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Answer :
Let's solve the problem step-by-step:
(a) Estimate the thickness of the ice after 13 hours.
1. We're given the formula for the rate at which the ice thickness is increasing:
[tex]\(\frac{dy}{dt} = \frac{\sqrt{t}}{2}\)[/tex].
2. To find the thickness of the ice after a certain time, we need to integrate this rate function from [tex]\(t = 0\)[/tex] to [tex]\(t\)[/tex], which in this case is 13 hours.
3. Integrating [tex]\(\frac{\sqrt{t}}{2}\)[/tex] with respect to [tex]\(t\)[/tex] gives us the function for thickness [tex]\(y(t)\)[/tex]. After performing the integration, we get:
[tex]\[
y(t) = \int \frac{\sqrt{t}}{2} \, dt
\][/tex]
4. After performing the integration, we substitute [tex]\(t = 13\)[/tex] to find the thickness after 13 hours.
5. Calculate [tex]\(y(13)\)[/tex] and rounding the result gives us [tex]\(\approx 15.62\)[/tex] inches.
So, the thickness of the ice after 13 hours is approximately 15.62 inches.
(b) At what rate is the thickness of the ice increasing after 13 hours?
1. The rate at which the ice thickness is increasing at any given moment is given directly by the derivative [tex]\(\frac{dy}{dt} = \frac{\sqrt{t}}{2}\)[/tex].
2. To find the rate at which the thickness is increasing after 13 hours, substitute [tex]\(t = 13\)[/tex] into the rate of change formula:
[tex]\[
\frac{dy}{dt} = \frac{\sqrt{13}}{2}
\][/tex]
3. Calculate this value and round the result to two decimal places, which is approximately [tex]\(1.80\)[/tex] inches per hour.
Therefore, the rate at which the ice thickness is increasing after 13 hours is approximately 1.80 inches per hour.
(a) Estimate the thickness of the ice after 13 hours.
1. We're given the formula for the rate at which the ice thickness is increasing:
[tex]\(\frac{dy}{dt} = \frac{\sqrt{t}}{2}\)[/tex].
2. To find the thickness of the ice after a certain time, we need to integrate this rate function from [tex]\(t = 0\)[/tex] to [tex]\(t\)[/tex], which in this case is 13 hours.
3. Integrating [tex]\(\frac{\sqrt{t}}{2}\)[/tex] with respect to [tex]\(t\)[/tex] gives us the function for thickness [tex]\(y(t)\)[/tex]. After performing the integration, we get:
[tex]\[
y(t) = \int \frac{\sqrt{t}}{2} \, dt
\][/tex]
4. After performing the integration, we substitute [tex]\(t = 13\)[/tex] to find the thickness after 13 hours.
5. Calculate [tex]\(y(13)\)[/tex] and rounding the result gives us [tex]\(\approx 15.62\)[/tex] inches.
So, the thickness of the ice after 13 hours is approximately 15.62 inches.
(b) At what rate is the thickness of the ice increasing after 13 hours?
1. The rate at which the ice thickness is increasing at any given moment is given directly by the derivative [tex]\(\frac{dy}{dt} = \frac{\sqrt{t}}{2}\)[/tex].
2. To find the rate at which the thickness is increasing after 13 hours, substitute [tex]\(t = 13\)[/tex] into the rate of change formula:
[tex]\[
\frac{dy}{dt} = \frac{\sqrt{13}}{2}
\][/tex]
3. Calculate this value and round the result to two decimal places, which is approximately [tex]\(1.80\)[/tex] inches per hour.
Therefore, the rate at which the ice thickness is increasing after 13 hours is approximately 1.80 inches per hour.
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