[tex]\rm \dfrac{d(f \circ c)}{dt} = 9 \;\;\;\;\;\;\;\;\;\; (when\; t =\pi )[/tex]
Step-by-step explanation:
Given :
[tex]\rm f:R_4 \to R[/tex]
[tex]\rm c(t) : R \to R_4[/tex]
[tex]\bigtriangledown f(1,1,\pi ,e^6) = (0,1,5,-2)\\c(\pi )= (1,1,\pi ,e^6)\\c'(\pi ) = (19,11,0,1)[/tex]
Solution :
[tex]\rm \dfrac{d(f \circ c)}{dt} = \dfrac {d}{dt} f(c(t)) = f'(c(t).c'(t))[/tex]
[tex]\rm \bigtriangledown f(c(t)) = c'(t)[/tex]
[tex]\rm \bigtriangledown f(c(t)) = \rm \bigtriangledown f((1,1,\pi,e^6). (19,11,0,1)) \;\;\;\;\;\;\;(when \;t=\pi )[/tex]
0 +11+0-2 = 9
Therefore,
[tex]\rm \dfrac{d(f \circ c)}{dt} = 9[/tex]
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https://brainly.com/question/14418346?referrer=searchResults
