Answer
a. $[2.56,5.76]ft$
b. $[3.24,4.84]ft$
Work Step by Step
a. Step 1. Given $y=\sqrt x/2$, we have $x=4y^2$. We are given a range in $y\in[y_0-\Delta y, y_0+\Delta y]$, and we need to find out the range of values in $x$ which satisfy the condition.
Step 2. With $y_0=1ft^3/min$ and $\Delta y=0.2 ft^3/min$, we have $y_1=y_0-\Delta y=1-0.2=0.8 ft^3/min$ and $y_2=y_0+\Delta y=1+0.2=1.2 ft^3/min$
Step 3. Using the relation $x=4y^2$, we can obtain corresponding x-values, $x_1=4y_1^2=4(0.8)^2=2.56ft$, and $x_2=4y_2^2=4(1.2)^2=5.76ft$.
Step 4. Thus, the range of depth that needs to be maintained is $[2.56,5.76]ft$
b. Step 1. Repeat the above process for $y_0=1ft^3/min$ and $\Delta y=0.1 ft^3/min$:
we have $y_1=y_0-\Delta y=1-0.1=0.9 ft^3/min$ and $y_2=y_0+\Delta y=1+0.1=1.1 ft^3/min$
Step 2. Using the relation $x=4y^2$, we can obtain corresponding x-values, $x_1=4y_1^2=4(0.9)^2=3.24ft$, and $x_2=4y_2^2=4(1.1)^2=4.84ft$.
Step 3. Thus, for this case, the range of depth that needs to be maintained is $[3.24,4.84]ft$
