Derivatives
Find the derivative of the function f(x)=6x4+8x2+12x
Answer: 24x3+16x+12
[Sol]
Since f'(x)=nxn-1,
f'(x)=4(6x3-1)+2(8x2-1)+1(12x1-1)
=24x2+16x+12
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?
Answer: 100πcm3/min
[Sol]
Organizing information:
dd/dt=2
Goal: Find dV/dt when d=10
Using the volume formula for a sphere, but with the diameter rewritten
V=4/3πr3
V=4/3π(d2)3
V=π/6d3
Differentiate both sides with respect to t:
dV/dt=3π/6d2 dd/dt
Plug in dd/dt=2 and d=10
dV/dt=3π/6(10)2(2)=100πcm3/min
A rectangular is to be made with one side being made of stone, and three being made of wood fences. Given that you can make about 100ft of fencing, determine the dimensions that would give you the maximum area for the garden.
Answer: 25ft, 25ft, 50ft
Solution:
Let x be the sides perpendicular to the stonewall, and y the side that’s parallel to the stonewall.
A=xy
The constraint equation will be 2xy = 100
To solve for y, we have y = 100 - 2x, thus we can write,
A(x) = x(100 - 2x) = 100x-2x2
Determine the domain, which we need it to be y > 0 and x >0. If y > 0, and x < 50. Then,
A(x)= 100x -2x2 over the interval [0, 50]
Then, A'(x) = 100 - 4x
We then determine that the max are occurs when x = 25
Then we have
y = 100-2x=100-2(25)=50
To maximize the area of the garden, let x = 25 and y = 50. The area of the garden is 1250 ft2
Differentiate f(x)=(1+5x)/ln(x)
Answer: f'(x)=[5ln(x)-(1/x)-5]/[ln(x)]2
[Sol]
Apply Quotient rule: d/dx(f(x)/g(x))=[g(x)f'(x)-f'(x)g(x)]/[g(x)]2
f'(x)={ln(x)[d/dx(1+5x)]-(1+5x)[d/dx(ln(x))]}/[ln(x)]2
=[5ln(x)-(1+5x)(1/x)]/[ln(x)]2
=[5ln(x)-(1/x)-5]/[ln(x)]2
A stone dropped in a pond sends out a circular ripple whose radius increases at a constant rate of 4 ft/sec. After 12 seconds, how rapidly is the area enclosed by the ripple increasing?
Answer: 384πft2/sec
[Sol]
Organizing information:
dr/dt=4
Goal: Find dA/dt when t=12
Using the area formula for a circle:
A=πr2
We differentiated both sides with respect to t:
dA/dt=2πr(dr/dt)
Plug in dr/dt=4. When t = 12 seconds, r = 4(12)=48
dA/dt=2π(48)4=384πft2/sec
A metal box without a top is to be constructed from a square sheet of metal that is 20cm on a side by cutting square pieces of the same size from the corners of the sheet and then folding up the sides. Find the dimensions with the largest volume that can be constructed.
Answer: 10/3 cm x 40/3 cm x 40/3 cm
[Sol]
V(x) = x(20 - 2x)(20 - 2x) = 400x - 80x2 +4x3
V’(x) = 400 -160x +12x2
400 - 160x+12x2= 0
4(100-40x+3x2) = 0
4(3x-10)(x-10) = x=10/3, 10
V”(x)= -160+24x V”(10/3) = -160+80<0 V”(10) = -160+240>0
By the 2nd derivative test, the dimensions would be 10/3 cm by 40/3 cm by 40/3 cm
Differentiate f(x)=3xlog(x)
Answer: f'(x)=[ln(3)](3x)[log(x)]+3x/ln(10)x
[Sol]
Apply Product rule:
f'(x)=d/dx(3x)[log(x)]+(3x)d/dx[log(x)]
=(3x)[ln(3)log(x)]+(3x)[1/ln(10)x]
=[ln(3)](3x)[log(x)]+3x/ln(10)x
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.
Answer: -2.25
[Sol]
Organizing Information:
dy/dt=3
Goal: Find dx/dt when 𝑦=30
Using the Pythagorean Theorem:
x2+302=502
x=40
And differentiating (see how the hypotenuse is constant):
2xx'+2yy'=0x' =-2yy'/2x=-yy'/x
Plugging in, x'=-30340=-2.25.
Note: x' is negative, meaning the distance x is decreasing — the ladder is slipping down the building.
You want to sell a certain number of your items to maximize your profit in a market. Market research tells you that if you set the price at $1.50 per piece, you will be able to sell 5000 items, and for every 10 cents you lower your price below $1.50, you would be able to sell another thousand items. Suppose that your fixed costs (start-up cost) total to $2000, and per cost of production(marginal cost) is $0.50, find the price set per item and the number of items sold in order to maximize your profit.
Answer: Max profit is $3625, when the price is set at $1.25 and sell 7500 items
[Sol]
P=nx - 2000 - 0.50n
n= 5000 + 1000(1.5 - x)/10 because (1.5 - x)
Substitute n in the function:
P(x) = (5000 + 1000(1.5 - x)/0.10) - 2000 - 0.5(5000 + 1000(1.5 - x)/0.10)
= -10000x2 + 25000x - 12000
P'(x) = -20000x + 25000, which is 0 when x = 1.25
Max profit is $3625, when the price is set at $1.25 and sell 7500 items