### Calculating the Derivative Using the Chain Rule Given: \[ f(x, y) = x^2 - 6xy \] \[ \mathbf{r}(t) = \langle \cos(t), \sin(t) \rangle \] #### Task: Use the Chain Rule to calculate \( \frac{d}{dt} f(\mathbf{r}(t)) \) at \( t = \frac{\pi}{2} \). (Use symbolic notation and fractions where needed.) #### Expression: \[ \left. \frac{d}{dt} f(\mathbf{r}(t)) \right|_{t = \frac{\pi}{2}} = \] (Please fill in the appropriate answer in the blank provided space.) --- To solve this problem, one would typically perform the following steps: 1. **Parameterize the function**: Substitute \( \mathbf{r}(t) = \langle \cos(t), \sin(t) \rangle \) into \( f \). So, \( x = \cos(t) \) and \( y = \sin(t) \). Thus, \( f(\cos(t), \sin(t)) = (\cos(t))^2 - 6(\cos(t))(\sin(t)) \). 2. **Find the partial derivatives** of \( f(x, y) \): \[ \frac{\partial f}{\partial x} = 2x - 6y \] \[ \frac{\partial f}{\partial y} = -6x \] 3. **Apply the Chain Rule** for multivariable functions: \[ \frac{d}{dt} f(\mathbf{r}(t)) = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} \] 4. **Evaluate the derivatives** at \( t = \frac{\pi}{2} \): - \( \cos\left(\frac{\pi}{2}\right) = 0 \) - \( \sin\left(\frac{\pi}{2}\right) = 1 \) - \( \frac{d}{dt}\cos(t) = -\sin(t) \) - \( \frac{d}{dt}\sin(t) = \
### Calculating the Derivative Using the Chain Rule Given: \[ f(x, y) = x^2 - 6xy \] \[ \mathbf{r}(t) = \langle \cos(t), \sin(t) \rangle \] #### Task: Use the Chain Rule to calculate \( \frac{d}{dt} f(\mathbf{r}(t)) \) at \( t = \frac{\pi}{2} \). (Use symbolic notation and fractions where needed.) #### Expression: \[ \left. \frac{d}{dt} f(\mathbf{r}(t)) \right|_{t = \frac{\pi}{2}} = \] (Please fill in the appropriate answer in the blank provided space.) --- To solve this problem, one would typically perform the following steps: 1. **Parameterize the function**: Substitute \( \mathbf{r}(t) = \langle \cos(t), \sin(t) \rangle \) into \( f \). So, \( x = \cos(t) \) and \( y = \sin(t) \). Thus, \( f(\cos(t), \sin(t)) = (\cos(t))^2 - 6(\cos(t))(\sin(t)) \). 2. **Find the partial derivatives** of \( f(x, y) \): \[ \frac{\partial f}{\partial x} = 2x - 6y \] \[ \frac{\partial f}{\partial y} = -6x \] 3. **Apply the Chain Rule** for multivariable functions: \[ \frac{d}{dt} f(\mathbf{r}(t)) = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} \] 4. **Evaluate the derivatives** at \( t = \frac{\pi}{2} \): - \( \cos\left(\frac{\pi}{2}\right) = 0 \) - \( \sin\left(\frac{\pi}{2}\right) = 1 \) - \( \frac{d}{dt}\cos(t) = -\sin(t) \) - \( \frac{d}{dt}\sin(t) = \
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Calculating the Derivative Using the Chain Rule
Given:
\[ f(x, y) = x^2 - 6xy \]
\[ \mathbf{r}(t) = \langle \cos(t), \sin(t) \rangle \]
#### Task:
Use the Chain Rule to calculate \( \frac{d}{dt} f(\mathbf{r}(t)) \) at \( t = \frac{\pi}{2} \).
(Use symbolic notation and fractions where needed.)
#### Expression:
\[ \left. \frac{d}{dt} f(\mathbf{r}(t)) \right|_{t = \frac{\pi}{2}} = \]
(Please fill in the appropriate answer in the blank provided space.)
---
To solve this problem, one would typically perform the following steps:
1. **Parameterize the function**: Substitute \( \mathbf{r}(t) = \langle \cos(t), \sin(t) \rangle \) into \( f \).
So, \( x = \cos(t) \) and \( y = \sin(t) \).
Thus, \( f(\cos(t), \sin(t)) = (\cos(t))^2 - 6(\cos(t))(\sin(t)) \).
2. **Find the partial derivatives** of \( f(x, y) \):
\[
\frac{\partial f}{\partial x} = 2x - 6y
\]
\[
\frac{\partial f}{\partial y} = -6x
\]
3. **Apply the Chain Rule** for multivariable functions:
\[
\frac{d}{dt} f(\mathbf{r}(t)) = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt}
\]
4. **Evaluate the derivatives** at \( t = \frac{\pi}{2} \):
- \( \cos\left(\frac{\pi}{2}\right) = 0 \)
- \( \sin\left(\frac{\pi}{2}\right) = 1 \)
- \( \frac{d}{dt}\cos(t) = -\sin(t) \)
- \( \frac{d}{dt}\sin(t) = \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8fe4eea6-b7ab-4471-b433-62327707440a%2F8f6f0a46-4050-4051-9074-7250095a7cef%2F1i8unqe_processed.png&w=3840&q=75)
Transcribed Image Text:### Calculating the Derivative Using the Chain Rule
Given:
\[ f(x, y) = x^2 - 6xy \]
\[ \mathbf{r}(t) = \langle \cos(t), \sin(t) \rangle \]
#### Task:
Use the Chain Rule to calculate \( \frac{d}{dt} f(\mathbf{r}(t)) \) at \( t = \frac{\pi}{2} \).
(Use symbolic notation and fractions where needed.)
#### Expression:
\[ \left. \frac{d}{dt} f(\mathbf{r}(t)) \right|_{t = \frac{\pi}{2}} = \]
(Please fill in the appropriate answer in the blank provided space.)
---
To solve this problem, one would typically perform the following steps:
1. **Parameterize the function**: Substitute \( \mathbf{r}(t) = \langle \cos(t), \sin(t) \rangle \) into \( f \).
So, \( x = \cos(t) \) and \( y = \sin(t) \).
Thus, \( f(\cos(t), \sin(t)) = (\cos(t))^2 - 6(\cos(t))(\sin(t)) \).
2. **Find the partial derivatives** of \( f(x, y) \):
\[
\frac{\partial f}{\partial x} = 2x - 6y
\]
\[
\frac{\partial f}{\partial y} = -6x
\]
3. **Apply the Chain Rule** for multivariable functions:
\[
\frac{d}{dt} f(\mathbf{r}(t)) = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt}
\]
4. **Evaluate the derivatives** at \( t = \frac{\pi}{2} \):
- \( \cos\left(\frac{\pi}{2}\right) = 0 \)
- \( \sin\left(\frac{\pi}{2}\right) = 1 \)
- \( \frac{d}{dt}\cos(t) = -\sin(t) \)
- \( \frac{d}{dt}\sin(t) = \
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