Chain rule (multivariable calculus)

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SUMMARY

The discussion focuses on the application of the chain rule in multivariable calculus, specifically evaluating the function derivatives f_x = 3*x^2 + y and f_y = 2*y + x. Participants explore the linear approximation of the function along the path defined by t, using points t=1 and t=1.1. The conversation highlights the importance of correctly formatting mathematical expressions and emphasizes the significance of the gradient and tangent plane in understanding changes along the path.

PREREQUISITES
  • Understanding of multivariable calculus concepts, particularly the chain rule.
  • Familiarity with derivatives and their applications in linear approximations.
  • Knowledge of gradient vectors and tangent planes in calculus.
  • Ability to format and interpret mathematical expressions correctly.
NEXT STEPS
  • Study the application of the chain rule in multivariable calculus.
  • Learn how to compute and interpret gradients in higher dimensions.
  • Explore the concept of tangent planes and their significance in calculus.
  • Practice formatting mathematical expressions using LaTeX for clarity.
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Students and educators in mathematics, particularly those focusing on calculus, as well as anyone interested in the practical applications of the chain rule and linear approximations in multivariable contexts.

Poetria
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Homework Statement
Let ##f(x,y)=x^3+y^2+x*y##
Suppose that a point is moving through the plane. At time t , the point is at ## (x(t), y(t))=(t^2, e^{t-1})##. Use linear approximation to estimate the change in f as t goes from 1 to 1.1 . In other words, approximate
Relevant Equations
Multivariable chain rule
##f_x=3*x^2+y##
##f_y=2*y+x##

##(3*(t^2)^2+e^{t-1})*2*t+(2*e^{t-1}+t^2)*e^{t-1}##
Well, I am not sure how to evaluate it.
I got a wrong result by multiplying by 0.1, i.e.
##((3*(t^2)^2+e^{t-1})*2*t+(2*e^{t-1}+t^2)*e^{t-1})*0.1##

I guess it is trivial but I am lost. :(
 
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We are interested in the change along ##t##. So why don't you write ##f(x,y)=f(x(t),y(t))=f(t)## in the first place? Then we have two points at ##t=1## and ##t=1.1##. The linear approximation is the secant through these points:
$$
y=\underbrace{\dfrac{f(1)-f(1.1)}{1-1.1}}_{=:m}\cdot x + b\;\wedge\;f(1)=m\cdot 1+b
$$
Finally, you can check the quality of this approximation by calculating the tangent at ##t=1##:
$$
y=\left. \dfrac{d}{dt}\right|_{t=1}f(t) \cdot x + c =f'(1)x+c\;\wedge\;f(1)=f'(1)\cdot 1 +c
$$
which is the linear approximation if only one point is given.
 
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Many thanks. But I can't read this:
"The linear approximation is the secant through these points:
You can't use 'macro parameter character #' in math mode ?

A nice quote from Leibniz. :) Émilie du Châtelet (my avatar) was Leibnizian. :)
 
Poetria said:
Many thanks. But I can't read this
You need to give me a second to correct my format. Please reload.

"He who hasn't tasted bitter things hasn't earned sweet things." (G. Leibniz)

"Let us choose for ourselves our path in life, and let us try to strew that path with flowers." (E. du Châtelet)

##t\longmapsto f(t)## is the path, and you should plant the flowers at ##t=1## and ##t=1.1.##
 
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fresh_42 said:
You need to give me a second to correct my format. Please reload.

"He who hasn't tasted bitter things hasn't earned sweet things." (G. Leibniz)

"Let us choose for ourselves our path in life, and let us try to strew that path with flowers." (E. du Châtelet)

##t\longmapsto f(t)## is the path, and you should plant the flowers at ##t=1## and ##t=1.1.##
I got:
-10.3022 + 13.3022*x = y
x=1.1
Approximation of the change:
4.33022-3=1.33022

A beautiful metaphor: ##t\longmapsto f(t)## is the path, and you should plant the flowers at ##t=1## and ##t=1.1.##
 
Perhaps a gradient would be also a good idea:
Gradient ##\vec (0.21, 0.105171)##
A tangent plane: 0.21*x + 0.105171*y-0.32

Slope of the gradient: 0.500814
 
Last edited:
Poetria said:
A beautiful metaphor: ##t\longmapsto f(t)## is the path, and you should plant the flowers at ##t=1## and ##t=1.1.##
Yes, but a little bit too fast posted. The path is actually ##t\longmapsto (x(t),y(t))## and ##f(t)## the function values along the path.
 
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