Related Rates
(Mathematical Terms)
(Mathematical Terms)
It is dy/dt, if x2 + y2 = 61, when x = 5 and dx/dt = 4
ANSWER: What is 10/3?
Solution:
Find y
x2 + y2 = 61
25 + y2 = 61
y2 = 36
y = 6
Find the derivative of the formula with respect to time.
x2 + y2 = 61 ; d/dt
d/dt [x2 + y2 = 61]
2x dx/dt + 2y dy/dt = 0
Plug in the given values.
2(5)(4) + 2(6) dy/dt = 0
40 + 12 dy/dt = 0
12 dy/dt = -40
dy/dt = 10/3
The radius of a circle is decreasing at a rate of 4cm/min. How fast is the area changing when the radius is 8cm?
Answer:
The area is changing at the speed of -64pi cm2/min.
Given:
dr/dt = -4
R = 8
Required:
dA/dt = ?
Equation:
A=pi r2
Solution:
A=pi r2
dA/dt= pi 2r2 dr/dt
dA/dt= pi 2(8) (-4)
dA/dt= -64pi
Source: Organic Chemistry "Related Rates"
It is the larger number between 2 numbers whose product is 10 and difference is a maximum.
ANSWER: What is -10?
Solution:
let x be the first number.
let y be the second number.
let M be the minimum product.
If xy = 10, and x - y = M, then we can find x by manipulating the constraint equation to y = 10/x and substituting the new equation to the objective equation.
M = x - 10/x
Finally, we can get the value of x by finding the derivative of the equation above and equating it to zero.
M' = 1 + 10/x2
0 = 1 + 10/x
0 = x + 10
x = -10
Source: Organic Chemistry Tutor (Optimization Problems - Calculus, YouTube)
A farmer has 2400 ft of fencing and wants to fence off a rectangular field that borders a straight river. If no fence is needed along the river, what are the dimensions of the field that has the largest area?
ANSWER: What is 600 ft deep and 1200 ft wide?
1. Draw a diagram.
2. Choose the best formulas that fits the problem. One will be used for substitution while the other will be used as our main equation. In this case, the formula for the area and the perimeter of a rectangle will be used. (P = 2x + 2y & A = xy)
[Note: Since no fence is needed along the river, the formula for its perimeter will be changed to: P=x + 2y]
3. Plug in the values.
2x + y = 2400
y = 2400 - 2x
4. Substitute this to the other formula. (A = xy)
A = x (2400 - 2x)
A = 2400x - 2x2
5. Differentiate.
A' = 2400 - 2x
6. Make A' into 0 and find x.
0 = 2400 - 2x
2x = 2400
x = 1200
7. Substitute 1200 to the equation (2x + y = 2400)
2(1200) + y = 2400
y = 600
Answer: 600 ft deep and 1200 ft wide
Source: Stewart, J. (2015). Calculus: Early Transcendentals (8th ed.). Cengage Learning.
A sheet of cardboard 3 ft. by 4 ft. will be made into a box by cutting equal-sized squares from each corner and folding up the four edges. What will be the dimensions of the box with largest volume?
ANSWER: What is 3.03 ft2?
1. Let variable x be the length of one edge of the square cut from each corner of the sheet of cardboard.
After removing the corners and folding up the flaps, we have an ordinary rectangular box.
2. Find the best formula for it.
V = LWH
V = (4 - 2x) (3 - 2x) (x)
3. Then do the following steps.
4. But x is not equal to 1.77 since variable x measures a distance. In addition, the short edge of the cardboard is 3 ft., so it follows that x is greater than equal to zero and lesser than equal to 1.50.
Therefore x must be .57ft which makes our V = 3.03 ft2.
Answer: The Largest volume will be V = 3.03 ft2.
Reference:
https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/maxminsoldirectory/MaxMinSol.html#SOLUTION%205
It is dz/dt, if x2 + y2 = z2, when dx/dt = 2, dy/dt = 4, x = 12, and y = 5
ANSWER = What is 44/13?
Solution:
Find z.
122 + 52 = z2
144 +25 = z2
z = 13
Find the derivative of the formula with respect to time.
x2 + y2 = z2 ; d/dt
d/dt [x2 + y2 = z2]
2x dx/dt + 2y dy/dt = 2z dz/dt
Plug in the given values.
2(12)(2) + 2(5)(4) = 2(13) dz/dt
48 + 40 = 26 dz/zt
88 = 26 dz/zt
dz/zt = 44/13
The surface area of a snowball at a rate of 6 square feet per hour, how fast is the diameter changing when the radius is 2 ft?
Answer:
The diameter is changing at the rate of 3/4 pi ft/hr.
Given:
r = 2
d = 4
dSA/dt = -6
Required:
dD/dt = ?
Equation:
SA=4 pi r2
D = 2r (looking for diameter) = r = D/2
SA=4 pi (d/2)2
=pi D2
Solution:
SA/dt = 4(2d)dD/dt
-6 = (2)(4)dD/dt
dD/dt =(-6/8)pi =(-3/4)pi
Source: Organic Chemistry "Related Rates"
It is the minimum product of 2 numbers whose sum is 20.
ANSWER: What is -30?
Solution:
let x be the first number.
let y be the second number.
let M be the minimum product.
if x + y = 20, and xy = M, then we can solve for M by manipulating the first equation to y = 20 - x and substituting it to the second equation.
M = x (20 - x)
M = 20x - x2
Now that we have our equation, we can get its first derivative and equate it to zero.
M' = 20 - 2x
0 = 20 - 2x
-2x = 20
x = -10
Great! Now that we have the value of x, we can use it to solve for y!
y = 20 - x
y = 20 - (-10)
y = 30
Now that we have our x and y values, we can solve for M by multiplying them!
M = xy
M = (-10)(30)
M = -30
Source: Organic Chemistry Tutor (Optimization Problems - Calculus, YouTube)
Find the point on the parabola y2 = 2x that is closest to the point (1, 4).
ANSWER: What is (2, 2)?
1. Visualize the problem.
2. Find the best formula that fits the question and plug in the points. (distance formula)
√(x - 1)2 + (y - 4)2
[Note: The whole thing is under a square root.]
3. Given the parabola y2 = 2x, x = 1/2 y2
d = √(1/2 y2 - 1)2 + (y - 4)2
d2 = (1/2 y2 - 1)2 + (y - 4)2
[This is taking off the square.]
[d2 is easier to work with and since there are no restrictions with y, it's okay to use this.]
4. Differentiate.
d' = 2 (1/2 y2 - 1)y + 2(y - 4)
d' = y3 - 8
5. Make d' into 0 and solve for the unknown.
0 = y3 - 8
y = 2
6. Substitute y to this equation (x = 1/2 y2)
x = 1/2 (4)
x = 2
Answer:
The point on y2 = 2x closest to (1, 4) is (2, 2).
The distance between them is √5 .
Source: Stewart, J. (2015). Calculus: Early Transcendentals (8th ed.). Cengage Learning.
It is dA/dt, if A = πr2, when r = 4 and dr/dt = 6
ANSWER: What is 48π?
Solution:
Find the derivative of the formula with respect to time.
A = πr2 ; d/dt
d/dt A = π d/dt r2
dA/dt = π x 2r x dr/dt
Plug in the given values.
dA/dt = π x 2(4) x 6
dA/dt = 48π
A spherical balloon is inflated with gas at a rate of 750 cubic centimeters per minute (cm^3/min), how fast is the radius of the balloon changing when the radius is 10 cm?
Answer:
The radius of the balloon is changing at 15/8 pi cm/min.
Given:
r = 10
dV/dt = 750 cm3/min
Required:
dr/dt = ?
Equation:
V =4/3 pi r3
Solution:
dV/dt =4/3 pi 3r2 dr/dt
750 = 4 pi 102 dr/dt
dr/dt = 750/400 pi = 15/8 pi
It is the 2 negative numbers, whose product is 50 and whose sum is a maximum.
ANSWER: What is -10 and -5?
Solution:
let x be the first number.
let y be the second number.
let M be the maximum product.
if xy = 50, and x + y = M, then we can solve for M by manipulating the first equation to y = 50/x and substituting it to the second equation.
M = x + (50/x)
M = x + 50x-1
Now that we have our equation, we can get its first derivative and equate it to zero.
M' = x + 50x-1
M' = 1 - 100x-2
0 = 1 - 100x-2
0 = 1 - 100/x2
0 = x2 - 100
x2 = 100
x = -10
Great! Now that we have the value of x, we can use it to solve for y!
y = 50/x
y = 50/(-10)
y = -5
Source: Organic Chemistry Tutor (Optimization Problems - Calculus, YouTube)
A cylindrical can is to be made to hold 1 liter of oil. Find the dimensions that will minimize the cost of the metal to manufacture the can.
ANSWER: What is 3√500/π and 2 x 3√500/π?
1. Draw a diagram.
2. Find the best equation / formula that fits the problem. (A = 2πr2 + 2πrh)
We would like to express "A" in terms of one variable, r. To eliminate h we use the fact that the volume is given at 1 liter, which is equivalent to 1000 cm3. Thus πr2h = 1000 and h = 1000 / πr2.
3. Substitute h to the equation, A = 2πr2 + 2πrh.
A = 2πr2 + 2πr (1000 / πr2)
A = 2πr2 + 1000 / r
Note: r > 0
4. Differentiate.
A' = 4πr - (2000 / r2)
A' = 4(πr3 - 500) / r2
5. Make A' = 0 and find r.
0 = 4(πr3 - 500) / r2
πr3 = 500
r = 3√500 / π
6. Substitute r to the original equation and find h.
h = 1000 / πr2
h = 2 x 3√500/π
Answer:
The radius is 3√500 / π.
The height is twice the radius.
Source: Stewart, J. (2015). Calculus: Early Transcendentals (8th ed.). Cengage Learning.
It is dr/dt, if V = 4/3π r3, when dV/vt = 5 and r = 3
ANSWER: What is 5/36π?
Solution:
Find the derivative of the formula with respect to time.
V = 4/3 πr3 ; d/dt
d/dt V = 4/3 π d/dt r3
dV/dt = 4/3 π x 3r2 x dr/dt
Plug in the values.
5 = 4/3 π x 3(3)2 x dr/dt
5 = 36π dr/dt
dr/dt = 5/36π
A 13-foot ladder leans against a house. The ladder slides down the wall at a rate of 3 ft/min. How fast is the ladder moving away from the base of the wall when the foot of the ladder is currently 5ft from the wall? How fast is the angle between the ground and ladder changing?
Answer: What is 36/5 ft/min and -⅗ rad/min?
The ladder moving away 36/5 ft/min from the base of the wall when the foot of the ladder is currently 5ft from the wall.
The angle between the ground and ladder is changing at the speed of -⅗ rad/min.
Given:
Ladder/L = 13
dL/dt = 0
dy/dt = -3
X = 5
Required:
Y = ?
dx/dt = ?
dteta/dt = ?
Equation:
L2 = x2 + y2
sin(teta)= y/L or cos(teta) x/L
Solution:
L2 = x2 + y2
132= 52+y2
y2= 132-52 = 169-25 = 144
y = 12
L2 = x2 + y2
2L dL/dt = 2x dx/dt + 2y dy/dt
0 = (5)dx/dt + (12)(-3)
-5 dx/dt = -36
dx/dt = 36/5
sin(teta) = y/L
cos(teta)d(teta)/dt= 1/L dy/dt (cosine= adj/hyp = x/L = 5/13)
(5/13) d(teta)/dt= (1/13)(-3)
5 d(teta)/dt = -3
d(teta)/dt =-⅗
Source: Organic Chemistry "Related Rates"
It is a positive number where the sum of that number and its reciprocal is a minimum.
ANSWER: What is 1?
Solution:
let x be the positive number.
let 1/x be the number's reciprocal.
let M be the minimum sum.
Since in the equation x + 1/x = M, there is only one unknown value, we can solve for x by getting its derivative and equating it to zero.
M = x + 1/x
M = x + x-1
M' = x + x-1
M' = 1 - x-2
M' = 1 - 1/x2
0 = 1 - 1/x2
0 = x2 - 1
x = 1
(we pick the positive number because its part of the given information)
Source: Organic Chemistry Tutor (Optimization Problems - Calculus, YouTube)
Find the area of the largest rectangle that can be inscribed in a semicircle of radius r.
ANSWER: What is r2?
There are two solutions to this problem.
Solution 1
1.1 Visualize the Problem.
1.2 Let (x, y) be the vertex that lies in the first quadrant. Then the rectangle has sides of lengths 2x and y, so its area is A = 2xy.
1.3 Use the equation x2 + y2 = r2 to isolate y and get y = √r2 - x2.
1.4 Plug y into the equation. (A = 2x √r2 - x2)
1.5 Differentiate.
A' = 2(r2 - 2x2) / √r2 - x2
1.6 A' becomes 0 when 2x2 = r2, that is, x = r/√2. This value of x gives a maximum value of A since A (0) = 0 and A (r) = 0. Therefore the area of the largest inscribed triangle, when x is applied, is r2.
Solution 2
2.1 Visualize the problem but use angles as a variable.
2.2 Let theta (θ) be the angle shown in the figure above. Then the area of the rectangle will be the given:
A (θ) = (2r cosθ) (r sinθ) = r2sin2θ
2.3 Since the maximum value of sin2θ is 1 and it occurs when 2θ = π/2, A (θ) has a maximum value of r2 and it occurs when θ = π/4.
Answer:
The answer for both solutions is that the maximum area is r2.
Source: Stewart, J. (2015). Calculus: Early Transcendentals (8th ed.). Cengage Learning.