Question

Under certain conditions, the velocity of a liquid in a pipe at distance r from the center of the pipe is given by V = 400(3.025 x 10-5--2) where Osrs5,5x10 -3. Writeras a function of V.r=where the domain is a compound inequality(Use scientific notation. Use integers or decimals for any numbers in the expression.)Le

Answer 1

Solving the equation for r:

`\begin{gathered} V=400(9.025\cdot10^(-5)-r^2) \\ r^2=9.025\cdot10^(-5)-(V)/(400) \\ r=\sqrt[]{9.025\cdot10^(-5)-(V)/(400)} \end{gathered}`

With the first equations, we can establish some limits for V:

With the lowest value for r (r=0):

`\begin{gathered} V=400(9.025\cdot10^(-5)-0^2) \\ V=400(9.025\cdot10^(-5)) \\ V=3.61\cdot10^(-2) \end{gathered}`

With the highest value for r (r=9.5x10^-3)

`\begin{gathered} V=400(9.025\cdot10^(-5)-(9.5\cdot10^(-3))^2) \\ V=400(9.025\cdot10^(-5)-9.025\cdot10^(-5)) \\ V=400(0) \\ V=0 \end{gathered}`

According to the radius range, velocity can be between 0 and 3.61x10^-2

It is also necessary to check the domain of the function considering it is a square root. The argument of an square root cannot be less than 0. Then:

`\begin{gathered} 9.025\cdot10^(-5)-(V)/(400)\ge0 \\ 9.025\cdot10^(-5)\ge(V)/(400) \\ V\leq400(9.025\cdot10^(-5)) \\ V\leq3.61\cdot10^(-2) \end{gathered}`

This is the same limit for velocity obtained before. Then, we can say for velocity that:

`0\leq V\leq3.61\cdot10^(-2)`