Engineers are draining a water reservoir until its depth is only 10 feet. The depth decreases exponentially as
shown in the graph below. The engineers measure the depth after 1 hour to be 64 feet and after 4 hours to be
28 feet. Develop an exponential equation in y=a(6)" to predict the depth as a function of hours draining.
Round a to the nearest integer and b to the nearest hundredth. Then, graph the horizontal line y = 10 and
find its intersection to determine the time, to the nearest tenth of an hour, when the reservoir will reach a
depth of 10 feet.

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Answer :

MRROYALIN MRROYALIN · Answer 1

The draining of the reservoir follows an exponential function

The reservoir will reach a depth of 10 feet at 7.6 hour

The given parameters are:

[tex]\mathbf{(x,y) = (1,64)(4,28)}[/tex]

An exponential equation is represented as:

[tex]\mathbf{y = ab^x}[/tex]

Substitute [tex]\mathbf{(x,y) = (1,64)(4,28)}[/tex] in [tex]\mathbf{y = ab^x}[/tex]

[tex]\mathbf{64 = ab^1}[/tex]

[tex]\mathbf{64 = ab}[/tex]

[tex]\mathbf{28 = ab^4}[/tex]

Divide [tex]\mathbf{64 = ab}[/tex] and [tex]\mathbf{28 = ab^4}[/tex]

[tex]\mathbf{\frac{ab^4}{ab} = \frac{28}{64}}[/tex]

[tex]\mathbf{b^3 = 0.4375}[/tex]

Take cube roots

[tex]\mathbf{b= 0.76}[/tex]

Substitute [tex]\mathbf{b= 0.76}[/tex] in [tex]\mathbf{64 = ab}[/tex]

[tex]\mathbf{64 = 0.76a}[/tex]

Solve for a

[tex]\mathbf{a = 84}[/tex]

So, the equation is:

[tex]\mathbf{y = 84(0.76)^x}[/tex]

See attachment for the graphs of [tex]\mathbf{y = 84(0.76)^x}[/tex] and y = 10

The point of intersection is: x = 7.6

Hence, the reservoir will reach a depth of 10 feet at 7.6 hour

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