Mathematics, 04.09.2019 00:30 wolfking800

Aladder 29 feet long leans against a wall and the foot of the ladder is sliding away at a constant rate of 3 feet/sec. meanwhile, a firefighter is climbing up the ladder at a rate of 2 feet/sec. when the firefighter has climbed up 6 feet of the ladder, the ladder makes an angle of ? /3 with the ground. answer the two related rates questions below. (hint: use two carefully labeled similar right ) if h is the height of the firefighter above the ground, at the instant the angle of the ladder with the ground is ? /3, find dh/dt= feet/sec.(b) if w is the horizontal distance from the firefighter to the wall, at the instant the angle of the ladder with the ground is ? /3, find dw/dt= feet/sec.

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Answer from: grahuntley31

dh/dt=1.373 ft/s

dw/dt=2.78 ft/s

Step-by-step explanation:απ

We need to find dh/dt. Looking at the diagram, we'll write h as a function of known parameters:

h=z sin(α) where h is the height of the firefighter from the ground, and z the distance along the ladder α is the angle of the ladder.

Deriving:

$\frac{dh}{dt} =\frac{dz}{dt}*sin(\alpha )+z*cos(\alpha )*\frac{d\alpha }{dt}$

We have $\frac{dz}{dt}$ given in the exercise but we need $\frac{d\alpha }{dt}$

Again, looking at the drawing we can write α in term of x, since we can calculated x and we have $\frac{dx }{dt}$

29 *cos(α)=x where x is the distance from the base of the ladder to the wall.

Deriving:

$-29*sin(α)\frac{d\alpha }{dt}=\frac{dx }{dt}$

$\frac{d\alpha }{dt}=\frac{dx }{dt}\frac{-1 }{29*sin(\alpha) }$

replacing out known values: α=π/3, $\frac{dx }{dt}=3$ , z=6, $\frac{dz}{dt}=2$

$\frac{d\alpha }{dt}=\frac{-2\sqrt{3}}{29}$

Now that we have dα/dt:

$\frac{dh}{dt} =\frac{dz}{dt}*sin(\alpha )+z*cos(\alpha )*\frac{d\alpha }{dt}=2*\frac{\sqrt{3}}{2}+6*0.5*\frac{-2\sqrt{3}}{29}=1.373 \frac { ft}{s}$

Now we do a similar process for dw/dt, looking at the drawing:

w=(29-z)*cos(α)

$\frac{dw}{dt}=-\frac{dz}{dt}*cos(\alpha )-(29-z)*sin(\alpha)*\frac{d\alpha }{dt}$

Replacing known values:

$\frac{dw}{dt}=2*.5-23*\frac{\sqrt{3}}{2}*\frac{-3\sqrt{3}}{29*2}=2.78 \ \frac { ft}{s}$

Aladder 29 feet long leans against a wall and the foot of the ladder is sliding away at a constant r

Answer from: Quest

i'm sorry mate, but i am not seeing any figures. i am unable to solve this for you.

Answer from: Quest

Answer: f(x) = -2x^3 +3x^2 +11x -6see attachedan infinite number. since the magnitude of the leading coefficient is not specified, it may be any negative number. (we have chosen the smallest magnitude integer that makes all coefficients be integers.)step-by-step explanation:

1. when "a" is a root of a polynomial, (x -a) is a factor of it. for the three roots given, the factors of the desired polynomial are (x +2)(x -1/2)(x -3).

in order to make the leading coefficient be negative, we need to multiply this product by a negative number. any negative number will do, but we choose a small (magnitude) value that will eliminate the fraction: -2.

then

f(x) = -2(x +2)(x -1/2)(x -3) = -(x +2)(2x -1)(x -3)

= -(2x² +3x -2)(x -3)

= -(2x³ -3x² -11x +6)

f(x) = -2x³ +3x² +11x -6

2. a graph created by the desmos on-line graphing calculator is shown, and the zeros are highlighted.

3. as indicated in part 1, the multiplier of this equation can be anything and the zeros will remain the same. you want a negative leading coefficient, so the "anything" is restricted to any of the infinite number of numbers that will make that be the case.

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Aladder 29 feet long leans against a wall and the foot of the ladder is sliding away at a constant r...