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!
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?)
#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.
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
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
Thank you both so much for taking your time to help me!!
1.
A. Your process for finding the area of region R is correct. You integrated the function and found the result to be 2.
B. Your process for finding the volume of the solid formed when region R is revolved about the x-axis is correct. You used the formula for the volume of a solid of revolution and integrated the function to find the result to be πlog(4).
C. Your process for finding the x-value where the plane intersects the x-axis is correct. You set up two integrals, one for each region, and equated them. Solving for the x-value, you found c to be 1.
2.
A. Your process for finding the values of a for which h(x) has a relative maximum at x = 1 is correct. You set up the conditions for a relative maximum and solved for a using numerical techniques and a graphing calculator. You found that a = 1.114 to 3 decimal places. Great job!
B. Your process for finding the values of a for which h(x) has an inflection point at x = 1 is correct. You set up the condition for an inflection point and found that a = pi/2 satisfies this condition. Well done!
C. Your explanation for why there is no solution to the question of which values of a make h(x) strictly decreasing is correct. For any value of a between 0 and 4, there will always be values of x where sin(ax) is negative, so h'(x) will not be negative for all x.
3.
A. Your process for estimating the definite integral using three trapezoids of equal width is correct. You divided the interval into three segments and used the formula for the area of a trapezoid to estimate the integral to be 29.6 million gallons of water used from t = 0 to t = 24.
B. Your explanation for why the estimate in Part A overestimates the value of the integral on the interval 0 < t < 8 is correct. The rate between 0 and 8 is larger than the rates at 0, 2, 4, 6, and 8.
C. Your process for estimating the average water usage over the 24-hour period is correct. You divided the answer from Part A by 24 to find the average water usage to be 1.2333 million gallons per hour.
D. Your process for estimating the slope of the usage rate curve at t = 12 is correct. You used the slope formula and the points (10, 1.4) and (14, 1.5) to find the slope to be -0.025.
4.
A. To write an equation for the line tangent to the curve y = f(x) at x = -1, you need to find the slope of the curve at that point and use the point-slope form of a line. The slope of the curve is given by dy/dx = (2x + 1)(y + 1), so plug in x = -1 to find the slope and then use the point (-1, f(-1)) to write the equation in point-slope form.
B. To estimate f(-0.9) using the tangent line from Part A, substitute x = -0.9 into the equation for the tangent line and solve for y.
C. To find an explicit or implicit formula for y = f(x) with no integrals remaining, you need to separate the variables and integrate both sides of the equation dy/dx = (2x + 1)(y + 1).
D. To find the limit of f(x) as x approaches infinity, you need to examine the behavior of the function as x becomes very large. You can use algebraic techniques like factoring or dividing by the highest power of x to determine the limit.
5.
A. To find the values of x in the interval (-2.6, 3.6) where f'(x) has a horizontal tangent, you need to look for points where the second derivative f''(x) equals zero.
B. To find the values of x in the interval (-2.6, 3.6) where f(x) is concave upwards, you need to look for points where the second derivative f''(x) is positive.
C. To classify the critical points at x = 1.37 and x = -0.97 as relative maxima or minima, you can use the second derivative test. Evaluate the second derivative at those points, and if it is positive, the point is a relative minimum, and if it is negative, the point is a relative maximum.
6.
A. Your expression for the rate of change of the water level with respect to time, dh/dt, in terms of h (the water's height in the tank) is correct. You used the formula for the volume of a cone to relate the rate of change of volume with the rate of change of height.
B. Your equation for h as a function of time, t, is correct. You integrated the rate of change of height equation and applied the initial condition to find the equation for h(t).
C. To find the rate of change of the radius of the cone with respect to time when the water is 8 ft deep, you first need to find the value of r at that height using the relationship r/h = 2. Then you can use the formula for the rate of change of volume to find the rate of change of radius, dr/dt.