y 2x 57. Let T(x, y) = be the temperature at points (x, y) in some region on the plane, and let r(t) = (g(t), t² + 2), where g(t) is a differentiable function such that g(2) = 1 and g'(2) = 3, parametrize a path in the plane. Let DT, denote the rate at which the temperature T changes per unit of distance traveled at position (x, y) = (1,6) in the direction of the vector v = (1,-1). Let DT, denote the rate at which the temperature a bug at position r(t) at time t experiences per unit change in time at t = 2. If possible, find DT, and DT₁.
y 2x 57. Let T(x, y) = be the temperature at points (x, y) in some region on the plane, and let r(t) = (g(t), t² + 2), where g(t) is a differentiable function such that g(2) = 1 and g'(2) = 3, parametrize a path in the plane. Let DT, denote the rate at which the temperature T changes per unit of distance traveled at position (x, y) = (1,6) in the direction of the vector v = (1,-1). Let DT, denote the rate at which the temperature a bug at position r(t) at time t experiences per unit change in time at t = 2. If possible, find DT, and DT₁.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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57 please
![### Calculus Problem: Temperature Rate of Change
**Problem Statement:**
57. Let \( T(x, y) = \frac{y}{2x} \) be the temperature at points \((x, y)\) in some region on the plane, and let \( \mathbf{r}(t) = \langle g(t), t^2 + 2 \rangle \), where \( g(t) \) is a differentiable function such that \( g(2) = 1 \) and \( g'(2) = 3 \), parametrize a path in the plane. Let \( DT_v \) denote the rate at which the temperature T changes per unit of distance traveled at position \((x, y) = (1, 6)\) in the direction of the vector \( \mathbf{v} = \langle 1, -1 \rangle \). Let \( DT_t \) denote the rate at which the temperature a bug at position \( \mathbf{r}(t) \) at time \( t \) experiences per unit change in time at \( t = 2 \). If possible, find \( DT_v \) and \( DT_t \).
**Solution:**
To find \( DT_v \) and \( DT_t \), we follow these steps:
1. **Calculate the Gradient of T:**
\[
\nabla T = \left( \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y} \right)
\]
Given \( T(x, y) = \frac{y}{2x} \):
\[
\frac{\partial T}{\partial x} = -\frac{y}{2x^2}, \quad \frac{\partial T}{\partial y} = \frac{1}{2x}
\]
Therefore,
\[
\nabla T = \left( -\frac{y}{2x^2}, \frac{1}{2x} \right)
\]
2. **Evaluate the Gradient at \((1, 6)\):**
\[
\nabla T (1, 6) = \left( -\frac{6}{2(1)^2}, \frac{1}{2(1)} \right) = \left( -3, \frac{1}{2}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdb24b717-8ee5-4ca9-8061-74ddb7e91c1a%2F0964efa0-5dab-43d3-873c-1f35b250de5d%2Fl0oxmxf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Calculus Problem: Temperature Rate of Change
**Problem Statement:**
57. Let \( T(x, y) = \frac{y}{2x} \) be the temperature at points \((x, y)\) in some region on the plane, and let \( \mathbf{r}(t) = \langle g(t), t^2 + 2 \rangle \), where \( g(t) \) is a differentiable function such that \( g(2) = 1 \) and \( g'(2) = 3 \), parametrize a path in the plane. Let \( DT_v \) denote the rate at which the temperature T changes per unit of distance traveled at position \((x, y) = (1, 6)\) in the direction of the vector \( \mathbf{v} = \langle 1, -1 \rangle \). Let \( DT_t \) denote the rate at which the temperature a bug at position \( \mathbf{r}(t) \) at time \( t \) experiences per unit change in time at \( t = 2 \). If possible, find \( DT_v \) and \( DT_t \).
**Solution:**
To find \( DT_v \) and \( DT_t \), we follow these steps:
1. **Calculate the Gradient of T:**
\[
\nabla T = \left( \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y} \right)
\]
Given \( T(x, y) = \frac{y}{2x} \):
\[
\frac{\partial T}{\partial x} = -\frac{y}{2x^2}, \quad \frac{\partial T}{\partial y} = \frac{1}{2x}
\]
Therefore,
\[
\nabla T = \left( -\frac{y}{2x^2}, \frac{1}{2x} \right)
\]
2. **Evaluate the Gradient at \((1, 6)\):**
\[
\nabla T (1, 6) = \left( -\frac{6}{2(1)^2}, \frac{1}{2(1)} \right) = \left( -3, \frac{1}{2}
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