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at (x, y) = (1, 2) and z² = x² + y² if. dt = 9 and dy dt = 2.

at (x, y) = (1, 2) and z² = x² + y² if. dt = 9 and dy dt = 2.

Calculus: Early Transcendentals
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 Statement:** Find \(\frac{dz}{dt}\) at \((x, y) = (1, 2)\) and \(z^2 = x^2 + y^2\) if \(\frac{dx}{dt} = 9\) and \(\frac{dy}{dt} = 2\). --- **Solution:** To find \(\frac{dz}{dt}\), we'll use the chain rule for derivatives. Given: - \(z^2 = x^2 + y^2\), Differentiate both sides with respect to \(t\): \[ 2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \] Simplify: \[ z \frac{dz}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt} \] Given values: - \(x = 1\) - \(y = 2\) - \(\frac{dx}{dt} = 9\) - \(\frac{dy}{dt} = 2\) First, find \(z\) using \(z^2 = x^2 + y^2\): \[ z^2 = 1^2 + 2^2 = 1 + 4 = 5 \] Thus, \[ z = \sqrt{5} \] Substitute into the differentiated equation: \[ \sqrt{5} \frac{dz}{dt} = 1 \cdot 9 + 2 \cdot 2 \] \[ \sqrt{5} \frac{dz}{dt} = 9 + 4 \] \[ \sqrt{5} \frac{dz}{dt} = 13 \] Solve for \(\frac{dz}{dt}\): \[ \frac{dz}{dt} = \frac{13}{\sqrt{5}} \] --- \(\frac{dz}{dt} = \frac{13}{\sqrt{5}}\) This is the derivative of \(z\) with respect to \(t\) at the given point and conditions.
Transcribed Image Text:**Problem Statement:** Find \(\frac{dz}{dt}\) at \((x, y) = (1, 2)\) and \(z^2 = x^2 + y^2\) if \(\frac{dx}{dt} = 9\) and \(\frac{dy}{dt} = 2\). --- **Solution:** To find \(\frac{dz}{dt}\), we'll use the chain rule for derivatives. Given: - \(z^2 = x^2 + y^2\), Differentiate both sides with respect to \(t\): \[ 2z \frac{dz}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \] Simplify: \[ z \frac{dz}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt} \] Given values: - \(x = 1\) - \(y = 2\) - \(\frac{dx}{dt} = 9\) - \(\frac{dy}{dt} = 2\) First, find \(z\) using \(z^2 = x^2 + y^2\): \[ z^2 = 1^2 + 2^2 = 1 + 4 = 5 \] Thus, \[ z = \sqrt{5} \] Substitute into the differentiated equation: \[ \sqrt{5} \frac{dz}{dt} = 1 \cdot 9 + 2 \cdot 2 \] \[ \sqrt{5} \frac{dz}{dt} = 9 + 4 \] \[ \sqrt{5} \frac{dz}{dt} = 13 \] Solve for \(\frac{dz}{dt}\): \[ \frac{dz}{dt} = \frac{13}{\sqrt{5}} \] --- \(\frac{dz}{dt} = \frac{13}{\sqrt{5}}\) This is the derivative of \(z\) with respect to \(t\) at the given point and conditions.
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