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

MATH110 December 2016

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Question 09 (b)

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. (b) At what rate (in kg per hr) does sugar dissolve $4{\text{ hr}}$ after it was first added to the water?

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Notice that ${\textstyle M'(t)}$ is the rate is change of the amount of sugar, which is negative and indicates a decrease. Therefore the rate of dissolution is ${\textstyle -M'(t)}$.

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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. Since ${\textstyle M(t)}$ is the undissolved amount, the rate (in kg per hr) of dissolution at time ${\textstyle t=4}$ is ${\textstyle -M'(4)}$. By part (a), $$M(t)=10\cdot 2^{-{\frac {t}{4}}}.$$ Differentiating using the formula ${\textstyle {\dfrac {d}{dx}}(a^{x})=a^{x}\ln a}$, we have $$M'(t)=10\cdot 2^{-{\frac {t}{4}}}\ln 2\cdot \left(-{\dfrac {1}{4}}\right)=-{\dfrac {5\ln 2}{2}}2^{-{\frac {t}{4}}}.$$ Putting ${\textstyle t=4}$, we have $$M'(4)=-{\dfrac {5\ln 2}{2}}2^{-{\frac {4}{4}}}=-{\dfrac {5\ln 2}{4}}$$ So the rate of dissolution is $$-M'(t)={\dfrac {5\ln 2}{4}}.$$ Answer: sugar dissolves at a rate of ${\textstyle {\color {blue}{\dfrac {5\ln 2}{4}}}}$ kg/hr.