/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 Graph the curve \(C\) that is de... [FREE SOLUTION] | 91Ó°ÊÓ

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Graph the curve \(C\) that is described by \(\mathbf{r}\) and graph \(\mathbf{r}^{\prime}\) at the indicated value of \(\boldsymbol{t}\). $$ \mathbf{r}(t)=t^{3} \mathbf{i}+t^{2} \mathbf{j} ; t=-1 $$

Short Answer

Expert verified
The curve is a parametric plot of \( t^3 \) and \( t^2 \); the tangent vector at \( t = -1 \) is \( 3\mathbf{i} - 2\mathbf{j} \).

Step by step solution

01

Understand the Vector Function

The given vector function is \( \mathbf{r}(t) = t^3 \mathbf{i} + t^2 \mathbf{j} \). This describes a position in space as a function of \( t \). The term \( t^3 \mathbf{i} \) represents the x-coordinate as \( t^3 \), and \( t^2 \mathbf{j} \) represents the y-coordinate as \( t^2 \).
02

Differentiate the Vector Function

To find \( \mathbf{r}'(t) \), differentiate each component of \( \mathbf{r}(t) \) with respect to \( t \):\[ \frac{d}{dt}(t^3 \mathbf{i} + t^2 \mathbf{j}) = 3t^2 \mathbf{i} + 2t \mathbf{j}. \]Thus, \( \mathbf{r}'(t) = 3t^2 \mathbf{i} + 2t \mathbf{j} \).
03

Evaluate the Derivative at the Given \( t \) Value

We need to evaluate \( \mathbf{r}'(t) \) at \( t = -1 \):\[ \mathbf{r}'(-1) = 3(-1)^2 \mathbf{i} + 2(-1) \mathbf{j} = 3 \mathbf{i} - 2 \mathbf{j}. \]
04

Graph the Curve and Vector

Graph the curve described by \( \mathbf{r}(t) = t^3 \mathbf{i} + t^2 \mathbf{j} \) for various values of \( t \) (e.g., \( -2 \leq t \leq 2 \)). Additionally, at the point on the curve where \( t = -1 \), draw the tangent vector \( \mathbf{r}'(-1) = 3 \mathbf{i} - 2 \mathbf{j} \), starting from the point \((-1^3, (-1)^2) = (-1, 1)\).

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Key Concepts

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

Vector Calculus
Vector calculus is a branch of mathematics that deals with vector fields and differential operators. It's an essential tool for describing physical phenomena like fluid flow, electromagnetic fields, or in our case, a curve in the plane. Unlike single-variable calculus, where you deal with derivatives and integrals of functions that return a single value, vector calculus deals with functions that output vectors.
In vector calculus, you often work with vector functions, which are functions of one or more variables that return a vector as an output. For instance, a vector function could describe a path or curve in 2D or 3D space by giving a position vector for each value of the parameter, commonly denoted as \( t \). These functions help to model real-world scenarios where direction and magnitude both matter, like paths of moving objects.
Here's why vector calculus is vital:
  • It allows for the calculation of physical quantities that have both direction and magnitude.
  • It is used for analyzing curves, surfaces, and their properties.
Derivative of Vector Functions
The derivative of a vector function is an extension of the derivative concept from single-variable calculus, applied to vector-valued functions. Here, instead of computing a rate of change of a single quantity, you compute the rate of change of vectors. This is crucial for understanding how a position vector changes and predicting the motion along a path.
Taking the derivative of a vector function involves differentiating each component of the vector individually. For example, if a vector function is given by \( \mathbf{r}(t) = f(t) \mathbf{i} + g(t) \mathbf{j} \), the derivative \( \mathbf{r}'(t) \) would be \( f'(t) \mathbf{i} + g'(t) \mathbf{j} \). Each derivative tells us the rate at which each component of the vector changes with respect to \( t \).
Key points:
  • When you differentiate a vector function, you're finding the vector's instantaneous rate of change.
  • This derivative can also be considered the tangent vector to the curve at a particular time \( t \).
Position Vector
The position vector is a fundamental concept in vector calculus used to specify a point's location in space. It is typically denoted as \( \mathbf{r}(t) \) and defined by the vector originating from the origin of the coordinate system to the point in space. In two-dimensional space, for example, \( \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} \) indicates that the x-coordinate is \( x(t) \) and the y-coordinate is \( y(t) \).
For our problem, the vector function \( \mathbf{r}(t) = t^3 \mathbf{i} + t^2 \mathbf{j} \) describes a path through points in 2D space as \( t \) varies. This vector changes as \( t \) changes, tracing a curve in the plane. When \( t = -1 \), the position vector is \( \mathbf{r}(-1) = -1 \mathbf{i} + 1 \mathbf{j} \), leading to the point (-1, 1) on the plane.
Key concepts about position vectors include:
  • They help in visualizing any object's path in space.
  • Position vectors are core components in defining curves and motion in physics and engineering.
Tangent Vector
A tangent vector provides essential information about the direction and rate of movement along a curve at a specific point. It helps to understand what "path" the curve will follow next when starting from that point. The tangent vector to a curve at any point is essentially the direction the curve is heading, with the magnitude representing the speed.
In our scenario, once we have the derivative \( \mathbf{r}'(t) = 3t^2 \mathbf{i} + 2t \mathbf{j} \), we find the tangent vector by evaluating this derivative at a specific t-value. Evaluating at \( t = -1 \) gives the tangent vector \( \mathbf{r}'(-1) = 3 \mathbf{i} - 2 \mathbf{j} \).
Important notes about tangent vectors:
  • Tangent vectors are crucial for predicting and understanding the motion's direction on any given curve.
  • They allow us to graphically represent the direction of a curve at any point.
  • The concept extends beyond mathematics, being vital in physics and engineering for understanding dynamic systems.

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