AP Calculus AB / Contextual Applications of Differentiation / Question No. 30

30. The number of minutes \( M \) that it takes to make a calculus lesson and the number of lessons \( L \) that are made per week satisfy the relationship:

\( L = \frac{k}{M} \), where \( k \) is a constant.

Which of the following best describes the relationship between the rate of change, with respect to time \( t \), of \( L \) and the rate of change, with respect to time \( t \), of \( M \)?

\( \frac{dL}{dt} = \frac{k}{\left(\frac{dM}{dt}\right)} \)

\( \frac{dL}{dt} = \frac{-k}{\left(\frac{dM}{dt}\right)} \)

\( \frac{dL}{dt} = \frac{k}{M^2} \left(\frac{dM}{dt}\right) \)

\( \frac{dL}{dt} = \frac{-k}{M^2} \left(\frac{dM}{dt}\right) \)

Answer is: option4

\( \frac{dL}{dt} = \frac{-k}{M^2} \left(\frac{dM}{dt}\right) \)

Solution:

We are given:

\( L = \frac{k}{M} \)

Differentiating both sides with respect to \( t \):

\( \frac{dL}{dt} = \frac{d}{dt} \left( \frac{k}{M} \right) \)

Using the derivative rule for a function of the form \( \frac{k}{M} \), where \( k \) is a constant:

\( \frac{d}{dt} \left( \frac{k}{M} \right) = -\frac{k}{M^2} \cdot \frac{dM}{dt} \)

\( \frac{dL}{dt} = -\frac{k}{M^2} \cdot \frac{dM}{dt} \)

Final Answer

\( \frac{dL}{dt} = -\frac{k}{M^2} \left(\frac{dM}{dt} \right) \)

Previous Next

AP Calculus AB Tags

AP Calculus BC Tags