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)
#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%
How did you get your answer?
#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?
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
6. To find the graph of the velocity function, you need to take the derivative of the motion function s(t). The velocity function v(t) is the derivative of the position function s(t). In this case, s(t) = 2e^(-1.5t) sin(2πt).
To find the derivative of s(t), you can apply the chain rule and the product rule. The chain rule states that if you have a function of a function, you need to multiply the derivative of the outer function with the derivative of the inner function. The derivative of the exponential function e^(-1.5t) is -1.5e^(-1.5t), and the derivative of sin(2πt) is 2πcos(2πt). When you multiply these derivatives, you get the velocity function v(t).
Therefore, v(t) = -3e^(-1.5t) sin(2πt) + 4πe^(-1.5t)cos(2πt).
Now, to determine the graph of the velocity function for 0 ≤ t ≤ 4, you can evaluate the function v(t) at different values of t within this interval and plot the corresponding points on a graph. The graph you provided in the link will show how the velocity changes over time.
7. When using the tangent line approximation to estimate the percentage error in volume, you can consider the formula for the percentage error:
Percentage error = (Change in volume / Actual volume) * 100.
In this case, the change in volume is equal to the length of the edge multiplied by the change in edge length, which is given as 0.1 cm. The actual volume is 125,000 cm^3.
Using the formula, the percentage error can be calculated as:
Percentage error = (0.1 cm / 50 cm) * 100 = 0.2%.
Therefore, the correct option is not listed. The percentage error in volume is 0.2%, not 1.2%.
8. To find the rate of change of the depth of the water when it is 10 feet deep, you can use the concept of related rates. The volume of a cone can be calculated using the formula V = (1/3)πr^2h, where r is the radius of the base and h is the height (or depth) of the cone.
In this case, the radius of the base is half of the diameter, which is 14 feet / 2 = 7 feet. The volume of the cone is given as 12 ft^3/min.
Differentiating the volume formula with respect to time, you get:
dV/dt = (1/3)π(2rh)(dh/dt),
where dV/dt represents the rate of change of volume, dh/dt represents the rate of change of depth, and 2r represents 14 feet.
Substituting the given values, you get:
12 ft^3/min = (1/3)π(14 feet)(dh/dt).
Now, you can solve for dh/dt:
dh/dt = (12 ft^3/min) / [(1/3)π(2)(14 feet)].
Simplifying this expression, you find:
dh/dt ≈ 0.449 ft/min.
Therefore, the rate of change of the depth of the water when it is 10 feet deep is approximately 0.449 ft/min.
9. To find the approximate location of the critical point of the function f(x) = ln(x)/x^2 in the interval (0, 5), you need to find the values of x where the derivative of f(x) is equal to zero.
First, find the derivative of f(x) using the quotient rule. The derivative of ln(x) is 1/x, and the derivative of x^2 is 2x. Applying the quotient rule, you get:
f'(x) = [(x^2)(1/x) - (ln(x))(2x)] / (x^2)^2
= (x - 2xln(x)) / x^4.
Setting f'(x) equal to zero and solving for x, you get:
x - 2xln(x) = 0.
This equation does not have an algebraic solution, so you can approximate the critical point by using numerical methods such as trial and error or a graphing calculator.
By inputting values within the given interval (0, 5) into the equation x - 2xln(x) = 0, you can determine which value(s) make the equation equal to zero. In this case, the approximate location of the critical point is (2, 0.173).
Therefore, the correct option is (2, 0.173).