91Ó°ÊÓ

Related rates problems involve finding the rate at which one quantity changes in relation to another quantity. These problems typically deal with quantities that are related by some geometric or algebraic relationship. The key to solving related rates problems is to correctly identify the variables and their relationships before differentiating with respect to time. In the context of this exercise, the supply of oranges and their wholesale price are related by an equation. To find out how fast the price is changing when the supply is changing at a certain rate, we use implicit differentiation with respect to time. First, we rewrite the equation involving the rates. The original equation is given as: \[ px + 7x + 8p = 328 \] By differentiating both sides of the equation with respect to time \( t \), and recognizing that both \( p \) and \( x \) are functions of time, we can determine the needed rate of change for the price.

The Chain Rule is a fundamental concept in calculus used for computing the derivative of the composite of two or more functions. When we have a function of a function, the Chain Rule allows us to differentiate indirectly. In this exercise, when we differentiate \( px \) with respect to time, we are essentially using the Chain Rule because both \( p \) and \( x \) are functions of time. This is expressed mathematically as: \[ \frac{d}{dt}(p \times x) = x \frac{dp}{dt} + p \frac{dx}{dt} \] This shows that to differentiate the product of \( p \) and \( x \), we need to take into account how both \( p \) and \( x \) change over time. Additionally, the other terms \( 7x \) and \( 8p \) in the original equation also require differentiation, each utilizing the Chain Rule, giving us: \[7 \frac{dx}{dt} + 8 \frac{dp}{dt} \] These derivative terms help us form a new equation that can be solved for the unknown rate \( \frac{dp}{dt} \).

The Product Rule is a technique used to find the derivative of the product of two functions. It states that if we have two functions, say \( u(t) \) and \( v(t) \), their derivative with respect to \( t \) is given by: \[ (uv)' = u'v + uv' \] In this exercise, the term \( px \) requires the Product Rule for proper differentiation, as both \( p \) and \( x \) are functions of time. Therefore, differentiating \( px \) with respect to time gives us: \[ \frac{d}{dt}(px) = x \frac{dp}{dt} + p \frac{dx}{dt} \] Here, \( x \) and \( p \) represent the values of supply and price, respectively. By accurately using the Product Rule, we account for how changes in both variables influence the overall rate of change. This step ensures that the relation between \( \frac{dp}{dt} \) and \( \frac{dx}{dt} \) is correctly computed, laying the groundwork to solve for the rate at which the price is changing given the rate of change of the supply.