/*! 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 2 If a curve \(C\) is given by \(\... [FREE SOLUTION] | 91影视

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If a curve \(C\) is given by \(\mathbf{r}(t)=\left\langle t, t^{2}\right\rangle,\) what is \(\left|\mathbf{r}^{\prime}(t)\right| ?\)

Short Answer

Expert verified
Answer: The magnitude of the derivative of the vector function 饾惈(饾憽) = 鉄潙, 饾憽虏鉄 is given by: |饾惈'(饾憽)| = 鈭(1 + 4t虏).

Step by step solution

01

Compute 饾惈'(饾憽)

First, we need to calculate the derivative of the vector function 饾惈(饾憽) with respect to 饾憽: \(\mathbf{r}(t) = \left\langle t, t^{2}\right\rangle\) To find the derivative, we take the derivative of each component of the vector with respect to 饾憽: \(\mathbf{r}^{\prime}(t) = \frac{d}{dt}\left\langle t, t^{2}\right\rangle = \left\langle \frac{d}{dt}t, \frac{d}{dt}t^{2}\right\rangle\) We then calculate the derivatives: - \((\frac{d}{dt}t)= 1\) - \((\frac{d}{dt}t^{2})= 2t\) So the derivative of 饾惈(饾憽) is: \(\mathbf{r}^{\prime}(t) = \left\langle 1, 2t \right\rangle\)
02

Compute the magnitude of 饾惈'(饾憽)

Now we need to calculate the magnitude of 饾惈'(饾憽). We will use the formula for the magnitude of a vector: \(|\mathbf{r}^{\prime}(t)| = \sqrt{(1)^{2} + (2t)^{2}}\) Simplify the expression: \(|\mathbf{r}^{\prime}(t)| = \sqrt{1 + 4t^{2}}\) So the magnitude of 饾惈'(饾憽) is: \(|\mathbf{r}^{\prime}(t)| = \sqrt{1 + 4t^{2}}\)

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Key Concepts

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

Vector Differentiation
Vector differentiation is the process of finding the derivative of a vector function. It鈥檚 similar to taking the derivative of a scalar function, but it involves multiple components. When dealing with vector calculus, it鈥檚 important to understand that each component of the vector function is differentiated separately with respect to a variable, usually time or a parameter like \( t \).

In the case of the function \( \mathbf{r}(t) = \left\langle t, t^2 \right\rangle \), we differentiate each component:
  • The derivative of \( t \) with respect to \( t \) is \( 1 \).
  • The derivative of \( t^2 \) with respect to \( t \) is \( 2t \).
This results in the vector derivative \( \mathbf{r}'(t) = \left\langle 1, 2t \right\rangle \).

Understanding this concept is crucial, especially when analyzing the motion of an object along a path, as it exposes the rate of change of position over time, often referred to in physics as velocity.
Vector Magnitude
Vector magnitude is a measure of the length or size of a vector. It is an essential concept in vector calculus, helping you quantify the extent of the vector without regard to its direction.

To compute the magnitude of a vector \( \mathbf{v} = \left\langle a, b \right\rangle \), the formula is:
  • \( |\mathbf{v}| = \sqrt{a^2 + b^2} \).
This principle is applied in the exercise to find the magnitude of the derivative \( \mathbf{r}'(t) = \left\langle 1, 2t \right\rangle \).

So, the magnitude \( |\mathbf{r}'(t)| \) is given by:
  • \( |\mathbf{r}'(t)| = \sqrt{1^2 + (2t)^2} = \sqrt{1 + 4t^2} \).
Grasping vector magnitude is vital because it aids in understanding physical quantities like speed, which is the magnitude of velocity.
Parametric Curves
Parametric curves describe a path in the plane or space using parameters. Unlike traditional functions that relate \( y \) directly to \( x \), parametric curves introduce a parameter, often \( t \), helping to trace complex shapes.

The given curve \( \mathbf{r}(t) = \left\langle t, t^2 \right\rangle \) is a parametric curve where each component describes the curve鈥檚 position in terms of \( t \).

Parametric equations are instrumental in representing motion where parameters can denote time, providing a clear depiction of an object's trajectory or pathway. They allow easy computation of velocity, acceleration, and other derivative aspects related to the motion.

Learning to work with parametric curves enhances your ability to describe dynamic systems and geometric figures beyond the constraints of simple functional relationships.

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Most popular questions from this chapter

Heat flux The heat flow vector field for conducting objects is \(\mathbf{F}=-k \nabla T,\) where \(T(x, y, z)\) is the temperature in the object and \(k > 0\) is a constant that depends on the material. Compute the outward flux of \(\mathbf{F}\) across the following surfaces S for the given temperature distributions. Assume \(k=1\) \(T(x, y, z)=100 e^{-x-y} ; S\) consists of the faces of the cube \(|x| \leq 1\) \(|y| \leq 1,|z| \leq 1\)

Heat flux The heat flow vector field for conducting objects is \(\mathbf{F}=-k \nabla T,\) where \(T(x, y, z)\) is the temperature in the object and \(k > 0\) is a constant that depends on the material. Compute the outward flux of \(\mathbf{F}\) across the following surfaces S for the given temperature distributions. Assume \(k=1\) $$T(x, y, z)=100 e^{-x^{2}-y^{2}-z^{2}} ; S \text { is the sphere } x^{2}+y^{2}+z^{2}=a^{2}$$

A scalar-valued function \(\varphi\) is harmonic on a region \(D\) if \(\nabla^{2} \varphi=\nabla \cdot \nabla \varphi=0\) at all points of \(D\). Show that the potential function \(\varphi(x, y, z)=|\mathbf{r}|^{-p}\) is harmonic provided \(p=0\) or \(p=1,\) where \(\mathbf{r}=\langle x, y, z\rangle .\) To what vector fields do these potentials correspond?

Green's Theorem as a Fundamental Theorem of Calculus Show that if the circulation form of Green's Theorem is applied to the vector field \(\left\langle 0, \frac{f(x)}{c}\right\rangle\), where \(c>0\) and \(R=\\{(x, y): a \leq x \leq b, 0 \leq y \leq c\\},\) then the result is the Fundamental Theorem of Calculus, $$ \int_{a}^{b} \frac{d f}{d x} d x=f(b)-f(a) $$

Streamlines and equipotential lines Assume that on \(\mathbb{R}^{2}\), the vector field \(\mathbf{F}=\langle f, g\rangle\) has a potential function \(\varphi\) such that \(f=\varphi_{x}\) and \(g=\varphi_{y},\) and it has a stream function \(\psi\) such that \(f=\psi_{y}\) and \(g=-\psi_{x}\). Show that the equipotential curves (level curves of \(\varphi\) ) and the streamlines (level curves of \(\psi\) ) are everywhere orthogonal.

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