Questions: Consider the formula: K=A^3 B^5/10. Given that both A and B vary with respect to t; find dK/dt. (A) dK/dt=3 A^2 B^5/10 dA/dt+A^3 B^4/2 dB/dt B dK/dt=3 A^2 B^5/10 C dK/dt=3 A^2 B^4/2 dA/dt dB/dt D dK/dt=A^3 B^5/10 dA/dt+A^3 B^5/10 dB/dt E dK/dt=3 A^3 B^5/10 dA/dt+A^3 B^5/2 dB/dt

Solution Steps

To find \(\frac{dK}{dt}\) for the given formula \(K = \frac{A^3 B^5}{10}\), we need to use the chain rule for differentiation. Since both \(A\) and \(B\) are functions of \(t\), we will differentiate \(K\) with respect to \(t\) by treating \(A\) and \(B\) as functions of \(t\).

1. Differentiate \(K\) with respect to \(A\) and multiply by \(\frac{dA}{dt}\).
2. Differentiate \(K\) with respect to \(B\) and multiply by \(\frac{dB}{dt}\).
3. Sum the results from steps 1 and 2.

We are given the formula: \[ K = \frac{A^3 B^5}{10} \] where both \( A \) and \( B \) are functions of \( t \).

To find \(\frac{dK}{dt}\), we need to apply the product rule for differentiation. The product rule states: \[ \frac{d}{dt}(uv) = u \frac{dv}{dt} + v \frac{du}{dt} \] Here, \( u = A^3 \) and \( v = \frac{B^5}{10} \).

First, differentiate \( u = A^3 \): \[ \frac{du}{dt} = 3A^2 \frac{dA}{dt} \]

Next, differentiate \( v = \frac{B^5}{10} \): \[ \frac{dv}{dt} = \frac{5B^4}{10} \frac{dB}{dt} = \frac{B^4}{2} \frac{dB}{dt} \]

Now, apply the product rule: \[ \frac{dK}{dt} = \frac{d}{dt} \left( A^3 \cdot \frac{B^5}{10} \right) \] \[ \frac{dK}{dt} = A^3 \cdot \frac{d}{dt} \left( \frac{B^5}{10} \right) + \frac{B^5}{10} \cdot \frac{d}{dt} \left( A^3 \right) \] \[ \frac{dK}{dt} = A^3 \cdot \frac{B^4}{2} \frac{dB}{dt} + \frac{B^5}{10} \cdot 3A^2 \frac{dA}{dt} \]

Combine the terms: \[ \frac{dK}{dt} = \frac{A^3 B^4}{2} \frac{dB}{dt} + \frac{3A^2 B^5}{10} \frac{dA}{dt} \]

Final Answer