Answer
a.) The graph of S(t) is shown below.
b.) S(t) is concave up on the interval $0 < t < \sqrt{\dfrac{65}{3}}$ and concave down on the interval $\sqrt{\dfrac{65}{3}} < x < \infty$
c.) Since $S'(t)$ is always positive and $S''(t)$ is negative after t ≈ 4.655, typing speed continually increases at a decreasing rate after t ≈ 4.655. This means that the typing speed will converge and according to the function will never increase above 100 words per minute (It will come very close to 100 but will never be greater).
Work Step by Step
a.) The graph of S(t) is shown below.
b.) First find $S''(t)$
$S(t) = \dfrac{100t^2}{65 + t^2}$
$S'(t) = \dfrac{(65 + t^2)(200t) - (100t^2)(2t)}{(65 + t^2)^2} = \dfrac{(13000t + 200t^3) - (200t^3)}{(65 + t^2)^2} = \dfrac{13000t}{t^4 + 130t^2 + 4225}$
$S''(t) = \dfrac{(t^4 + 130t^2 + 4225)(13000) - (13000t)(4t^3 + 260t)}{(t^4 + 130t^2 + 4225)^2}$
$S''(t) = \dfrac{(13000t^4 + 1690000t^2 + 54925000) - (52000t^4 + 3380000t^2)}{(65 + t^2)^4}$ $S''(t) = \dfrac{54925000 -39000t^4 - 1690000t^2}{(65 + t^2)^4} = \dfrac{−13000(t^2+65)(3t^2−65)}{(65 + t^2)^4} = \dfrac{−13000(3t^2−65)}{(65 + t^2)^3}$
Now find when $S''(t) = 0$
$0 =\dfrac{−13000(3t^2−65)}{(65 + t^2)^3} \rightarrow 0 =−13000(3t^2−65) \rightarrow 0 = 3t^2 -65$
$65 = 3t^2$
$t = \sqrt{\dfrac{65}{3}} ≈ 4.655$
Determine the concavity of $S(t)$
S''(4) > 0
S''(5) < 0
Therefore S(t) is concave up on the interval $0 < t < \sqrt{\dfrac{65}{3}}$ and concave down on the interval $\sqrt{\dfrac{65}{3}} < x < \infty$
c.) $S'(t)$ > 0 for $t > 0$
Since $S'(t)$ is always positive and $S''(t)$ is negative after t ≈ 4.655, typing speed continually increases at a decreasing rate after t ≈ 4.655. This means that the typing speed will converge and according to the function will never increase above 100 wpm.
