Question

two variable quantities a and b are found to be related by the equation given below. what is the rate of change da/dt at the moment when a = 3 and db/dt = 1?
a³ + b³ = 152

\\( \frac { d a } { d t } = \square \\) when a = 3 and db/dt = 1
(simplify your answer)

Explanation:

Step1: Differentiate the equation with respect to \(t\)

Differentiate \(A^{3}+B^{3}=152\) term - by - term using the chain rule.
The derivative of \(A^{3}\) with respect to \(t\) is \(3A^{2}\frac{dA}{dt}\), and the derivative of \(B^{3}\) with respect to \(t\) is \(3B^{2}\frac{dB}{dt}\), and the derivative of the constant \(152\) is \(0\). So, \(3A^{2}\frac{dA}{dt}+3B^{2}\frac{dB}{dt}=0\).

Step2: Solve for \(\frac{dA}{dt}\)

First, factor out the \(3\): \(3(A^{2}\frac{dA}{dt}+B^{2}\frac{dB}{dt}) = 0\), then \(A^{2}\frac{dA}{dt}+B^{2}\frac{dB}{dt}=0\).
Solve for \(\frac{dA}{dt}\): \(\frac{dA}{dt}=-\frac{B^{2}}{A^{2}}\frac{dB}{dt}\).

Step3: Find the value of \(B\) when \(A = 3\)

Substitute \(A = 3\) into the original equation \(A^{3}+B^{3}=152\).
\(3^{3}+B^{3}=152\), \(27 + B^{3}=152\), \(B^{3}=152 - 27=125\), so \(B = 5\).

Step4: Substitute \(A = 3\), \(B = 5\), and \(\frac{dB}{dt}=1\) into the \(\frac{dA}{dt}\) formula

\(\frac{dA}{dt}=-\frac{5^{2}}{3^{2}}\times1\).

Answer:

\(-\frac{25}{9}\)