I'd appreciate corrections for the ones I get wrong. :) Thanks!

6. A point on a damped spring has motion given by s(t)=2e^(-1.5t) s in2(pi)t, where s is measured in centimeters and t is measured in seconds. Choose which of these is the graph for the velocity function for 0 < or equal to t < or equal to 4
gyazo.com/3f1404ac65a1bd108cbe2b3efb5edcb8
(not sure about how to solve this one)

7. The edge of a cube was found to have a length of 50 cm with a possible error in measurement of 0.1 cm. Based on the measurement, you determine that the volume is 125,000 cm^3. Use tangent line approximation to estimate the percentage error in volume.

0.6%
0.9%
*1.2%
1.5%
1.8%

8. An inverted conical tank (with vertex down) is 14 feet across the top and 24 feet deep. If water is flowing in at a rate of 12 ft^3/min, find the rate of change of the depth of the water when the water is 10 feet deep.

0.007 ft/min
*0.449 ft/min
0.018 ft/min
0.051 ft/min
0.065 ft/min

9. For the function f(x)=Inx/x^2, find the approximate location of the critical point in the interval (0, 5).

(0.5, −2.773)
(1, 0)
(1.649, 0.184)
*(2, 0.173)
(0.778, −1.813)

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1. #6
s(t)=2e^(-1.5t) sin2πt
so the period is 2π/2π = 1
You want the graph that starts out at (0,0) and increases, and you want 4 periods.
The first max will be at t=1/4, so s(1/4) = 2e^(-3/8) ≈ 1.37
Looks like B to me

#7
Hmmm.
dv = 3s^2 ds = 3*2500*0.1 = 750
750/125000 = 0.006 = 0.6%

#8
using similar triangles, when the water has a depth of y, the radius of the surface is 7/24 y. So, the volume of the water is
V = 1/3 π r^2y = π/3 (7/24)^2 * y^3
dV/dt = π/3 (7/24)^2 * 3y^2 dy/dt
Thus, when y=10,
12 = (7/24)^2 * π * 100 dy/dt
dy/dt = 1728/(1225π) ≈ 0.449 ft/min
good work

#9
f(x) = lnx/x^2
f'(x) = (1-2lnx)/x^3
f'(x)=0 when 1-2lnx=0, or x=√e ≈ 1.649
How did you arrive at x=2?

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2. Thank you so much for your time and help!
#9 I resolved and got the answer: (1.649, 0.184) but it wouldnt let me edit the post.
#7 was a guess because I was quite confused on how to solve it

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