Question

in a certain memory experiment, subject A is able to memorize words that are right of given by m’(t)=-0.012t^2+0.2t (words per min). in the same memory experiment, subject B is able to memorize at the rate given by M’(t)=-0.006t^2+0.2t (words per min) how many more words does subject B memorize from t=0 to t=20 (during the first 20 minutes) round to the nearest word

Answer 1

Given the question in the image, the following are the solution steps to answer the question.

STEP 1: Write the given functions

`\begin{gathered} m^(\prime)(t)=-0.012t^2+0.2t \\ M^(\prime)(t)=-0.006t^2+0.6t^2 \end{gathered}`

STEP 2: Find m'(20) for subject A

`\begin{gathered} When\text{ t=20,} \\ \int_0^(20)-0.012t^2+0.2t \\ Apply\text{ the sum/difference rule} \\ =-31.999999992+40=8.0000008 \\ \\ \therefore M^(\prime)(20)=8.00000008 \end{gathered}`

STEP 3: Find M'(20) for subject B

`\begin{gathered} When\text{ t=20} \\ \int_0^(20)-0.006t^2+0.2t \\ =-15.999999996+40=24.00000004 \end{gathered}`

STEP 4: Find how many more words subject B memorizes more than subject A

`\begin{gathered} Difference=Subject\text{ B - Subject A} \\ =24.00000004-8..00000008\approx16 \end{gathered}`

Hence, Answer is approximately 16 words