Evaluating a function by its derivatives values

lukas_

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Joined
Feb 17, 2019
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Here's the question:

Knowing that function [FONT=MathJax_Math-italic]f[FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math-italic]y[/FONT][FONT=MathJax_Main])[/FONT]f(x,y)[/FONT] satisfies f_x(48,8) = -3 , f_y (48,8) = -4

And that function g(s,t) = f(3s^2*t^2, -2*s*t)

What are g_s(-2,2) and g_t(-2,2) ?

What I've done so far is conclude that g_s(s,t) = f(6*s*t^2, -2*t) and g_t(s,t) = f(6*s^2*t, -2s)
And g_s(-2,2) = f(48,4) , g_t(-2,2) = f(-48,-4)

Am I on the right path? Can tangent plane equation help ? I'm stuck, thanks.


 

Dr.Peterson

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Joined
Nov 12, 2017
Messages
3,605
Knowing that function [FONT=MathJax_Math-italic]f[FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main],[/FONT][FONT=MathJax_Math-italic]y[/FONT][FONT=MathJax_Main])[/FONT][/FONT] satisfies f_x(48,8) = -3 , f_y (48,8) = -4

And that function g(s,t) = f(3s^2*t^2, -2*s*t)

What are g_s(-2,2) and g_t(-2,2) ?

What I've done so far is conclude that g_s(s,t) = f(6*s*t^2, -2*t) and g_t(s,t) = f(6*s^2*t, -2s)
And g_s(-2,2) = f(48,4) , g_t(-2,2) = f(-48,-4)

Am I on the right path? Can tangent plane equation help ? I'm stuck, thanks.
No, what you have done is wrong. It is not true that g_s(s,t) = f(6*s*t^2, -2*t).

Are you trying to apply the chain rule given as Case 2 here? http://tutorial.math.lamar.edu/Classes/CalcIII/ChainRule.aspx

Your "g" plays the role of z = f(x, y) = f(g(s,t), h(s,t)) there, where their g(s,t) = 3s^2*t^2 and h(s,t) = -2*s*t.

You don't replace the arguments of f with their partial derivatives; rather, you multiply each f's partial derivatives by the partial derivatives of the corresponding arguments.

Give it another try.
 
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