Question

A ladder 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 triangles.)(a) 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.

Answer 1

Answer:

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:

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

Answer 2

Part(a): The required value is $\frac{dh}{dt} =1.3737 ft/sec$

Part(b): The required value is $\frac{dw}{dt} =1.3793 ft/sec$

Differentiation:

Differentiation, in calculus, can be applied to measure the function per unit change in the independent variable.

Now, the given part is,

Let $x$ be the distance from the will to the base of the ladder.

$y$ be the distance along the ladder that the firewall climbed $w$ to be the distance from the firefighter.

$h$ be the height from firefighter $\theta$ be the angle of the base of the ladder with the ground

So, using the information we draw the below-attached diagram as,

Part(a): From the diagram,

$sin\theta=\frac{h}{y} \\h=y sin\theta\\\frac{dh}{dt}=sin \theta\frac{dy}{dt}+ycos \theta\frac{d\theta}{dt} \\\theta=\frac{\pi}{3}\\ \frac{dh}{dt}=sin (\frac{\pi}{3} )\frac{dy}{dt}+ycos\theta (\frac{\pi}{3} )\frac{d\theta}{dt}\\\frac{dy}{dt}=2,y=6,\frac{d\theta}{dt}=-\frac{2 \sqrt{3} }{29} \\\frac{dh}{dt} =\sqrt{3}-\frac{6\sqrt{3} }{29} \\=1.3737 ft/sec$

Part(b): From the diagram,

$cos\theta=\frac{z}{29-y}\\ z=29cos\theta-ycos\theta$

Now, differentiating with respect to $t$ then,

$\frac{dw}{dt}=-29 sin\theta\frac{d\theta}{dt} -\frac{dy}{dt}cos\theta+ysin\theta\frac{d\theta}{dt} \\=(y-29)sin\theta\frac{d\theta}{dt} -\frac{dy}{dt}cos\theta\\\theta=\frac{\pi}{3} ,\frac{dw}{dt}=(y-29)sin(\frac{\pi}{3}) \frac{d\theta}{dt}-\frac{dy}{dt}cos(\frac{\pi}{3} ) \\\frac{dw}{dt} =(y-29)\frac{\sqrt{3} }{2} \frac{d\theta}{dt}- \frac{dy}{dt}(\frac{1}{2} )\\ \frac{d\theta}{dt}=-\frac{2\sqrt{3} }{29} ,\frac{dy}{dt} =2,y=6\\\frac{dw}{dt} =\frac{(-23)(-3)}{29}\\ \frac{dw}{dt} =1.3793 ft/sec$

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