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this assignment is on related rates. I believe questions 3/4 are on linear approximation and relate back to the last unit.
My problem is that I am entirely unconfident on this work, and am going to be taking a test on the material soon. If someone could check my work, I'd be delighted.
Homework Assignment
1. Fabio stands atop his 16 foot ladder when he realizes that the ladder
is slipping down the side of the building. He decides that the base of
the ladder is moving away from the bottom of the building at a rate
of 2 feet per second when it is 3 feet from the bottom of the building.
How fast is Fabio falling at that instant?
To do this problem I construct a triangle. Height is H Base is L and the distance between the two is 16.
H^2 + L^2 = 256. Therefore (dh/dt)2h +(dl/dt)2l=0 At this instant, h=3, which means l = sqrt(256-9) = sqrt(247). I am going to divide both sides of my equation by 2 to obtain (dh/dt)h +(dl/dt)l = 0. plugging in gives (x)(3) +(-2)(sqrt(247))=0; x = -2/3 sqrt (247)
2. A decaying nerf ball shrinks in such a way that its radius decreases
by 1/6th inch per month. How fast is the volume changing when the
radius is one quarter of an inch?
V=(pi)(4/3)(r^3)
dv/dt = (4)(pi)(r^2)(dr/dt)=4(pi)((1/4)^2)(-1/6)
For the following two questions, suppose that y = 2x^2 − 3x + 1.
3. Find and simplify a formula for the y increment, ∆y.
∆y = f(x+ ∆x) - f(x) = 2x^2 +4x(∆x)+2(∆x)^2-3x-3(∆x)+1-2x^2+3x-1= 4x(∆x)+2(∆x)^2 -3(∆x)
4. Find a formula for the y differential dy.
dy=f'(x)dx= (4x-3)dx
5. When a stone is dropped in a pool, a circular wave moves out from
the point of impact at a rate of six inches per second. How fast is the
area enclosed by the wave increasing when the wave is two inches in
radius?
A=(pi)(r^2)
dA/dt = dr/dt (2r)(pi) = (6)(2)(2)(pi) =24(pi) in^2/sec
6. The electric resistance of espresso as a function of its temperature is
given by
R = 6.000 + 0.002T 2
where R is measured in Ohms and T in degrees Celsius. If the tem-
perature is decreasing at the rate of 0.2 degrees per second, find the
rate of change of the resistance when T = 38 ◦ C.
dR/dt = 0.004 T (dT/dt)= 0.004 x 38 x -0.2 = -.0304
My problem is that I am entirely unconfident on this work, and am going to be taking a test on the material soon. If someone could check my work, I'd be delighted.
Homework Assignment
1. Fabio stands atop his 16 foot ladder when he realizes that the ladder
is slipping down the side of the building. He decides that the base of
the ladder is moving away from the bottom of the building at a rate
of 2 feet per second when it is 3 feet from the bottom of the building.
How fast is Fabio falling at that instant?
To do this problem I construct a triangle. Height is H Base is L and the distance between the two is 16.
H^2 + L^2 = 256. Therefore (dh/dt)2h +(dl/dt)2l=0 At this instant, h=3, which means l = sqrt(256-9) = sqrt(247). I am going to divide both sides of my equation by 2 to obtain (dh/dt)h +(dl/dt)l = 0. plugging in gives (x)(3) +(-2)(sqrt(247))=0; x = -2/3 sqrt (247)
2. A decaying nerf ball shrinks in such a way that its radius decreases
by 1/6th inch per month. How fast is the volume changing when the
radius is one quarter of an inch?
V=(pi)(4/3)(r^3)
dv/dt = (4)(pi)(r^2)(dr/dt)=4(pi)((1/4)^2)(-1/6)
For the following two questions, suppose that y = 2x^2 − 3x + 1.
3. Find and simplify a formula for the y increment, ∆y.
∆y = f(x+ ∆x) - f(x) = 2x^2 +4x(∆x)+2(∆x)^2-3x-3(∆x)+1-2x^2+3x-1= 4x(∆x)+2(∆x)^2 -3(∆x)
4. Find a formula for the y differential dy.
dy=f'(x)dx= (4x-3)dx
5. When a stone is dropped in a pool, a circular wave moves out from
the point of impact at a rate of six inches per second. How fast is the
area enclosed by the wave increasing when the wave is two inches in
radius?
A=(pi)(r^2)
dA/dt = dr/dt (2r)(pi) = (6)(2)(2)(pi) =24(pi) in^2/sec
6. The electric resistance of espresso as a function of its temperature is
given by
R = 6.000 + 0.002T 2
where R is measured in Ohms and T in degrees Celsius. If the tem-
perature is decreasing at the rate of 0.2 degrees per second, find the
rate of change of the resistance when T = 38 ◦ C.
dR/dt = 0.004 T (dT/dt)= 0.004 x 38 x -0.2 = -.0304
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