Toothpaste

Calculus Level 4

Marcia will use less and less toothpaste when the tube is close to being empty. We will model this irrational behavior as follows: r = r 0 ⋅ ( 1 − α 2 x 2 ) , r = r_0\cdot \big(1-\alpha^2x^2\big), where

  • 0 ≤ x ≤ 1 0 \leq x \leq 1 is the fraction of the toothpaste that has been used up;
  • r r is the amount of toothpaste used per day;
  • r 0 r_0 is the amount of toothpaste used per day when the tube has just been opened;
  • 0 < α < 1 0 < \alpha < 1 is a parameter measuring how strong this behavior is. For Marcia, α = 0.9 \alpha = 0.9 , which means that in the end, she uses only 19 % 19\% of the amount she used at the beginning.

By what factor does Marcia increase the lifetime of a tube of toothpaste?

That is, letting t 0 t_0 be the number of days it takes to empty the tube when it is used at a constant rate r 0 r_0 , and t t the number of days it takes for Marcia to finish it, calculate t t 0 . \frac{t}{t_0}.


The answer is 1.635799433.

Arjen Vreugdenhil by Arjen Vreugdenhil , 44, USA

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1 solution

Arjen Vreugdenhil
Oct 27, 2017

If the toothpaste is used at a constant rate r 0 r_0 , the time needed to finish a tube of amount 1 is t 0 = 1 r 0 . t_0 = \frac{1}{r_0}. With Marcia's behavior, it is t = ∫ 0 1 d t = ∫ 0 1 d x r = 1 r 0 ∫ 0 1 d x 1 − α 2 x 2 t = \int_0^1 dt = \int_0^1 \frac{dx}{r} = \frac 1{r_0} \int_0^1 \frac{dx}{1 - \alpha^2x^2} = t 0 ∫ 0 1 1 2 ( 1 1 + α x + 1 1 − α x ) d x = t_0 \int_0^1 \frac 1 2\left(\frac{1}{1 + \alpha x} + \frac{1}{1 - \alpha x} \right)\:dx = t 0 2 α [ ln ⁡ ( 1 + α x ) − ln ⁡ ( 1 − α x ) ] 0 1 = \frac {t_0}{2\alpha}\left[\ln (1+\alpha x) - \ln (1 - \alpha x)\right]_0^1 = t 0 2 α ( ln ⁡ ( 1 + α ) − ln ⁡ ( 1 − α ) = \frac {t_0}{2\alpha}(\ln(1 + \alpha) - \ln(1 - \alpha) = ln ⁡ 1.9 − ln ⁡ 0.1 2 ⋅ 0.9 t 0 ≈ 1.636 t 0 . = \frac {\ln 1.9 - \ln 0.1}{2\cdot 0.9}t_0 \approx \boxed{1.636}\:t_0. Thus, Marcia makes the tube of toothpaste last 64 % 64\% longer.

Note : It would be a fallacy to reason as follows: r average = ∫ 0 1 r d x = r 0 ∫ 0 1 ( 1 − α 2 x 2 ) d x = r 0 [ x − 1 3 α 2 x 3 ] 0 1 = r 0 ( 1 − 1 3 α 2 ) = 0.73 r 0 , r_{\text{average}} = \int_0^1 r\:dx = r_0 \int_0^1 (1 - \alpha^2x^2)\:dx = r_0 \left[x - \tfrac13\alpha^2x^3\right]_0^1 = r_0(1 - \tfrac13\alpha^2) = 0.73r_0, and then t = 1 r average = 1 0.73 r 0 ≈ 1.37 t 0 . t = \frac 1{r_{\text{average}}} = \frac 1{0.73\:r_0} \approx 1.37\:t_0. Why is this incorrect?

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