Calculus derivative equation

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dear2009

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Dear Math help Participants,



Suppose that A, B, C are all time varying quantities, which at all times satisfy the equation:

A=3
B=2 dB/dt= -1
C=7 dC/dt= 5

This is what i did to get the derivative
A^1 dA/dt +B[2C^1]dC/dt+C^1 dB/dt=0


The only problem is, I dont know if i got the derivative, because after that it is simple plug in, can anybody show me how to do the proper derivative of the given equation?
 
dear2009 said:
Dear Math help Participants,



Suppose that A, B, C are all time varying quantities, which at all times satisfy the equation:

A=3
B=2 dB/dt= -1
C=7 dC/dt= 5

This is what i did to get the derivative
A^1 dA/dt +B[2C^1]dC/dt+C^1 dB/dt=0


The only problem is, I dont know if i got the derivative, because after that it is simple plug in, can anybody show me how to do the proper derivative of the given equation?
Click to expand...

As posted - your problem does not make sense to me.

Please post the EXACT problem as it was presented to you.
 
Dear Subhostosh Khan,



I am sorry for not writing the problem completely, here it is:

Suppose that A, B, C are all time varying quantities, which at all times satisfy the equation:

B(A + C^2) = 1

A=3
B=2
dB/dt= -1
C=7
dC/dt= 5

This is what i did to get the derivative
A^1 dA/dt +B[2C^1]dC/dt+C^1 dB/dt=0

I am not sure if this derivative is correct
 
dear2009 said:
Dear Subhostosh Khan,



I am sorry for not writing the problem completely, here it is:

Suppose that A, B, C are all time varying quantities, which at all times satisfy the equation:

B(A + C^2) = 1

A=3
B=2
dB/dt= -1
C=7
dC/dt= 5

I do not see a question here.

This is what i did to get the derivative
A^1 dA/dt +B[2C^1]dC/dt+C^1 dB/dt=0

I am not sure if this derivative is correct
Click to expand...
 
Dear Subhostosh Khan,



Sorry to write again but here is the problem

Suppose that A, B, C are all time varying quantities, which at all times satisfy the equation:

B(A + C^2) = 1


Further suppose that at a particular instant,
A=3
B=2
dB/dt= -1
C=7
dC/dt= 5

then dA/dt = ?
 
\(\displaystyle B(A+C^{2}) \ = \ 1, \ \implies \ AB+BC^{2} \ = \ 1\)

\(\displaystyle \frac{dA}{dt}B+A\frac{dB}{dt}+\frac{dB}{dt}C^{2}+2BC\frac{dC}{dt} \ = \ 0\)

\(\displaystyle \frac{dA}{dt}(2)+(3)(-1)+(-1)(49)+(2)(2)(7)(5) \ = \ 0\)

\(\displaystyle 2\frac{dA}{dt} \ = \ -140+52 \ = \ -88\)

\(\displaystyle Hence, \ \frac{dA}{dt} \ = \ \frac{-88}{2} \ = \ -44\)
 
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