Find dz dt · at (x, y) = (−2, −1) for z² = x² + y² if de = dt = help (numbers) 10, dt 5, and x > 0.

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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**Problem: Calculating the Derivative \(\frac{dz}{dt}\)**

Given the function \(z^2 = x^2 + y^2\), calculate \(\frac{dz}{dt}\) at the point \((x, y) = (-2, -1)\). The rates of change are given as \(\frac{dx}{dt} = 10\) and \(\frac{dy}{dt} = 5\), with the condition \(z > 0\).

### Step-by-Step Solution:

1. **Differentiate with respect to \(t\):**

   Start with the equation:
   \[
   z^2 = x^2 + y^2
   \]

   Differentiate both sides with respect to \(t\):
   \[
   \frac{d}{dt}(z^2) = \frac{d}{dt}(x^2 + y^2)
   \]

2. **Apply the chain rule:**

   The chain rule states:
   \[
   2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt}
   \]

3. **Substitute the given values:**

   - At point \((x, y) = (-2, -1)\), substitute \(x = -2\) and \(y = -1\).
   - Also use \(\frac{dx}{dt} = 10\) and \(\frac{dy}{dt} = 5\).

   \[
   2z \frac{dz}{dt} = 2(-2)(10) + 2(-1)(5)
   \]

   Simplify:
   \[
   2z \frac{dz}{dt} = -40 - 10 = -50
   \]

4. **Solve for \(\frac{dz}{dt}\):**

   If \(z > 0\), solve for \(\frac{dz}{dt}\):
   \[
   \frac{dz}{dt} = \frac{-50}{2z}
   \]

5. **Find \(z\) using the equation \(z^2 = x^2 + y^2\):**

   \[
   z^2 = (-2)^2 + (-1)^2 = 4 + 1 = 5
Transcribed Image Text:**Problem: Calculating the Derivative \(\frac{dz}{dt}\)** Given the function \(z^2 = x^2 + y^2\), calculate \(\frac{dz}{dt}\) at the point \((x, y) = (-2, -1)\). The rates of change are given as \(\frac{dx}{dt} = 10\) and \(\frac{dy}{dt} = 5\), with the condition \(z > 0\). ### Step-by-Step Solution: 1. **Differentiate with respect to \(t\):** Start with the equation: \[ z^2 = x^2 + y^2 \] Differentiate both sides with respect to \(t\): \[ \frac{d}{dt}(z^2) = \frac{d}{dt}(x^2 + y^2) \] 2. **Apply the chain rule:** The chain rule states: \[ 2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \] 3. **Substitute the given values:** - At point \((x, y) = (-2, -1)\), substitute \(x = -2\) and \(y = -1\). - Also use \(\frac{dx}{dt} = 10\) and \(\frac{dy}{dt} = 5\). \[ 2z \frac{dz}{dt} = 2(-2)(10) + 2(-1)(5) \] Simplify: \[ 2z \frac{dz}{dt} = -40 - 10 = -50 \] 4. **Solve for \(\frac{dz}{dt}\):** If \(z > 0\), solve for \(\frac{dz}{dt}\): \[ \frac{dz}{dt} = \frac{-50}{2z} \] 5. **Find \(z\) using the equation \(z^2 = x^2 + y^2\):** \[ z^2 = (-2)^2 + (-1)^2 = 4 + 1 = 5
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