Certainly! Let's determine the rate of change [tex]\(\frac{dA}{dt}\)[/tex] at the moment when [tex]\(A = 3\)[/tex] and [tex]\(\frac{dB}{dt} = 1\)[/tex] for the given equation:
[tex]\[
A^3 + B^3 = 91
\][/tex]
First, we start by differentiating both sides of the equation with respect to time [tex]\(t\)[/tex]. Since [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are functions of [tex]\(t\)[/tex], we'll use implicit differentiation.
The given equation is:
[tex]\[
A^3 + B^3 = 91
\][/tex]
Differentiating both sides with respect to [tex]\(t\)[/tex], we get:
[tex]\[
\frac{d}{dt}(A^3 + B^3) = \frac{d}{dt}(91)
\][/tex]
On the left side, apply the chain rule. Differentiating [tex]\(A^3\)[/tex] and [tex]\(B^3\)[/tex] with respect to [tex]\(t\)[/tex] gives:
[tex]\[
3A^2 \frac{dA}{dt} + 3B^2 \frac{dB}{dt} = 0
\][/tex]
We can simplify this to:
[tex]\[
3A^2 \frac{dA}{dt} + 3B^2 \frac{dB}{dt} = 0
\][/tex]
Divide the entire equation by 3:
[tex]\[
A^2 \frac{dA}{dt} + B^2 \frac{dB}{dt} = 0
\][/tex]
Now, we substitute the given values [tex]\(A = 3\)[/tex] and [tex]\(\frac{dB}{dt} = 1\)[/tex] into the equation:
[tex]\[
(3)^2 \frac{dA}{dt} + B^2 (1) = 0
\][/tex]
This simplifies to:
[tex]\[
9 \frac{dA}{dt} + B^2 = 0
\][/tex]
Solving for [tex]\(\frac{dA}{dt}\)[/tex], we obtain:
[tex]\[
9 \frac{dA}{dt} = -B^2
\][/tex]
[tex]\[
\frac{dA}{dt} = -\frac{B^2}{9}
\][/tex]
So the rate of change [tex]\(\frac{dA}{dt}\)[/tex] at the moment when [tex]\(A = 3\)[/tex] and [tex]\(\frac{dB}{dt} = 1\)[/tex] is:
[tex]\[
\frac{dA}{dt} = -\frac{B^2}{9}
\][/tex]
