Please check my answers! I'm bad at explaining/showing my process and I'm trying to improve so if there's anything you think could help to add, please tell me!

Question

1. Region R is bounded by the x-axis, y-axis, x = 3 and y = (1)/(sqrt(x+1))

A. Find the area of region R
∫[0,3] 1/√(x+1) dx
= 2√(x+1) [0,3]
= 2

B. Find the volume of the solid formed when region R is revolved about the x-axis
using discs, v=∫πy^2 dx
= π∫[0,3] 1/(x+1) dx
= πlog(x+1) [0,3]
= πlog4

C. The solid formed in Part B is divided into two solids of equal volume by a plane perpendicular to the
x-axis. Find the x-value where this plane intersects the x-axis.
c where
π∫[0,c] 1/(x+1) dx = π∫[c,3] 1/(x+1) dx
log(c+1)-log(0+1) = log(3+1)-log(c+1)
2log(c+1) = log(4)
log(c+1) = log(2) (and not because 4/2=2)
c = 1

2. In this problem, you will investigate the family of functions h(x) = x + cos(ax), where a is a positive constant such that 0 < a < 4.

A. For what values of a will h(x) have a relative maximum at x = 1?
I need to find a such that h'(1) = 0 and h''(1) < 0.
Now h'(x) = 1 - a*sin(ax) and h''(x) = -a^2*cos(ax), so I need a such that
h'(1) = 1 - a*sin(a) = 0 ----> sin(a) = 1/a, and cos(a) > 0, since a > 0.
I can't solve sin(a) = 1/a algebraically, so I need to use numerical
techniques and/or a graphing calculator. Using my calculator I find possible solutions a = 1.114 and a = 2.773 to 3 decimal places. However, only for
a = 1.114 is h''(x) < 0, so the only value for a such that h(x) has a relative
maximum at x = 1 is a = 1.114 to 3 decimal places.

B. For what value(s) of a will h(x) have an inflection point at x = 1?
I need to find a such that h''(x) = 0 at x = 1. Now h''(x) = -a^2*cos(ax),
so h''(1) = -a^2*cos(a). With a not 0, I require that cos(a) = 0, which is the
case when a = (pi/2) + n*pi for any integer n. For 0 < a < 4 the only potential solution is a = pi/2. Now with a = pi/2 I have
h''(0.5) = -(pi/2)^2 * cos(pi/4) = -(pi)^2 / (4*sqrt(2)) < 0 and
h''(2) = -(pi/2)^2 * cos(pi) = (pi)^2 /4 > 0, so there is in fact an inflection point
at x = 1 when a = pi/2.

C. Which values of a make h(x) strictly decreasing? Justify your answer.
It requires that h'(x) < 0 for all real x.
Now h'(x) = 1 - a*sin(ax), so I need a such that
(1 - a*sin(ax)) < 0 ----> a*sin(ax) > 1 for all x.
This is not possible since for any a such that 0 < a < 4 there will be values x
such that sin(ax) < 0, so there is no solution to this question. Any help on this question?

3. The water usage rate for Seattle in a given 24 h period is represented in the table and graph below
(Note: t = 0 corresponds to midnight). Time is measured in hours, and the rate of water usage is measured in millions of gallons per hour.
(table and graph: gyazo.com/d50055e0132b4aee114b14f432b971ce)

A. Using three trapezoids of equal width, estimate ∫(24 on top, 0 on bottom)rate(t) dt.
Explain the physical meaning of this definite integral.
Three groups (0,8) (8,16) (16,24)
Use the trapezoid equation => ½h(b₁+b₂)
since ½ and h (h = 8) stay the same for every trapezoid factor them out
½(8) (.5 + 1.7 +1.7 + 1.5+ 1.5 + .5) = 29.6 million gallons of water were used from t =0 to t=24

B. Does the estimate in Part A overestimate or underestimate the value of ∫(b on top, a on bottom)rate(t) dt.
on the interval 0 < t < 8? Explain your reasoning.
Overstimates, because the rate between 0 and 8 is larger than the rates at 0, at 2 at 4 at 6 and at 8

C. Using your answer from Part A, estimate the average water usage over the 24 h period.
To find the average water usage just divide the answer in part (a) by 24 hrs
Average water usage = 29.6/24 = 1.2333 million gallons per hour

D. Estimate the slope of the usage rate curve at t = 12. Describe the method that you use.
using the slope formula (points around 12) use x = 10 and x = 14
slope = (1.4-1.5)/(14-10) = -.025

(help on 4 and 5 if it's not too much trouble. If you could show the work it'd be really helpful for me)
4. The function f(x) has the value f(−1) = 1. The slope of the curve y = f(x) at any point is given by the expression dy/dx = (2x + 1)(y + 1).

A. Write an equation for the line tangent to the curve y = f(x) at x = − 1.

B. Use the tangent line from Part A to estimate f(−0.9)

C. Use separation of variables to find an explicit or implicit formula for y = f(x), with no integrals remaining.

D. Find lim f(x).
x->∞

5. The figure at right is the graph of f ′′(x), the second derivative of a function f(x). The domain of the function f(x) is all real numbers, and the graph shows f ′′(x) for −2.6 ≤ x ≤ 3.6.
(graph: gyazo.com/5d0dc89b23539eeb95294947b85f4a69)

A. Find all values of x in the interval (−2.6, 3.6) where f ′(x) has a horizontal tangent.

B. Find all values of x in the interval (−2.6, 3.6) where f(x) is concave upwards. Explain your answer.

C. Suppose it is known that in the interval (−2.6, 3.6), f(x) has critical points at x =1.37 and x = −0.97. Classify these points as relative maxima or minima of f(x). Explain your answer.

6. Water is flowing at the rate of 50 m^3/min into a holding tank shaped like a cone, sitting vertex down. The tank’s base diameter is 40 m and the tank is 10 m high.

A. Write an expression for the rate of change of the water level with respect to time, in terms of h (the water’s height in the tank).
using similar triangles, the radius of the water surface at height h can be found

r/h = (40/2)/10 = 2
so, r = 2h

v = 1/3 π r^2 h
= π/3 (2h)^2 h
= 4π/3 h^3

dv/dt = 4π h^2 dh/dt
50 = 4π h^2 dh/dt

dh/dt = 25/(2πh^2)

B. Assume that at t = 0, the tank is empty. Find the water level, h, as a function of the time, t.
25/(2π) t= ∫ h² dh
25/(2π) t= 1/3 h³ +C
But at t=0, h=0 so C=0.
So h(t)=(75/(2π) t)^(1/3)

C. What is the rate of change of the radius of the cone with respect to time when the water is 8 ft deep?
r(t)/h(t)=2 so when the water is 8 ft. deep (2.44m) we have r/2.44=2, r=4.88 m at this time. So I used r/h=2 in V(t) = 1/3 π r² h to get:
V(t) = 1/6 π r³
V'(t) = 1/2 π r² dr/dt
50 = 1/2 π (4.88)² dr/dt (should I solve further?)

Answer 1

#1 ok
#2 ok. for 2C, you would have to restrict the domain, which they did not do.
#3B kinda right. You might also consider that the graph is concave up for the first two trapezoids, so the overestimate the area. For the 3rd area, the trapezoid fits more nearly the actual graph.
#4
A. y' = (2x+1)(y+1)
so, at (-1,1), y' = (2*-1+1)/(1+1) = -1/2
so the line with slope -1/2 through (-1,1) is
y-1 = -1/2 (x+1)
B. f(-0.9) ≈ f(-1) + (-1/2)(0.1) = 1 - .05 = 0.95
C.
dy/dx = (2x+1)(y+1)
dy/(y+1) = (2x+1) dx
ln(y+1) = x^2+x+c
y = c*e^(x^2+x) - 1
y(-1) = 1, so
c*e^0 - 1 = 1
c = 2
so, y = 2e^(x^2+x) - 1
D y(∞) = ∞

#5 A. f' has a horizontal tangent where f" = 0
So, that would be at x ≈ -0.9, 2.1
B. f is concave up where f" > 0
So, that would be (-2,1)U(3,3.6)
C. max if f" < 0, so x = -0.97
min if f" > 0, so x = 1.37

#6 ok
I'd take your final answer as ok, but you might want to check with your teacher on just how "numeric" the answer needs to be.

good work there.

Answer 2

good job on #1
#2, I agree with what you did
in #2 c, I agree there is no solution for 0<x<4
I plotted the curve in Wolfram, and by changing the value of a, you can see
that as a ---> 0 , eg. a = .001 , we approach a line with a positive slope
for values of a >0 , e.g. a = 5, there will always be a maximum point for 0<x<4
and that maximum point is "creeking" closer to (0,1).
I changed a to 1.114 and it showed a max at x = 1, confirming your answer to b)

#3, your method is correct, you will just have to trust your arithmetic.

#4 dy/dx = (2x + 1)/(y + 1)
(y+1) dy = (2x+1) dx , think cross-multiplying
divide by dx
(y+1) dy/dx = (2x+1) dx/dx
integrate both sides:
(1/2)(y+1)^2 = (1/4)(2x+1)^2
2(y+1)^2 = (2x+1)^2 <---- check by differentiating this implicitly
solve this equation for y and you have d)
when x = 1, 2(y+1)^2 = 9
y+1 = 3/√2 , y = 3/√2 - 1
when x = 1 , dy/dx = (2x + 1)/(y + 1) = 3/(3/√2) = √2
tangent equation: y - 3/√2 + 1 = √2(x - 1)
simplify to the required form

leaving #5 go

#6. good
in 6c) since we know that dh/dt = 25/(2πh^2)
when h = 8, dh/dt = 25/(128π)
and we know r = 2h, so
dr/dt = 2dh/dt
so when h=8, dr/dt = 25/(64π)

So far that is all I looked at

Answer 3

forgot the link to Wolfram for #2c
https://www.wolframalpha.com/input/?i=h(x)+%3D+x+%2B+cos(ax)++for+a+%3D+20

Also noticed that oobleck did the others

Answer 4

Thank you both so much for taking your time to help me!!