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A version of the following problem first appeared in print in the 16th Century, A frog and a toad fall to the bottom of a well that is 50 feet deep - Scottish Highers Maths - Question 3 - 2015
Question 3
A version of the following problem first appeared in print in the 16th Century, A frog and a toad fall to the bottom of a well that is 50 feet deep. Each day, the f... show full transcript
Worked Solution & Example Answer:A version of the following problem first appeared in print in the 16th Century, A frog and a toad fall to the bottom of a well that is 50 feet deep - Scottish Highers Maths - Question 3 - 2015
Step 1
Calculate $f_2$, the height of the toad at the end of the second day.
Answer
To calculate , use the recurrence relation defined for the frog: f_{n+1} = rac{1}{3}f_n + 32 Substituting into the equation: f_2 = rac{1}{3}(32) + 32 = rac{32}{3} + 32 = rac{32}{3} + rac{96}{3} = rac{128}{3} Thus, f_2 ext{ (the height of the toad after 2 days)} = rac{128}{3} ext{ feet} ext{ or approximately } 42.67 ext{ feet}.
Step 2
Determine whether or not either of them will eventually escape from the well.
Answer
To determine if either the frog or the toad will escape, we analyze their heights over time using their respective recurrence relations:
For the frog: f_{n+1} = rac{1}{3}f_n + 32 Setting the limit:
L - rac{1}{3}L = 32 \ rac{2}{3}L = 32 \ L = 48$$ Since 48 feet is less than 50 feet, the frog will not escape. For the toad: $$g_{n+1} = rac{3}{4}g_n + 13$$ Setting the limit: $$L = rac{3}{4}L + 13 \ L - rac{3}{4}L = 13 \ rac{1}{4}L = 13 \ L = 52$$ Since 52 feet is greater than the depth of the well (50 feet), the toad will escape. In conclusion: - The frog will not escape as its limit is 48 feet. - The toad will escape as its limit is 52 feet.