Chapter 13: Problem 44
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 \(d f / d s=\operatorname{grad} f \cdot \mathbf{T}\). \(f=x y \quad x=t^{2}+1 \quad y=3\)
Short Answer
Expert verified
Velocity: \((2t, 0)\), Tangent: \((\text{sgn}(t), 0)\). Rate \(df/dt = 6t\), slope \(df/ds = 3 \text{sgn}(t)\).
Step by step solution
01
Find the gradient of f
The function given is \(f = xy\). Since \(y = 3\) is constant, the gradient of \(f\) is \(abla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)\). Calculating this for \(f = (t^2 + 1) \cdot 3\), we have \(abla f = (3, t^2 + 1)\), a vector.
02
Determine the velocity vector v
The velocity vector \(\mathbf{v}\) is the derivative of the position vector \(\mathbf{r}(t) = (x(t), y(t))\) with respect to time. We have \(x(t) = t^2 + 1\) and \(y(t) = 3\), so \(\mathbf{r}(t) = (t^2 + 1, 3)\). Differentiating, \(\dot{x}(t) = 2t\) and \(\dot{y}(t) = 0\), thus \(\mathbf{v} = (2t, 0)\).
03
Compute the tangent vector T
The tangent vector \(\mathbf{T}\) is the unit vector in the direction of \(\mathbf{v}\), given by \(\mathbf{T} = \frac{\mathbf{v}}{||\mathbf{v}||}\). The magnitude \(||\mathbf{v}|| = \sqrt{(2t)^2 + 0^2} = 2|t|\). Therefore, \(\mathbf{T} = \left(\frac{2t}{2|t|}, 0\right) = (\text{sgn}(t), 0)\).
04
Calculate the rate of change \(\frac{df}{dt}\)
The rate of change of \(f\) with respect to time is \(\frac{df}{dt} = abla f \cdot \mathbf{v}\). With \(abla f = (3, t^2 + 1)\) and \(\mathbf{v} = (2t, 0)\), we find \(\frac{df}{dt} = 3 \cdot 2t + (t^2+1) \cdot 0 = 6t\).
05
Compute the slope \(\frac{df}{ds}\)
The slope \(\frac{df}{ds}\) is defined as \(abla f \cdot \mathbf{T}\). With \(abla f = (3, t^2 + 1)\) and \(\mathbf{T} = (\text{sgn}(t), 0)\), the slope is \(\frac{df}{ds} = 3 \cdot \text{sgn}(t) + (t^2 + 1) \cdot 0 = 3 \cdot \text{sgn}(t)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gradient
In the realm of multivariable calculus, the gradient is a crucial concept. It consists of partial derivatives of a multivariable function, bundling them into a vector. Consider a function like \( f(x, y) = xy \). By taking the partial derivatives with respect to \( x \) and \( y \), we find the gradient \( abla f \).
The partial derivative \( \frac{\partial f}{\partial x} \) measures how \( f \) changes as \( x \) changes, with \( y \) held constant, and vice versa.
For our exercise, substituting \( y = 3 \), we compute the gradient:
The partial derivative \( \frac{\partial f}{\partial x} \) measures how \( f \) changes as \( x \) changes, with \( y \) held constant, and vice versa.
For our exercise, substituting \( y = 3 \), we compute the gradient:
- \( \frac{\partial f}{\partial x} = 3 \)
- \( \frac{\partial f}{\partial y} = t^2 + 1 \)
Velocity Vector
The velocity vector is a fundamental concept in physics and calculus that represents the rate of change of a position vector with respect to time. This vector shows how an object's position changes over time.
To find the velocity vector \( \mathbf{v} \), differentiate the position vector \( \mathbf{r}(t) = (x(t), y(t)) \) with respect to time.
To find the velocity vector \( \mathbf{v} \), differentiate the position vector \( \mathbf{r}(t) = (x(t), y(t)) \) with respect to time.
- Given \( x(t) = t^2 + 1 \), we find \( \dot{x}(t) = 2t \).
- Since \( y(t) = 3 \) is constant, \( \dot{y}(t) = 0 \).
Tangent Vector
The tangent vector provides the direction and rate at which a curve progresses at a particular point. It is aligned along the path of the curve, showing the trajectory of movement. The tangent vector \( \mathbf{T} \) can be derived from the velocity vector \( \mathbf{v} \) by normalization.
Normalization ensures that \( \mathbf{T} \) is a unit vector, which preserves the direction of \( \mathbf{v} \) but sets its length to 1.
To normalize:
Normalization ensures that \( \mathbf{T} \) is a unit vector, which preserves the direction of \( \mathbf{v} \) but sets its length to 1.
To normalize:
- Calculate the magnitude of \( \mathbf{v} \): \( ||\mathbf{v}|| = \sqrt{(2t)^2 + 0^2} = 2|t| \).
- Normalize \( \mathbf{v} \) to get \( \mathbf{T} = \left(\frac{2t}{2|t|}, 0\right) = (\text{sgn}(t), 0) \).
Rate of Change
The rate of change in calculus is an essential concept that quantifies how a quantity alters over time or along a path. In our context, we're examining two specific rate of changes:
1. **With respect to time (\( \frac{df}{dt} \))**: This reflects how the function \( f \) changes as time progresses. It's calculated as the dot product of the gradient \( abla f \) and the velocity vector \( \mathbf{v} \). For our function, this results in \( \frac{df}{dt} = abla f \cdot \mathbf{v} = 6t \). This signifies that the rate of change of \( f \) over time is proportional to \( t \).
2. **With respect to space (\( \frac{df}{ds} \))**: Here, we look at how \( f \) changes along a specific path, parallel to the tangent vector \( \mathbf{T} \). Again, using a dot product, \( \frac{df}{ds} = abla f \cdot \mathbf{T} = 3 \cdot \text{sgn}(t) \). This provides the slope, which describes \( f \)'s steepness along the curve.
By understanding these rates, one can discern how \( f \) evolves over time and space, providing insight into both motion and change.
1. **With respect to time (\( \frac{df}{dt} \))**: This reflects how the function \( f \) changes as time progresses. It's calculated as the dot product of the gradient \( abla f \) and the velocity vector \( \mathbf{v} \). For our function, this results in \( \frac{df}{dt} = abla f \cdot \mathbf{v} = 6t \). This signifies that the rate of change of \( f \) over time is proportional to \( t \).
2. **With respect to space (\( \frac{df}{ds} \))**: Here, we look at how \( f \) changes along a specific path, parallel to the tangent vector \( \mathbf{T} \). Again, using a dot product, \( \frac{df}{ds} = abla f \cdot \mathbf{T} = 3 \cdot \text{sgn}(t) \). This provides the slope, which describes \( f \)'s steepness along the curve.
By understanding these rates, one can discern how \( f \) evolves over time and space, providing insight into both motion and change.