Respuesta :
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:
[tex]\frac{dh}{dt} =\frac{dz}{dt}*sin(\alpha )+z*cos(\alpha )*\frac{d\alpha }{dt}[/tex]
We have [tex]\frac{dz}{dt}[/tex] given in the exercise but we need [tex]\frac{d\alpha }{dt}[/tex]
Again, looking at the drawing we can write α in term of x, since we can calculated x and we have [tex]\frac{dx }{dt}[/tex]
29 *cos(α)=x where x is the distance from the base of the ladder to the wall.
Deriving:
[tex]-29*sin(α)\frac{d\alpha }{dt}=\frac{dx }{dt}[/tex]
[tex]\frac{d\alpha }{dt}=\frac{dx }{dt}\frac{-1 }{29*sin(\alpha) }[/tex]
replacing out known values: α=π/3, [tex]\frac{dx }{dt}=3[/tex], z=6,[tex]\frac{dz}{dt}=2[/tex]
[tex]\frac{d\alpha }{dt}=\frac{-2\sqrt{3}}{29}[/tex]
Now that we have dα/dt:
[tex]\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}[/tex]
Now we do a similar process for dw/dt, looking at the drawing:
w=(29-z)*cos(α)
[tex]\frac{dw}{dt}=-\frac{dz}{dt}*cos(\alpha )-(29-z)*sin(\alpha)*\frac{d\alpha }{dt}[/tex]
Replacing known values:
[tex]\frac{dw}{dt}=2*.5-23*\frac{\sqrt{3}}{2}*\frac{-3\sqrt{3}}{29*2}=2.78 \ \frac { ft}{s}[/tex]
Part(a): The required value is [tex]\frac{dh}{dt} =1.3737 ft/sec[/tex]
Part(b): The required value is [tex]\frac{dw}{dt} =1.3793 ft/sec[/tex]
Differentiation:
Differentiation, in calculus, can be applied to measure the function per unit change in the independent variable.
Now, the given part is,
Let [tex]x[/tex] be the distance from the will to the base of the ladder.
[tex]y[/tex] be the distance along the ladder that the firewall climbed [tex]w[/tex] to be the distance from the firefighter.
[tex]h[/tex] be the height from firefighter [tex]\theta[/tex] 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,
[tex]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[/tex]
Part(b): From the diagram,
[tex]cos\theta=\frac{z}{29-y}\\ z=29cos\theta-ycos\theta[/tex]
Now, differentiating with respect to [tex]t[/tex] then,
[tex]\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[/tex]
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