two variable quantities a and b are found to be related by the equation given below. what is the rate of…

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?\na³ + b³ = 152\n\n\\( \\frac { d a } { d t } = \\square \\) when a = 3 and db/dt = 1\n(simplify your answer)

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?\na³ + b³ = 152\n\n\\( \\frac { d a } { d t } = \\square \\) when a = 3 and db/dt = 1\n(simplify your answer)

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})