3t)\mathbf{j} \).

Step by step solution

01

Understand the Problem Statement

The goal is to find the vector equation of the line tangent to the graph of the vector function \( \mathbf{r}(t) = 4 \cos t \mathbf{i} - 3t \mathbf{j} \) at the point \( P_{0}(2, -\pi) \). The tangent line at a point is defined by the point and the direction of the derivative of the vector function at that point.
02

Match the Point with the Parameter \( t \)

Find the parameter \( t \) such that the position vector \( \mathbf{r}(t) \) corresponds to the point \( P_{0}(2, -\pi) \). This is done by setting up equations from the components of \( \mathbf{r}(t) \).\[ 4 \cos t = 2 \quad \text{and} \quad -3t = -\pi \]By solving these, \( t = \frac{\pi}{3}.\)
03

Differentiate the Vector Function

Find the derivative of \( \mathbf{r}(t) \), which gives the tangent (direction) vector. Differentiate each component:\[ \frac{d}{dt}(4 \cos t) = -4 \sin t \quad \text{and} \quad \frac{d}{dt}(-3t) = -3 \].So, \( \mathbf{r}'(t) = -4 \sin t \mathbf{i} - 3 \mathbf{j} \).
04

Evaluate the Derivative at \( t = \frac{\pi}{3} \)

Substitute \( t = \frac{\pi}{3} \) into \( \mathbf{r}'(t) \) to find the tangent vector at \( P_0 \):\[ \mathbf{r}'(\frac{\pi}{3}) = -4 \sin(\frac{\pi}{3}) \mathbf{i} - 3 \mathbf{j} = -4 \cdot \frac{\sqrt{3}}{2} \mathbf{i} - 3 \mathbf{j} = -2\sqrt{3} \mathbf{i} - 3 \mathbf{j} \].
05

Write the Tangent Line Equation

Use the point \( P_0 \) and the direction vector \( \mathbf{r}'(\frac{\pi}{3}) \) to form the vector equation of the tangent line:\[ \mathbf{l}(t) = (2, -\pi) + t(-2\sqrt{3}, -3) \].This simplifies to:\[ \mathbf{l}(t) = (2 - 2\sqrt{3}t)\mathbf{i} + (-\pi - 3t)\mathbf{j} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Tangent Line
A tangent line "touches" a curve at just one point without crossing it. It's like the straight path a curve would follow if it stopped curving for a moment. The tangent line to a graph of a vector function at a specific point also has the same direction as the curve at that point.

So, to find the equation of a tangent line, we need two things:
  • The point of tangency, which we get from the curve itself.
  • The direction of the line, which we find by calculating the derivative of the vector function.
In this exercise, the point of tangency is given as \( P_{0}(2, -\pi) \). The direction is found from the derivative of the vector function.
Vector Function
Vector functions are functions where the outputs are vectors. Imagine you throw a ball into the air; at any moment, its position can be expressed as a vector from the ground's origin to its current location. That's like what a vector function does. Here, it's expressed as \( \mathbf{r}(t) = 4 \cos t \mathbf{i} - 3t \mathbf{j} \).

The components \( 4 \cos t \mathbf{i} \) and \(-3t \mathbf{j} \) are the building blocks. The "\( \mathbf{i} \)" and "\( \mathbf{j} \)" represent unit vectors in the x and y directions. This essentially tells you how much you move along each axis as \( t \) changes. The function maps a single number \( t \) to a 2D position (vector).
Derivative of Vector Function
The derivative of a vector function gives us the rate at which the vector's components change with respect to \( t \). It helps in finding how fast and in which direction you're moving along the curve.

For the vector function \( \mathbf{r}(t) = 4 \cos t \mathbf{i} - 3t \mathbf{j} \), calculating derivatives component-wise involves:
  • Differentiate the x-component: \( \frac{d}{dt}(4 \cos t) = -4 \sin t \).
  • Differentiate the y-component: \( \frac{d}{dt}(-3t) = -3 \).
Thus, the derivative or tangent vector is \( \mathbf{r}'(t) = -4 \sin t \mathbf{i} - 3 \mathbf{j} \). It points in the direction the curve is heading at any point \( t \).
Parameterization
Parameterization refers to expressing a line, curve, or surface using parameters. It's like giving every point on the curve a specific "time stamp" with a parameter value. In the case of vector functions, \( t \) often takes this role.

The goal is to find which value of \( t \) corresponds to the given point \( P_0 \). We match the components of the position vector \( \mathbf{r}(t) = 4 \cos t \mathbf{i} - 3t \mathbf{j} \) to the coordinates \( (2, -\pi) \), resulting in two equations:
  • \( 4 \cos t = 2 \)
  • \( -3t = -\pi \)
Solving these equations gives \( t = \frac{\pi}{3} \). This tells us exactly "where" on the curve the point \( P_0 \) is located in terms of the parameter \( t \).

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Most popular questions from this chapter

Solve the vector initial-value problem for \(y(t)\) by integrating and using the initial conditions to find the constants of integration. $$y^{\prime}(t)=\cos t \mathbf{i}+\sin t \mathbf{j}, \quad y(0)=\mathbf{i}-\mathbf{j}$$

Evaluate the indefinite integral. $$\int\left(t^{2} \mathbf{i}-2 t \mathbf{j}+\frac{1}{t} \mathbf{k}\right) d t$$

Use a graphing utility to generate the graph of \(\mathbf{r}(t)\) and the graph of the tangent line at \(t_{0}\) on the same screen. $$\mathbf{r}(t)=\sin \pi t \mathbf{i}+t^{2} \mathbf{j}: t_{0}=\frac{1}{2}$$

Find the displacement and the distance traveled over the indicated time interval. $$\mathbf{r}=(1-3 \sin t) \mathbf{i}+3 \cos t \mathbf{j} ; 0 \leq t \leq 3 \pi / 2$$

Find the velocity, speed, and acceleration at the given time \(t\) of a particle moving along the given curve. $$\mathbf{r}(t)=e^{t} \sin t \mathbf{i}+e^{t} \cos t \mathbf{j}+t \mathbf{k} ; t=\pi / 2$$
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