An ice cube is freezing in such a way that the side length [tex]$s$[/tex], in inches, is [tex]$s(t)=\frac{1}{2} t+4$[/tex], where [tex]$t$[/tex] is in hours. The surface area of the
Part A: Write an equation that gives the volume at [tex]$t$[/tex] hours after freezing begins. (2 points)
Part B: Find the surface area as a function of time, using composition, and determine its range. (4 points)
Part C: After how many hours will the surface area equal 294 square inches? Show all necessary calculations, and check for extra
[tex]$\varsigma \rightarrow q_{\sim}$[/tex] Paragraph [tex]$\sim I_x B I \subseteq \underline{u} \quad x^2 \Omega \vee-$[/tex]
[tex]$\begin{array}{l}
v(s)=\left(\frac{1}{2} t+4\right)^3 \\
\left(\frac{1}{n}+4\right)\left(\frac{1}{n}+4\right)\left(\frac{1}{-}+4\right)
\end{array}$[/tex]
Answer (1)
The volume of the ice cube as a function of time is V ( t ) = ( 2 1 t + 4 ) 3 .
The surface area of the ice cube as a function of time is A ( t ) = 6 ( 2 1 t + 4 ) 2 , with a range of [ 96 , ∞ ) .
Setting the surface area equal to 294 square inches, 6 ( 2 1 t + 4 ) 2 = 294 , and solving for t .
The time at which the surface area equals 294 square inches is 6 hours.
Explanation
Problem Analysis We are given that the side length of an ice cube is changing with time according to the equation s ( t ) = 2 1 t + 4 , where s is in inches and t is in hours. We need to find the volume as a function of time, the surface area as a function of time and its range, and the time when the surface area equals 294 square inches.
Volume as a Function of Time The volume of a cube is given by V = s 3 . Substituting the given expression for s ( t ) , we get the volume as a function of time:
Volume Equation V ( t ) = ( 2 1 t + 4 ) 3
Surface Area as a Function of Time The surface area of a cube is given by A = 6 s 2 . Substituting the given expression for s ( t ) , we get the surface area as a function of time:
Surface Area Equation A ( t ) = 6 ( 2 1 t + 4 ) 2
Finding the Range To find the range of A ( t ) , we consider the domain of t . Since time t must be non-negative, t ≥ 0 . When t = 0 , we have:
Initial Surface Area A ( 0 ) = 6 ( 2 1 ( 0 ) + 4 ) 2 = 6 ( 4 ) 2 = 6 ( 16 ) = 96
Range of Surface Area As t increases, A ( t ) also increases. Since there is no upper bound on t , the surface area can increase indefinitely. Therefore, the range of A ( t ) is [ 96 , ∞ ) .
Finding the Time We are asked to find the time t when the surface area equals 294 square inches. We set A ( t ) = 294 and solve for t :
Setting up the Equation 6 ( 2 1 t + 4 ) 2 = 294
Simplifying Dividing both sides by 6, we get:
Further Simplification ( 2 1 t + 4 ) 2 = 49
Square Root Taking the square root of both sides, we have:
Isolating t 2 1 t + 4 = ± 7
Solving for t Solving for t , we get two possible values:
Two Possible Solutions 2 1 t = − 4 ± 7
Positive Solution t = 2 ( − 4 + 7 ) = 2 ( 3 ) = 6
Negative Solution t = 2 ( − 4 − 7 ) = 2 ( − 11 ) = − 22
Final Time Since time t must be non-negative, we discard the negative solution. Therefore, t = 6 hours.
Summary of Results The volume at t hours after freezing begins is V ( t ) = ( 2 1 t + 4 ) 3 . The surface area as a function of time is A ( t ) = 6 ( 2 1 t + 4 ) 2 , and its range is [ 96 , ∞ ) . The surface area will equal 294 square inches after 6 hours.
Examples
Imagine you're designing a cooling system for a server room. You need to predict how the surface area of ice blocks changes over time as they melt, to ensure efficient heat absorption. By modeling the side length of the ice as a function of time, similar to this problem, you can determine the surface area and volume at any given time, optimizing the cooling process and preventing overheating.
