Question

A student must learn $M$ unfamiliar words for an upcoming test. The rate at which the student learns is proportional to the number of items remaining to be learned, with constant of proportionality equal to $k$. Initially, the student knows none of the words. Let $y(t)$ stand for the number of words that the student knows at time $t$. (a) Write down the right hand side of the differential equation satisfied by $y$. (Your answer should be given in terms of $y$.) $\frac{dy}{dt} = -k(M - y)$ (b) What is the initial value of $y$? (Give an exact answer.) y(0) = 0 (c) Suppose that there are 140 words to be learned and that $k = 0.3$ per hour for this student. Recall that the solution to $y' = k(M - y)$ with $y(0) = 0$ is given by $y(t) = M(1 - e^{-kt})$. (Give your answers correct to at least three decimal places.) How many hours would it take the student to learn the first 22 percent of the words? 0.828 hours How long would it take to learn the next 22 percent of the words? 1.105 hours How long would it take to learn the next 22 percent of the words? 2.492 hours (d) The student has just finished learning the first 22 percent of the words. (Give your answers correct to at least three decimal places.) How long would it take to learn 22 percent of the remaining words? hours After doing this, how long would it take to learn 22 percent of the words that then remain? hours

          A student must learn $M$ unfamiliar words for an upcoming test. The rate at which the student learns is proportional to the number of items remaining to be learned, with constant of proportionality equal to $k$. Initially, the student knows none of the words. Let $y(t)$ stand for the number of words that the student knows at time $t$.
(a) Write down the right hand side of the differential equation satisfied by $y$. (Your answer should be given in terms of $y$.)
$\frac{dy}{dt} = -k(M - y)$
(b) What is the initial value of $y$? (Give an exact answer.)
y(0) = 0
(c) Suppose that there are 140 words to be learned and that $k = 0.3$ per hour for this student. Recall that the solution to $y' = k(M - y)$ with $y(0) = 0$ is given by $y(t) = M(1 - e^{-kt})$. (Give your answers correct to at least three decimal places.)
How many hours would it take the student to learn the first 22 percent of the words?
0.828 hours
How long would it take to learn the next 22 percent of the words?
1.105 hours
How long would it take to learn the next 22 percent of the words?
2.492 hours
(d) The student has just finished learning the first 22 percent of the words. (Give your answers correct to at least three decimal places.)
How long would it take to learn 22 percent of the remaining words?
 hours
After doing this, how long would it take to learn 22 percent of the words that then remain?
 hours

A student must learn M unfamiliar words for an upcoming test. The rate at which the student learns is proportional to the number of items remaining to be learned, with constant of proportionality equal to k. Initially, the student knows none of the words. Let y(t) stand for the number of words that the student knows at time t.
(a) Write down the right hand side of the differential equation satisfied by y. (Your answer should be given in terms of y.)
(dy)/(dt) = -k(M - y)
(b) What is the initial value of y? (Give an exact answer.)
y(0) = 0
(c) Suppose that there are 140 words to be learned and that k = 0.3 per hour for this student. Recall that the solution to y' = k(M - y) with y(0) = 0 is given by y(t) = M(1 - e^-kt). (Give your answers correct to at least three decimal places.)
How many hours would it take the student to learn the first 22 percent of the words?
0.828 hours
How long would it take to learn the next 22 percent of the words?
1.105 hours
How long would it take to learn the next 22 percent of the words?
2.492 hours
(d) The student has just finished learning the first 22 percent of the words. (Give your answers correct to at least three decimal places.)
How long would it take to learn 22 percent of the remaining words?
hours
After doing this, how long would it take to learn 22 percent of the words that then remain?
hours

Added by Samuel F.

Question

Please give Ace some feedback