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Chapter 13: Problem 41

Find the velocity \(v\) and the tangent vector \(T\). Then compute the rate of change \(d f / d t=\operatorname{grad} f \cdot \mathbf{v}\) and the slope \(d f / d s=\operatorname{grad} f \cdot \mathbf{T}\). \(f=x^{2}+y^{2} \quad x=t \quad y=t^{2}\)

Short Answer

Expert verified
Velocity \(\mathbf{v} = \langle 1, 2t \rangle\) and tangent \(\mathbf{T} = \langle \frac{1}{\sqrt{1 + 4t^2}}, \frac{2t}{\sqrt{1 + 4t^2}} \rangle\). Rate \(\frac{df}{dt} = 2t + 4t^3\), Slope \(\frac{df}{ds} = \frac{2t + 4t^3}{\sqrt{1 + 4t^2}}\).

Step by step solution

01

Express f in terms of t

Given that \(f = x^2 + y^2\), substitute \(x = t\) and \(y = t^2\). Thus, \(f = t^2 + (t^2)^2 = t^2 + t^4\).
02

Calculate the partial derivatives of f

First, find the gradient \(\operatorname{grad} f = \left( 2x, 2y \right)\) in terms of \(t\). This becomes \(\operatorname{grad} f = \left( 2t, 2t^2 \right)\).
03

Find the velocity vector \(v\)

The velocity vector \(\mathbf{v}\) is given by the derivative of the position vector \(\mathbf{r}(t) = \langle x, y \rangle = \langle t, t^2 \rangle\). So, \(\mathbf{v} = \frac{d}{dt} \langle t, t^2 \rangle = \langle 1, 2t \rangle\).
04

Compute the rate of change \( \frac{df}{dt} \)

Calculate \(\frac{df}{dt} = \operatorname{grad} f \cdot \mathbf{v} = \langle 2t, 2t^2 \rangle \cdot \langle 1, 2t \rangle = 2t + 4t^3\).
05

Find the tangent vector \(T\)

The tangent vector \(\mathbf{T}\) is the unit vector in the direction of \(\mathbf{v}\). Thus, \(\mathbf{T} = \frac{\mathbf{v}}{\|\mathbf{v}\|}\). First, find \(\|\mathbf{v}\| = \sqrt{1^2 + (2t)^2} = \sqrt{1 + 4t^2}\). Hence, \(\mathbf{T} = \left\langle \frac{1}{\sqrt{1 + 4t^2}}, \frac{2t}{\sqrt{1 + 4t^2}} \right\rangle\).
06

Compute the slope \( \frac{df}{ds} \)

The slope \(\frac{df}{ds} = \operatorname{grad} f \cdot \mathbf{T} = \langle 2t, 2t^2 \rangle \cdot \left\langle \frac{1}{\sqrt{1 + 4t^2}}, \frac{2t}{\sqrt{1 + 4t^2}} \right\rangle = \frac{2t}{\sqrt{1 + 4t^2}} + \frac{4t^3}{\sqrt{1 + 4t^2}} = \frac{2t + 4t^3}{\sqrt{1 + 4t^2}}\).

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Key Concepts

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

Velocity
In calculus, velocity is a critical vector that represents the rate of change of an object's position with respect to time. For any parametric curve defined by a position vector \( \mathbf{r}(t) = \langle x(t), y(t) \rangle \), the velocity vector \( \mathbf{v} \) is obtained by differentiating \( \mathbf{r} \) with respect to \( t \). This gives \( \mathbf{v} = \frac{d}{dt} \langle x(t), y(t) \rangle \).
In our exercise, the position vector is \( \langle t, t^2 \rangle \), so the velocity vector \( \mathbf{v} \) is the derivative, \( \langle 1, 2t \rangle \).
  • The first component represents change in the \( x \) direction.
  • The second component handles changes in the \( y \) direction.
Tangent Vector
The tangent vector is an important concept when understanding how curves behave at any point. It represents the direction and "speed" of the curve as it passes through a particular point.
To obtain a tangent vector \( \mathbf{T} \), you normalize the velocity vector \( \mathbf{v} \), giving a unit vector in the same direction.
Normalization involves dividing \( \mathbf{v} \) by its magnitude \( \| \mathbf{v} \| \).For our case, the tangent vector \( \mathbf{T} \) is:
  • \( \| \mathbf{v} \| = \sqrt{1 + 4t^2} \),
  • \( \mathbf{T} = \left\langle \frac{1}{\sqrt{1+4t^2}}, \frac{2t}{\sqrt{1+4t^2}} \right\rangle \).
Rate of Change
In the context of multivariable calculus, the rate of change \( \frac{df}{dt} \) describes how a function \( f \) behaves as its variables change over time.
One powerful tool is using the dot product of the gradient \( \operatorname{grad} f \) and the velocity vector \( \mathbf{v} \) to calculate this rate.
For our function \( f(x, y) = x^2 + y^2 \), with paths \( x = t \) and \( y = t^2 \), the derivative becomes:
  • \( \operatorname{grad} f = \langle 2t, 2t^2 \rangle \)
  • \( \frac{df}{dt} = \operatorname{grad} f \cdot \mathbf{v} = 2t + 4t^3 \)
This results in a comprehensive view of function changes over time.
Gradient
A gradient is a vector field pointing in the direction of the steepest ascent of a function. It indicates how a function changes most rapidly and is visualized as a multi-dimensional generalization of a derivative.
For a function \( f(x, y) = x^2 + y^2 \), the gradient is \( \operatorname{grad} f = \langle 2x, 2y \rangle \) but expressed in terms of \( t \): \( \operatorname{grad} f = \langle 2t, 2t^2 \rangle \).
  • Provides crucial information about the rate and direction of change.
  • Central to calculating both the rate of change and slope in our problem.
Slope
In calculus, a slope refers to how steep a line or a curve is at a particular point. The slope of a function along a curve can be found by taking the dot product of the gradient and a unit tangent vector, giving \( \frac{df}{ds} \).For our function:
  • Use the tangent vector \( \mathbf{T} = \left\langle \frac{1}{\sqrt{1+4t^2}}, \frac{2t}{\sqrt{1+4t^2}} \right\rangle \).
  • Calculate the slope: \( \frac{df}{ds} = \frac{2t + 4t^3}{\sqrt{1 + 4t^2}} \).
  • This value tells us how sharply the function changes at any point \( t \).

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Most popular questions from this chapter

Write down the level direction \(\mathbf{u}\) for \(\theta=\tan ^{-1}(y / x)\) at the point (3,4) . Then compute \(\operatorname{grad} \theta\) and check \(D_{\bullet} \theta=0\).

If \(f_{x x}+f_{y y}=0,\) show that \(f(x, y)\) cannot have an interior maximum or minimum (only saddle points).

(a) Find \(d f / d s\) and \(d f / d t\) for the roller-coaster \(f=x y\) along the path \(x=\cos 2 t, y=\sin 2 t\) (b) Change to \(f=x^{2}+y^{2}\) and explain why the slope is zero.

Find grad \(f=\left(f_{x}, f_{y}, f_{z}\right)\) for the functions from physics. \(1 / \sqrt{(x-1)^{2}+y^{2}+z^{2}}-1 / \sqrt{(x+1)^{2}+y^{2}+z^{2}}\) (dipole)

True or false: (a) If we know grad \(f,\) we know \(f\). (b) The line \(x=y=-z\) is perpendicular to the plane \(z=\) \(x+y\) (c) The gradient of \(z=x+y\) lies along that line.
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