Science:Math Exam Resources/Courses/MATH110/December 2016/Question 09 (c)

MATH110 December 2016

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Question 09 (c)

Sugar dissolves in water at a rate proportional to the amount still undissolved. Suppose the amount $M$ of undissolved sugar present in a water solution $t$ hours after the sugar was added to the water is modelled using an exponential decay function $M(t)=Ce^{-kt}$, where $C$ and $k$ are positive constants. Suppose initially $10{\text{ kg}}$ of sugar are added to the water. Assume it takes $4{\text{ hr}}$ for half of the initial amount of sugar to dissolve. Answer the questions below. You may leave your answers unsimplified, i.e. in ”calculator-ready” form. (c) If 10 more kg of sugar is added to the solution $4{\text{ hr}}$ after the solution was first mixed, when will there be $1{\text{ kg}}$ of undissolved sugar left in the solution?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
We need to determine the unknown constants again.

Hint 2
The constant ${\textstyle k}$ remains unchanged because it indicates how fast sugar dissolves.

Hint 3
The constant ${\textstyle C}$ is changed. It is now determined by the amount of sugar at time ${\textstyle t=4}$.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Following the Hints, let us denote the amount of undissolved sugar after ${\textstyle t=4}$ by $$N(t)=C\cdot 2^{-{\frac {t}{4}}},\quad t\geq 4.$$ According to the description, we have that $$N(4)=M(4)+10=5+10=15.$$ Putting ${\textstyle t=4}$, $$C\cdot 2^{-{\frac {4}{4}}}=15,$$ $$C\cdot 2^{-1}=15,$$ $$C=30.$$ So the new formula for the amount of undissolved sugar is $$N(t)=30\cdot 2^{-{\frac {t}{4}}},\quad t\geq 4.$$ Now we solve for ${\textstyle t}$ if ${\textstyle N(t)=1}$: $$30\cdot 2^{-{\frac {t}{4}}}=1,$$ $$2^{\frac {t}{4}}=30.$$ Taking the natural logarithm on both side, $${\dfrac {t}{4}}\ln 2=\ln 30,$$ $$t=4{\dfrac {\ln 30}{\ln 2}}.$$ Answer: there will be ${\textstyle 1}$ kg of sugar at time ${\textstyle t={\color {blue}4{\dfrac {\ln 30}{\ln 2}}}}$ hours.