MHB 2.2.294 Euler's method word problem

[karush]

George makes a paper boat with his brother Bill and goes out in the rain to play with it. It falls in the stream along the curb, racing towards the sewer.
Let t be measured in seconds, p be the velocity of the paper boat in meters/second and g be George’s velocity, measured in meters/seconds.
At $t = 28$, if George has not retrieved his boat, it will fall in the sewer, Let the position of George and his boat both be 0 at $t = 0$
$$\begin{array}{rrrrrrr}
x&0&7&14&21&28\\
g(t)&0&5&8&6&4
\end{array}$$
a. George's velocity is modeled by the function $g(t)$. Certain values of g are given in the table. Estimate the value of $g'(10.5)$, including units of measure.

the table points are on this Desmos plot. the first 3 pts are not a straight line but could represent a section of a parabola
then take the d/dx?

View attachment 9314

ok actually not sure what to do with this and its due tom

 

[karush]

c. The velocity of the paper boat is modeled by the function
$\displaystyle p(t)=(e^{0.05t}-\cos(0.35t))\left[6-\dfrac{t}{5}\right]$
what is the value of $p'(7)$
$$p'=\frac{0.2\left(-t+30\right)\left(e^{0.05\left(t\right)}-\cos\left(0.35\left(t\right)\right)\right)}{t}$$
then
$$p'(7)=\frac{0.2\left(-t+30\right)\left(e^{0.05\left(t\right)}-\cos\left(0.35\left(t\right)\right)\right)}{t}=1.43868207031$$

 

[skeeter]

$t=10.5$ is midway between $t=7$ and $t=14$

best estimate with the given data ...

$g'(10.5) \approx \dfrac{g(14)-g(7)}{14-7} = \dfrac{3}{7} \, m/s^2$

 

[karush]

$t=10.5$ is midway between $t=7$ and $t=14$

best estimate with the given data ...

$g'(10.5) \approx \dfrac{g(14)-g(7)}{14-7} = \dfrac{3}{7} \, m/s^2$

thanks

 

[karush]

c. The velocity of the paper boat is modeled by the function
$\displaystyle p(t)=(e^{0.05t}-\cos(0.35t))\left[6-\dfrac{t}{5}\right]$
what is the value of $p'(7)$
$$p'=\frac{0.2\left(-t+30\right)\left(e^{0.05\left(t\right)}-\cos\left(0.35\left(t\right)\right)\right)}{t}$$
then
$$p'(7)=\frac{0.2\left(-t+30\right)\left(e^{0.05\left(t\right)}-\cos\left(0.35\left(t\right)\right)\right)}{t}=1.43868207031$$

done

 

[skeeter]

$p(t) = [e^{0.05t}-\cos(0.35t)] \cdot \left(6-\dfrac{t}{5}\right)$

$p'(t) = [e^{0.05t}-\cos(0.35t)] \cdot \left(-\dfrac{1}{5}\right) + \left(6-\dfrac{t}{5}\right) \cdot [0.05e^{0.05t}+0.35\sin(0.35t)]$

$p'(7) = [e^{0.35}-\cos(2.45)] \cdot \left(-\dfrac{1}{5}\right) + \left(\dfrac{23}{5}\right) \cdot [0.05e^{0.35}+0.35\sin(2.45)] = 0.9153$