What is the rate of change of the width (in ft/sec) when the height is 10 feet, if the height is decreasing at that moment at the rate of 1 ft/sec.A rectangle has both a changing height and a changing width, but the height and width change so that the area of the rectangle is always 60 square feet?

Question

Ezra Adams · Accepted Answer

The rate of change of the width with time #(dW)/(dt)# = #0.6"ft/s"# #(dW)/(dt)=(dW)/(dh)xx(dh)/dt# #(dh)/(dt)=-1"ft/s"# So #(dW)/(dt)=(dW)/(dh)xx-1=-(dW)/(dh)# #Wxxh=60# #W=60/h# #(dW)/(dh)=-(60)/(h^2)# So #(dW)/(dt)=-(-(60)/(h^2))=(60)/(h^2)# So when #h=10#: #rArr#   #(dW)/(dt)=(60)/(10^2)=0.6"ft/s"#

Evelyn Agard · Answer

undefined Let $ w $ represent the width of the rectangle and $ h $ represent the height. The given condition is that the area $ A $ of the rectangle is always 60 square feet, so $ A = wh = 60 $. To find the rate of change of the width ($ \frac{{dw}}{{dt}} $) when the height is 10 feet and decreasing at a rate of 1 ft/sec ($ \frac{{dh}}{{dt}} = -1 $), we differentiate the area equation with respect to time using the product rule:

$$ \frac{{d}}{{dt}}(wh) = \frac{{dA}}{{dt}} = w \frac{{dh}}{{dt}} + h \frac{{dw}}{{dt}} $$

Substituting the given values ($ A = 60 $, $ h = 10 $, $ \frac{{dh}}{{dt}} = -1 $), we have:

$$ \frac{{dA}}{{dt}} = 60 = w(-1) + 10 \frac{{dw}}{{dt}} $$

Solving for $ \frac{{dw}}{{dt}} $, we get:

$$ \frac{{dw}}{{dt}} = \frac{{60 + w}}{{10}} $$

At the moment when $ h = 10 $, we can find $ w $ using the equation $ A = wh = 60 $. Thus, $ w = \frac{{60}}{{h}} = \frac{{60}}{{10}} = 6 $. Substituting $ w = 6 $ into the rate equation, we get:

$$ \frac{{dw}}{{dt}} = \frac{{60 + 6}}{{10}} = \frac{{66}}{{10}} = 6.6 \, \text{ft/sec} $$

So, the rate of change of the width is 6.6 ft/sec when the height is 10 feet and decreasing at a rate of 1 ft/sec.