Question

Calculus Related Rates

Answer 1

Answer:

-2000pi cm^2/h (obviously lol)

Explanation:

Okay so the main way I solved these types of questions when I took Calc 1 was to first literally draw out the puddle. So as you can picture a puddle would form something circular (also mentioned in the problem) and the area of this circle is decreasing over time. I would encourage you to draw out a puddle with radius "r". So to find the area of this puddle it would simply be the area of a circle or A = pir^2( $A = \pi r^{2}$ ). Now that we have considered this there is one other thing to consider: related rates or rates in general are functions of time. This means that any change to a value will occur with respect to time! So when we differentiate this expression to find an expression for the rate of change of area we are going to do this with respect to time. This means that we will have to differentiate implicitly (basically chain rule). So lets do that:

$\frac{d}{dt} (A=\pi r^{2})$

$\frac{dA}{dt} = \pi (2r) (\frac{dr}{dt})$

So there's our expression. The key thing to notice here is how we differentiated. We treated r as a function, not as a variable because our differentiating variable is not r its t. If we were to, for example, take (d/dr) to this entire expression this would just be a simple power rule without the added chain rule. However, because this is in terms of time we are treating it differently.

So now lets consider our problem.

The radius is decreasing by 10cm/h this means that dr/dt = -10cm/h or the rate of change of the radius of the puddle is -10cm/h.

The problem also says when the radius is 100cm so r = 100cm

Lets plug that in:

$\frac{dA}{dt} = \pi 2(100)(-10)$

Simplifying that you get:

dA/dt = -2000pi cm^2/h (the units should be square units due to area)

Hopefully this makes sense!