Q. 3 Given z = x²y, x = 2t+s, and y = 1 - st². dz We have to find dt s=1, t=2. We first find dz dt

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
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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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Step
b)
Q. 3
Given z = x²y, x = 2t+s, and y = 1 – st².
dz
We have to find
dt s=1, t=2.
We first find
dt
By chain rule of partial differentiation, we
dz
dt
=
dz dx
dx dt
+
dz
Now, = 2xy;
dx
From (1), we have
dz
dz dy
dy dt
dz
dt s=1,t=2
dz
dt
at s = 1, t = 2,
Therefore,
dz
dy
=
=(2xy) (2)+(x²) (−2 st)= 4xy – 2x²st
= 4(2t + s) (1 − st²) −2 (2t+s)²
..(1)
dt
dz
dt s=1,t=2
= 2 and
=
: 4(2 × 2 + 1) (1 − 1. 2²) — 2(2 × 2 + 1
= 4(5)(-3)-2(25)
= -60 - 50 = -110
= -110
11
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dy
dt
√x
DO
Transcribed Image Text:10:26 ◄ Search Step b) Q. 3 Given z = x²y, x = 2t+s, and y = 1 – st². dz We have to find dt s=1, t=2. We first find dt By chain rule of partial differentiation, we dz dt = dz dx dx dt + dz Now, = 2xy; dx From (1), we have dz dz dy dy dt dz dt s=1,t=2 dz dt at s = 1, t = 2, Therefore, dz dy = =(2xy) (2)+(x²) (−2 st)= 4xy – 2x²st = 4(2t + s) (1 − st²) −2 (2t+s)² ..(1) dt dz dt s=1,t=2 = 2 and = : 4(2 × 2 + 1) (1 − 1. 2²) — 2(2 × 2 + 1 = 4(5)(-3)-2(25) = -60 - 50 = -110 = -110 11 Not what you're looking for? Continue submitting your question. Submit Your Question dy dt √x DO
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