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Exercise 4
Math Animated™
Mathematical Introduction for Physics and Engineering
by Samuel Dagan (Copyright © 2007-2020)
Chapter 3: Many Variables; Section 1: Differentiation; page 2
Partial Derivatives, Exercise 4
Question
The function z of two variables x and y is presented in the implicit form:
- Calculate
- What are
- What is the value of z for the point (x, y)=(1, 0) ?
- Obtain the tangent plane at
Reminder
We define the partial derivative of a function
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(3.1.2.1)
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with respect to the variable xk, as the limit
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(3.1.2.2)
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In the case of a function of two variables,
the first approximation is the tangent plane:
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(3.1.2.35)
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(3.1.2.37)
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With the help of (3.1.2.37), one can transform the expression of the tangent plane (3.1.2.35) in the following symmetric form:
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(3.1.2.38)
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Parts 1-3
Solution of question 1
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-
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Parts 4-5
Solution of question 2
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-
Part 6
Solution of question 3
- The substitution of x=1 and y=0 in the implicit form of the function yields:
therefore
Parts 7-9
Solution of question 4
- The required tangent plane should be obtained by the substitution of
in (3.1.2.38)
- The partial derivatives become
- and the tangent plane:
or
Score
By parts.
Parts 1, 2, 3, 4, 5, 7, 8, 9 are worth 1 point each.
Part 6 is worth 2 points.
By questions.
Questions 1 and 4 are worth 3 points each.
Questions 2 and 3 are worth 2 points each.