Related Rates: lengths of sides of square increasing so that

[BobertGeraldo]

Good evening heres a question that I need help with.

The lengths of the sides of a square are increasing such that the area doubles every minute.

So, the area A at time t can be expressed as A(t)=A(naught)e^kt. If initially the area is 1 cm squared, how fast is the length of an edge of the square increasing after one minute?

Okay, so A naught is equal to 1, A'(t)=k(A(t)). A=x^2 so dA/dT= 2x dx/dt. I really just dont know where to go from here. What kind of throws me off is K. i guess since I know what da/dt is i could substitute it into the other equation, which makes the most sense, but then what? Thanks,

 

[Deleted member 4993]

Re: Related Rates.

Good evening heres a question that I need help with.

The lengths of the sides of a square are increasing such that the area doubles every minute.

\(\displaystyle \frac {A_2}{A_1} = e^k = 2\)

Now solve for 'k'.

So, the area A at time t can be expressed as A(t)=A(naught)e^kt. If initially the area is 1 cm squared, how fast is the length of an edge of the square increasing after one minute?


Okay, so A naught is equal to 1, A'(t)=k(A(t)). A=x^2 so dA/dT= 2x dx/dt. I really just dont know where to go from here. What kind of throws me off is K. i guess since I know what da/dt is i could substitute it into the other equation, which makes the most sense, but then what? Thanks,