Derivative and Maximum of a exponential function (study time and learning rate)

[Axl_Adler]

So below is the question:

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3. Sammy is studying for an exam in which there are 40 questions. Let \(\displaystyle Q(t)\) be the number of questions to which Sammy has learnt the answers, in time \(\displaystyle t\) hours.

(a) Sammy assumes that \(\displaystyle Q(t)\, =\, 40\, (1\, -\, e^{-t}).\) Show that \(\displaystyle \dfrac{dQ}{dt}\, =\, 40\, -\, Q.\)

(b) A friend points out to Sammy that she will forget some of what she has learnt. Sammy then estimates that \(\displaystyle Q(t)\) can now be given by:

. . . . .\(\displaystyle Q(t)\, =\, 40\, \left(1\, -\, e^{-t}\, -\, t\, e^{-t}\right)\)

(i) Show that the learning rate is \(\displaystyle \dfrac{dQ}{dt}\, =\, 40\, t\, e^{-t}\) for this new \(\displaystyle Q(t).\)

(ii) Hence, determine when Sammy's learning rate is a maximum.

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So I already successfully managed to do b(i) of the question. My troubles are with part (a) and part b(ii)

For (a) I found the derivative of the function to be 40e^-t. I cant seem to understand however how that is equivalent 40-Q. Could anyone explain the rules or what it means being that this is my first year doing calculus.

Also, For part b(ii) I found the derivative of function (plus it was given but we just had to prove it) which was 40te^-t. So to find the maximum i assumed i had to set it to 0 which in fact i did. I'm not sure if I'm doing the algebra wrong but I got t=0. Wouldn't that indicate no maximum? How would I got about solving this question?

Help is much appreciated, thank you

 

[Deleted member 4993]

3. Sammy is studying for an exam in which there are 40 questions. Let \(\displaystyle Q(t)\) be the number of questions to which Sammy has learnt the answers, in time \(\displaystyle t\) hours.

(a) Sammy assumes that \(\displaystyle Q(t)\, =\, 40\, (1\, -\, e^{-t}).\) Show that \(\displaystyle \dfrac{dQ}{dt}\, =\, 40\, -\, Q.\)

(b) A friend points out to Sammy that she will forget some of what she has learnt. Sammy then estimates that \(\displaystyle Q(t)\) can now be given by:

. . . . .\(\displaystyle Q(t)\, =\, 40\, \left(1\, -\, e^{-t}\, -\, t\, e^{-t}\right)\)

(i) Show that the learning rate is \(\displaystyle \dfrac{dQ}{dt}\, =\, 40\, t\, e^{-t}\) for this new \(\displaystyle Q(t).\)

(ii) Hence, determine when Sammy's learning rate is a maximum.

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So I already successfully managed to do b(i) of the question. My troubles are with part (a) and part b(ii)

For (a) I found the derivative of the function to be 40e^-t. I cant seem to understand however how that is equivalent 40-Q. Could anyone explain the rules or what it means being that this is my first year doing calculus.

Also, For part b(ii) I found the derivative of function (plus it was given but we just had to prove it) which was 40te^-t. So to find the maximum i assumed i had to set it to 0 which in fact i did. I'm not sure if I'm doing the algebra wrong but I got t=0. Wouldn't that indicate no maximum? How would I got about solving this question?

a) After you calculated dQ/dt - it is an algebra problem! Can you "simplify" the term "40 - Q", from the knowledge Q = 40*[1-e^(-t)].

 

[Steven G]

3. Sammy is studying for an exam in which there are 40 questions. Let \(\displaystyle Q(t)\) be the number of questions to which Sammy has learnt the answers, in time \(\displaystyle t\) hours.

(a) Sammy assumes that \(\displaystyle Q(t)\, =\, 40\, (1\, -\, e^{-t}).\) Show that \(\displaystyle \dfrac{dQ}{dt}\, =\, 40\, -\, Q.\)

(b) A friend points out to Sammy that she will forget some of what she has learnt. Sammy then estimates that \(\displaystyle Q(t)\) can now be given by:

. . . . .\(\displaystyle Q(t)\, =\, 40\, \left(1\, -\, e^{-t}\, -\, t\, e^{-t}\right)\)

(i) Show that the learning rate is \(\displaystyle \dfrac{dQ}{dt}\, =\, 40\, t\, e^{-t}\) for this new \(\displaystyle Q(t).\)

(ii) Hence, determine when Sammy's learning rate is a maximum.

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So I already successfully managed to do b(i) of the question. My troubles are with part (a) and part b(ii)

For (a) I found the derivative of the function to be 40e^-t. I cant seem to understand however how that is equivalent 40-Q. Could anyone explain the rules or what it means being that this is my first year doing calculus.

Also, For part b(ii) I found the derivative of function (plus it was given but we just had to prove it) which was 40te^-t. So to find the maximum i assumed i had to set it to 0 which in fact i did. I'm not sure if I'm doing the algebra wrong but I got t=0. Wouldn't that indicate no maximum? How would I got about solving this question?

Help is much appreciated, thank you

To do part a, you need to compute dQ/dt AND simplify 40-Q and then OBSERVE that they are the same (just as Subhotosh Khan has pointed out)

Also (as already pointed out) you just need algebra to see that the derivative (which you claim you found) and 40-Q are equal (Just compute 40 - (40(1-e^-t) ). If this subtraction is difficult for you to compute then you need to review your algebra as good algebra skills is mandatory to succeed in Calculus. Please notice that you were able to do the calculus (that is you found dQ/dt) but you can't show the two are equal, which is purely algebra.