Help checking my work please!
The region R is a rectangle with vertices P (a,ln a), Q (a, 0), S (3, 0), and T (3, ln a), where 1 < a < 3
(graph: gyazo.com/9e52cd3cd10fad297232efadb9f94d87)
A. Write an expression that gives the area of the rectangle as a function of a.
Area of a rectange = length x breadth. 1 < a < 3 so length = 3-a. The breadth = ln a.
So area, A = (3-a)•(ln a)
B. The area of the rectangle is maximized for some c between 1 and 3. Write the expression you would need to solve in order to find c. You don’t need to find c.
3/c - 1 - ln(c) = 0 such that -3/(c^2) - 1/c < 0.
C. What is the rate of change of the area when a =3, if a is decreasing at a rate of da/dt = − 0.5 units / sec? You don’t need to simplify, of course. Use appropriate units.
A = (3 - a)•ln(a)
dA/dt = -ln(a)(da/dt) + (3 - a)(da/dt)/a
= -ln(3)(-.5)un/sec + (3 - 3)(-.5)un/(3sec)
= .5•ln(3)un/sec
= ln(√3) units/sec
D. Again, a is decreasing at the rate of da/dt = − 0.5 units / sec. At a = 2.9, is the area of the rectangle increasing or decreasing? How do you know? Use appropriate units in your answer.
dA/dt = -ln(a)(da/dt) + (3 - a)(da/dt)/a
= -ln(2.9)(-.5)un/sec + (3 - 2.9)(-.5)un/(2.9sec)
= .5•ln(2.9)un/sec - .05un/(2.9sec)
≈ 0.515units/sec
> 0 units/sec
Hence, the area of the rectangle is increasing at a = 2.9
2. The shaded regions R1 and R2, shown below, are enclosed by the graphs of f(x)= −x^2 and g(x)= −2^x. (gyazo.com/7bcf02392ff69e2a1280588308342e8e)
A. Find the x- and y-coordinates of the three points of intersection of the graphs of f and g
I need help on this one!
B. Estimate the area of R1 to three reasonable places past the decimal.
The area of R1 is the integral from x ~ -0.76667 to x = 2 of
(-x^2+2^x) dx. Integrating these functions gives
-(1/3)x^3 + 2^x/(ln 2)
At the upper limit, the value is -8/3 + 4 ln(2)
At the lower limit, the value is about 0.55762
So area R1 is about -2.66667 + 2.77250 - 0.55762 = about 0.4517
C. Find the EXACT area of R2. (No calculator approximations; don’t simplify.)
The area of R2 is the integral from x = 2 to 4 of
(x^2-2^x) dx. Integrating these functions gives
(1/3)x^3 - 2^x/(ln 2).
area of R2 is exactly (64/3) - 16 ln(2) - (8/3) + 4 ln(2)
(56/3) - 12 ln(2)
D. Find the exact volume of the solid formed by revolving R2 about the x-axis. (No calculator approximations; don’t simplify.)
Area of each annulus = pi x^4 - pi 2^(2x)
Thickness of each annulus = dx
Integrate A dx from x = 2 to x = 4
The integrated function is (pi/5) x^5 - (pi/ln 2) [ 2^(2x) ]
The volume is (1024 pi/5) - (256 pi/ln 2) - (32 pi/5) + (16 pi/ln 2).
pi [ 992/5 - (240 pi/ln 2) ]
Hey! I think you made a mistake in 2B, remember that the derivative is (2^x/ln2), you multiplied by ln 2.
Also, for 2A, graph the two functions and write down the three coordinates where the two graphs meet (-0.7667,-0.588), (2,-4), and (4,-16)
I think you're good
Do you think you could help me on 2A?
A. Your expression for the area of the rectangle is correct: A = (3-a) * ln(a)
B. The expression you would need to solve in order to find c is correct: (3/c) - 1 - ln(c) = 0
C. The rate of change of the area when a = 3 can be calculated as follows:
dA/dt = -ln(a) * (da/dt) + (3 - a) * (da/dt) / a
Plugging in the values, we get:
dA/dt = -ln(3) * (-0.5) + (3 - 3) * (-0.5) / 3
Simplifying, we get:
dA/dt = 0.5 * ln(3) units/sec
So the rate of change of the area when a = 3 is ln(sqrt(3)) units/sec.
D. To determine if the area of the rectangle is increasing or decreasing at a = 2.9, we need to calculate dA/dt using the given rate of change of da/dt:
dA/dt = -ln(a) * (da/dt) + (3 - a) * (da/dt) / a
Plugging in the values, we get:
dA/dt = -ln(2.9) * (-0.5) + (3 - 2.9) * (-0.5) / 2.9
Simplifying, we get:
dA/dt ≈ 0.515 units/sec
Since the value is greater than 0, the area of the rectangle is increasing at a = 2.9.
For the second problem:
A. To find the points of intersection of the graphs of f(x) and g(x), we need to solve the equation f(x) = g(x):
−x^2 = −2^x
One intersection point is clearly x = 0.
To find the other two points, we need to solve the equation numerically or graphically.
B. Your estimation for the area of R1 seems reasonable based on the given information.
C. The exact area of R2 can be found by integrating the function (x^2-2^x) from x = 2 to 4:
∫(x^2-2^x) dx from x = 2 to 4
This integral needs to be evaluated analytically to find the exact area.
D. The exact volume of the solid formed by revolving R2 about the x-axis can be found by integrating the function pi * (x^2-2^x)^2 from x = 2 to 4:
∫(pi * (x^2-2^x)^2) dx from x = 2 to 4
This integral needs to be evaluated analytically to find the exact volume.
A. The expression for the area of the rectangle is correct. The length of the rectangle is given by 3-a, and the breadth is ln(a). So, the area A = (3-a)·(ln(a)).
B. To find the value of c at which the area of the rectangle is maximized, you need to solve the equation 3/c - 1 - ln(c) = 0. Since you don't need to find the value of c, you can leave the equation as it is.
C. To find the rate of change of the area when a = 3, you can use the formula for the derivative of the area function with respect to time:
dA/dt = -ln(a) · (da/dt) + (3 - a) · (da/dt)/a
Substituting the given values: a = 3 and da/dt = -0.5, we get:
dA/dt = -ln(3) · (-0.5) + (3 - 3) · (-0.5)/3
Simplifying further, we get:
dA/dt = 0.5 · ln(3)
So, the rate of change of the area when a = 3 is ln(√3) units/sec.
D. To determine if the area of the rectangle is increasing or decreasing when a = 2.9, you can again use the formula for the derivative of the area function:
dA/dt = -ln(a) · (da/dt) + (3 - a) · (da/dt)/a
Substituting the given values: a = 2.9 and da/dt = -0.5, we get:
dA/dt = -ln(2.9) · (-0.5) + (3 - 2.9) · (-0.5)/2.9
Simplifying further, we get:
dA/dt ≈ 0.515 units/sec
Since the value is positive, the area of the rectangle is increasing at a = 2.9.
#1 looks ok
#2 looks ok
good work