Derivative as a Rate of Change
A rectangular water tank is being filled at the constant rate of 20 liters / second. The base of the tank has dimensions w = 1 meter and L = 2 meters. What is the rate of change of the height of water in the tank?(express the answer in cm / sec).
The volume V of water in the tank is given by V = w*L*H. We know the rate of change of the volume dV/dt = 20 liter /sec. We need to find the rate of change of the height H of water dH/dt. V and H are functions of time. We can differentiate both side of the above formula to obtain dV/dt = W*L*dH/dt note W and L do not change with time and are therefore considered as constants in the above operation of differentiation.
We now find a formula for dH/dt as follows. dH/dt = dV/dt / W*L
We need to convert liters into cubic cm and meters into cm as follows: 1 liter = 1 cubic decimeter = 1000 cubic centimeters = 1000 cm 3 and 1 meter = 100 centimeter.
We now evaluate the rate of change of the height H of water. dH/dt = dV/dt / W*L= ( 20*1000 cm 3 / sec) / (100 cm * 200 cm) = 1 cm / sec.
Final Answer: 1 cm / sec.
A 50ft ladder is placed against a large building. The base of the ladder is resting on an oil spill, and it slips at the rate of 3 ft. per minute. Find the rate of change of the height of the top of the ladder above the ground at the instant when the base of the ladder is 30 ft. from the base of the building.
Organizing information: dy/dt=3, Goal: Find dx/dt when y = 30.
*use Pythagorean Theorem again: x2+302 = 502 ⇒ x = 40
And differentiating (notice how the hypotenuse is constant):
2xx’ + 2yy’ = 0x’ = -2yy'/2x= -yy'/x
Plugging in, x’= (-30)(3)40 = -2.25
Note: x’ is negative. It means that the distance x is decreasing—the ladder is slipping down the building.
Final Answer: -2.25
Find two positive numbers whose sum is 300 and whose product is a maximum.
Let’s call the two numbers x and y and we are told that the sum is 300, so x+y=300. We are also being asked to maximize the product, so A=xy.
We now need to solve the constraint for x or y and plug this into the product equation. y=300−x ⇒ A(x)=x(300−x)=300x-x2
The next step is to determine the critical points for this equation. A′(x)=300-2x → 300-2x=0 → x=150
In this case we can quickly see that A′′(x)=-2. From this we can see that the second derivative is always negative and so A(x) will always be concave down and so the single critical point we got must be a relative maximum and hence must be the value that gives a maximum product.
We already have x so we need to determine y and that is easy to do from the constraint. y=300−150=150
The final answer is then, x=150 y=150
If there are 200 bacteria after 2 hours and 800 bacteria after 5 hours?
Use the equation y = Cekt, where C is the initial amount, and k is the proportionality constant. We have two unknowns (C and k) and two (t,y) data points: (2,200) and (5,800).
Plug one of the points in and solve for C:
y=Cekt ; 200=Ce2k ; C = 200/e2k ; 800=Ce5k ; C = 800/e5k
Set the C’s together and cross-multiply to solve for k:
200/e2k = 800/e5k ; 200e2k = 800e2k ; e5k=4e2k ; ln(e5k)=ln(4e2k)
5k=ln4+2k ; 3k=ln4 ; k=ln4/3 = 0.462
Final Answer:0.462
Find the area of the region bounded by the graphs of y = x2 + 1 and y = x3 and the vertical lines x = 1 and x = 1.
After quickly plotting the graphs we see that x2 + 1 lies above x3 on the interval. So let f(x) = x2 + 1 and g(x) = x3. Since both are continuous (polynomials)
Area between f and g = ab∫[ f (x) - g (x) ] dx = -11∫[(x2 + 1) x3] dx
= x3/3+ x - x4/4
= ( 1/3+ 1 - 1/4) - (-1/3- 1 -14/) = 83
Final answer: 83
If s(t) = t2 - 5t, what is the position, velocity and acceleration at t = 2? Assume s is in feet and t is in seconds, and interpret these results.
s(t) = t2 - 5t + 3
v(t) = s'(t) = 2t - 5
a(t) = v'(t) = 2
s(2) = 2
(the object is located at +2 feet from the start)
v(2) = - 1
(headed in the negative direction, and it is moving backwards at a rate of one foot per second)
Final Answer:
a(2) = 2
(at that instant, its velocity is increasing by a rate of 2 feet per second each second.)
A spherical balloon is being inflated so that its diameter is increasing at a rate of 2 cm/min. How quickly is the volume of the balloon increasing when the diameter is 10 cm?
Organizing information: dd/dt=2, Goal: Find dV/dt when d=10.
*We use the volume formula for a sphere, but rewrite it with the diameter
V = 4/3πr3
V = 4/3π(d/2)3
V = π/6(d3)
Differentiate both sides with respect to t:
dV/dt= 3(π/6)d2(dd/dt)
Plug in dd/dt = 2 and d = 10,
dV/dt= 3(π/6)(10)2 *2 = 100πcm3/min
Final Answer:100πcm3/min
Find two positive numbers whose product is 750 and for which the sum of one and 10 times the other is a minimum.
Let’s call the two numbers x and y and we are told that the product is 750 or, xy=750.
We are then being asked to minimize the sum of one and 10 times the other, S=x+10y
We now need to solve the constraint for x or y and plug this into the product equation. x=750/y ⇒ S(y)=(750/y)+10y
The next step is to determine the critical points for this equation. Because we are told that y must be positive, we can eliminate the negative value. S’(y)=(-750/y^2)+10 → (-750/y^2)+10=0 → y=±√75=5√3
In this case we can quickly see that, S′′(y)=1500/y^3. From this we can see that, provided we recall that y is positive, then the second derivative will always be positive. Therefore, S(y) will always be concave up and so the single critical point that we can use must be a relative minimum and hence must be the value that gives a minimum sum.
We already have y so we need to determine x and that is easy to do from the constraint. x=750/5√3=50√3
The final answer is then, x=50√3 y=5√3
A city had a population of 2 million people in 2000, and a population of 2.5 million in 2010. Find the exponential growth model y=Cekt for the population growth of this city and use this model to predict its population in the year 2030.
For the equation y=Cekt, we already have y=2ekt (in millions), since we can begin counting at year 2000 (make t=0); this (t,y) data point is (0,2).
Let’s use the other (t,y) data point (10, 2.5) to solve for k (the population growth rate, or proportionally constant)
y=2ekt ; 2.5=2e10k ; 1.25=e10k ; k=ln(1.25)/10 = 0.0223
So the exponential growth model for the population is y=2e0.0223t
Find the population in the year 2030 (t=30)
y=2e0.0223(30) = 3.9 million
Final Answer: 3.9 million
Determine the area below f (x) = 3x + 2x - x2 and above the x-axis.
Let’s start off with getting a sketch of the region we want to find the area of.
We are assuming that, at this point, you are capable of graphing most of the basic functions that we’re dealing with in these problems and so we won’t be showing any of the graphing work here.
It should be clear from the graph the upper function is the parabola (i.e. y = 3 + 2x - x2) and the lower function is the x-axis (i.e. y = 0).
Since we weren’t given any limits on x in the problem statement we’ll need to get those. From the graph it looks like the limits are (probably) -1≤x≤3. However, we should never just assume that our graph is accurate or that we were able to read it accurately. For all we know the limits are close to those we guessed from the graph but are in fact slightly different.
So, to determine if we guessed the limits correctly from the graph let’s find them directly. The limits are where the parabola crosses the x-axis and so all we need to do is set the parabola equal to zero (i.e. where it crosses the line y = 0) and solve. Doing this gives...
3 + 2x - x2= 0 -( x + 1 ) ( x - 3 ) = 0 x = -1, x = 3
So, we did guess correctly, but it never hurts to be sure. That is especially true here where finding them directly takes almost no time.
At this point there isn’t much to do other than step up the integral and evaluate it.
We are assuming that you are comfortable with basic integration techniques so we’ll not be including any discussion of the actual integration process here we will be skipping some of the intermediate steps.
The area is…
A = ∫-133 + 2x - x2dx = ( 3x + x2-1/3x3) |-13 = 32/3
Final Answer: 32/3
A ball is dropped from a height of 64 feet. Its Height above ground (in feet) t seconds later is
given s(t) = -16t2 + 64. What is the instantaneous velocity of the ball when it hits the ground?
v(2)
v(t) = s’(t) = -16t2 + 64
= -32t
Final Answer:
v(t) = -64 ft./s
One car leaves a given point and travels north at 30 mph. Another car leaves 1 hour later, and travels west at 40 mph. At what rate is the distance between the cars changing at the instant the second car has been traveling for 1 hour?
Set up the problem by extracting information in terms of the variables x, y, and z, as pictured on the triangle:
First sentence: dx/dt= 30 and x(t) = 30t
Second sentence: dy/dt= 40 and y(t) = 40(t - 1) (start an hour later)
Goal: Find dz/dt at t = 2.
The property that combines the sides of a triangle is Pythagorean Theorem: x2 + y2 = z2
At √ t = 2, x(2) = 60 and y(2) = 40,
Use the Pythagorean Theorem: z(2) = √602 + 402 ≈ 72.111
Taking the derivative in t:
2x(dx/dt) + 2y (dy/dt) = 2z (dz/dt)
Plug in: 2*60*30 + 2*40*40 = 2*72.111(dz/dt)
Thus, dz/dt≈ 47.150
Final Answer: 47.150
Let x and y be two positive numbers such that x+2y=50 and (x+1)(y+2) is a maximum.
In this case we were given the constraint in the problem, x+2y=50. We are also told the equation to maximize, f=(x+1)(y+2).
So, let’s just solve the constraint for x or y and plug this into the product equation. x=50−2y ⇒ f(y) = (50−2y+1)(y+2) = (51−2y)(y+2) = 102+47y−2y^2
The next step is to determine the critical points for this equation. f′(y)=47−4y → 47−4y=0 → y=47/4
In this case we can quickly see that, f′′(y)=−4. From this we can see that the second derivative is always negative and so f(y) will always be concave down and so the single critical point we got must be a relative maximum and hence must be the value that gives a maximum.
We already have y so we need to determine x and that is easy to do from the constraint. x=50−2(47/4)=53/2
The final answer is then, x=53/2 y=47/4
Consider the population of bacteria described earlier. This population grows according to the function f(t)=200e0.02t, where t is measured in minutes. How many bacteria are present in the population after 5 hours (300 minutes)? When does the population reach 100,000 bacteria?
We have f(t)=200e0.02t. Then 𝑓(300)=200𝑒0.02(300) ≈80,686.
There are 80,686 bacteria in the population after 5 hours.
To find when the population reaches 100,000 bacteria, we solve the equation
100,000=200e0.02t
500=e0.02t
ln500=0.02t
t=ln5000.02≈310.73.
Final Answer:
The population reaches 100,000 bacteria after 310.73 minutes.
Determine the area of the region bounded by y = x2+ 2, y = sin(x), x = -1 and x = 2.
Let’s start off with getting a sketch of the region we want to find the area of.
We are assuming that, at this point, you are capable of graphing most of the basic functions that we’re dealing with in these problems and so we won’t be showing any of the graphing work here.
It should be clear from the graph that the upper function is y = x2+ 2 and the lower function is y = sin(x).
Next, we were given limits on x in the problem statement and we can see that the two curves do not intersect in that range. Now that this is something that we can’t always guarantee and so we need the graph to verify if the curves intersect or not. We should never just assume that because limits were given in the problem statement that the curves will not intersect anywhere between the given limits.
So, because the curves do not intersect we will be able to find the area with a single integral using the limits :
A = ∫-12x2 + 2 - sin(x) dx = (1/3x3+ 2x + cos(x) ) |-12 = 9 + cos (2) - cos (1) = 8.04355
Final Answer: = 8.04355
Determine where the following function is increasing and decreasing.
A(t) = 27t5 - 45t4 - 130t3 + 150
Derivative of function:
A’(t) = 135t4 - 180t3 - 390t2
= 15t2 (9t2 - 12t - 26)
t = 0
t = 12 ± √144 - 4(9)(-26) ÷ 18
= 12 ± √1080 ÷ 18
= 12 ± 6 √30 ÷ 18
= 2 ± √30 ÷ 3
= -1.159, 2.492
Final Answer:
Increasing: - ∞ < t < -1.159, 2.492 < t < ∞
Decreasing: -1.159 < t < 0, 0< t < 2.492
A cylindrical fountain is filled with juice. The can has a 10 cm radius. How fast does the height of the juice in the can drop when the drink is being drained at 5 cm3/sec?
V= πr2h
dV/dt= π(2rh(dr/dt) + (dh/dt)r2)
-5 = π(2(10)(0) + dh/dt(10)2)
-5 = π(100 dh/dt)
-120= dhdt
Final Answer: -120= dhdt
We are going to fence in a rectangular field. If we look at the field from above the cost of the vertical sides are $10/ft, the cost of the bottom is $2/ft and the cost of the top is $7/ft. If we have $700 determine the dimensions of the field that will maximize the enclosed area.
We are told that we have $700 to spend and so the cost of the material will be the constraint for this problem. The cost for the material is then, 700=10y+2x+10y+7x=20y+9x. We are being asked to maximize the area so that equation is, A=xy.
Now, let’s solve the constraint for y, which is y=35−(9/20)x. Plugging this into the area formula gives, A(x)=x(35−(9/20)x)=35x−(9/20)x^2
Finding the critical point(s) for this shouldn’t be too difficult at this point so here is that work. A′(x)=35−(9/10)x → 35−(9/10)x=0 → x=350/9
The second derivative of the area function is, A′′(x)=−9/10. From this we can see that the second derivative is always negative and so A(x) will always be concave down and so the single critical point we got must be a relative maximum and hence must be the value that gives a maximum area.
Now, let’s finish the problem by getting the second dimension. y=35−9/20(350/9)=35/2
The final dimensions are then, x=350/9 y=35/2
Assume a population of fish grows exponentially. A pond is stocked initially with 500 fish. After 6 months, there are 1000 fish in the pond. The owner will allow his friends and neighbors to fish on his pond after the fish population reaches 10,000. When will the owner’s friends be allowed to fish?
We know it takes the population of fish 6 months to double in size. So, if 𝑡 represents time in months, by the doubling-time formula, we have
6=(ln2)/𝑘. Then, 𝑘=(ln2)/6. Thus, the population is given by y=500e((ln2)/6)t. To figure out when the population reaches 10,000 fish, we must solve the following equation:
10,000=500e(ln2/6)t
20=e(ln2/6)t
ln20=(ln2/6)t
t=6(ln20)/ln2 ≈25.93.
Final Answer:
The owner’s friends have to wait 25.93 months (a little more than 2 years) to fish in the pond.
Use the method of disks/rings to determine the volume of the solid obtained by rotating the region bounded by y = √x , y = 3, and the y-axis about the y-axis.
We need to start the problem somewhere so let’s start “simple”.
Knowing what the bounded region looks like will definitely help for most of these types of problems since we need to know how all the curves relate to each other when we go to set up the area formula and we’ll need limits for the integral which the graph will often help with.
We now need to find a formula for the area of the disk. Because we are using disks that are centered on y-axis we know that the area formula will need to be in terms of y. This in turn means that we’ll need to rewrite the equation of the boundary curve to get into terms of y.
In other words,
Radius = y2
The area of the disk is then,
A (y) = π(Radius) 2= (y2)2= πy4
The final step is to then set up the integral for the volume and evaluate it.
For the limits on the integral we can see that the “first” disk in the solid would occur at y = 0 and the “last” disk would occur at y = 3. Our limits are then: 0≤y≤3.
The volume is then,
V = ∫03πy4dy = 1/5πy5|03 = 243/5π
Final Answer: 243/5π
An airplane is flying in a straight direction and at a constant height of 5000 meters. The angle of elevation of the airplane from a fixed point of observation is a. The speed of the airplane is 500 km / hr. What is the rate of change of angle a when it is 25 degrees? (Express the answer in degrees / second and round to one decimal place)
S:
dx/dt = 500 km/hr
tan a = h/x
d(tan a)/dt = d(h/x)/dt
d(tan a)/dt = (sec2a) da/dt
d(h/x)/dt = h*(-1 / x2) dx/dt.
(note: height h is constant)
(sec2a) da/dt = h*(-1 / x2) dx/dt
da/dt = [ h*(-1 / x2) dx/dt ] / (sec2a)
x = h / tan a
da/dt = [ h*(- tan2a / h2) dx/dt ] / (sec2a)
= [ (- tan2a/h) dx/dt ] / (sec2a)
= (- sin2a / h) dx/dt
Final Answer:
da/dt = [- sin2(25 deg)/5000 m]*[500 000 m/3600 sec]
= -0.005 radians/sec
= -0.005 * [ 180 degrees / Pi radians] /sec
= -0.3 degrees/sec
You are looking at the New York ball drop on New Year’s Eve at a distance of 100 m away from the base of the structure. If the ball drops at a constant rate of 2 m/s, what is the rate of change of the angle between you and the ball when the angle is /4?
x/100= tanθ
1/100(dx/dt)= sec2(θ) dθ/dt
1/100(2) = (2)dθ/dt
dθ/dt=1/100rad/s
Final Answer: 1/100rad/s
We have 45 m^2 of material to build a box with a square base and no top. Determine the dimensions of the box that will maximize the enclosed volume.
We are told that we have 45 m^2 of material to build the box and so that is the constraint.
The amount of material that we need to build the box is, 45=lw+2(lh)+2(wh)=w^2+2wh+2wh=w^2+4wh
We are being asked to maximize the volume so that equation is, V=lwh=w^2h
Note as well that we went ahead and used the fact that l=w in both of these equations to reduce the three variables in the equation down to two variables.
Now, let’s solve the constraint for h, h=(45-w^2)/4w
Plugging this into the volume formula gives, V(w)=w^2(45-w^2)/4w) = (1/4)w(45-w^2) = (1/4)(45w-w^3)
Plugging this into the volume formula gives, V(w)=w^2(45-w^2)/4w) = (1/4)w(45-w^2) = (1/4)(45w-w^3)
now, let's Find the critical point(s) V′(w)=1/4(45-3w^2) → 1/4(45-3w^2)=0 → w=±√45/3=√15
Because we are dealing with the dimensions of a box, the negative width doesn’t make any sense, and so the only critical point that we can use here is: w=√15.
The second derivative of the volume function is, V′′(w)=−(3/2)w. Provided w is positive, the single critical point we got must be a relative maximum.
Now, let’s finish the problem by getting the remaining dimensions. l=w=√15=3.8730 h=(45-15)/4√15=1.9365
The final dimensions are then, l=w=3.8730 h=1.9365
A colony of bacteria is growing exponentially. At time t=0 it has 10 bacteria in it, and at time t=4 it has 2000. At what time will it have 100,000 bacteria?
Even though it is not explicitly demanded, we need to find the general formula for the number f(t) of bacteria at time t, set this expression equal to 100,000, and solve for t. Again, we can take a little shortcut here since we know that c=f(0) and we are given that f(0)=10. (This is easier than using the bulkier more general formula for finding c). And use the formula for k
k:
k=lnf(t1)−lnf(t2)/t1−t2
=ln10−ln2,000/0−4
=ln(10/2,000)/−4
= ln200/4
Therefore, we have
f(t)=10⋅e(ln(200/4)t)=10⋅200(t/4)
as the general formula. Now we try to solve
100,000=10⋅e(ln(200/4))t
for t: divide both sides by the 10 and take logarithms, to get
ln10,000=ln(200/4)t
Thus,
t=4(ln10,000/ln200)
Final Answer: 6.953407835.
Use the method of disks/rings to determine the volume of the solid obtained by rotating the region bounded by y = 7 - x2, x = -2, x = 2 and the x-axis about the x-axis.
We now need to find a formula for the area of the disk. Because we are using disks that are centered on the x-axis we know that the area formula will need to be in terms of x.Therefore, the equation of the curve will need to be in terms of x.
The radius of the disk is the distance from the x-axis to the curve defining the edge of the solid. In other words,
Radius = 7 - x2
The are of the disk is then,
A (x) = (Radius)2= (7 - x2 )2= (49 -14x2+x4)
The final step is to then set up the integral for the volume and evaluate it.
For the limits on the integral we can see that the “first” disk in the solid would occur at x = -2 and the “last” disk would occur at x = 2. Our limits are then : -2 x 2.
The volume is then,
V = ∫-22 π(49 -14x2+x4) dx = π(49 -14/3x3+1/5x5)|-22 = 2012/15π
Final Answer: 2012/15π