Answer:(a)10 ft/s
(b)4 ft/s
Explanation:
Given
height of light [tex]=15 feet[/tex]
height of man[tex]=6 feet[/tex]
[tex]\frac{\mathrm{d} x}{\mathrm{d} t}=6 ft/s[/tex]
From diagram
[tex]\frac{15}{y}=\frac{6}{y-x}[/tex]
[tex]5(y-x)=2y[/tex]
[tex]3y=5x[/tex]
differentiate both sides
[tex]3\times \frac{\mathrm{d} y}{\mathrm{d} t}=5\times \frac{\mathrm{d} x}{\mathrm{d} t}[/tex]
Tip of shadow is moving at the rate of
[tex]\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{5}{3}\times 6=10 ft/s[/tex]
(b)rate at which length of his shadow is changing
Length of shadow is [tex]y-x[/tex]
differentiating w.r.t time
[tex]\frac{\mathrm{d} (y-x)}{\mathrm{d} t}=\frac{\mathrm{d} y}{\mathrm{d} t}-\frac{\mathrm{d} x}{\mathrm{d} t}[/tex]
[tex]\frac{\mathrm{d} (y-x)}{\mathrm{d} t}=10-6=4 ft/s[/tex]
A) When the man is 10 ft from the base of the light, the rate at which the tip of his shadow is moving is; dy/dt = 10 ft/s
B) When the man is 10 ft from the base of the light, at what rate is the length of his shadow changing is; d(BD)/dt = 4 ft/s
We are given;
Height of man = 6 ft
Height of light = 15 ft
Rate of walk by man; dx/dt = 6 ft/s
I have attached an image showing the triangle formed by the man and the light.
From the diagram;
AB is height of light = 15 ft
CD is height of man = 6 ft
A) From the attached diagram, if BE = y, then we can use ratio of similar triangles to get;
15/y = 6/(y - x)
Cross multiply to get;
15(y - x) = 6y
15y - 15x = 6y
15y - 6y = 15x
9y = 15x
Divide both sides by 3 to get;
3y = 5x
Differentiating both sides with respect to t gives;
3(dy/dt) = 5(dx/dt)
dy/dt = (5/3)(dx/dt)
Putting 6 ft/s for dx/dt yields;
dy/dt = (5/3)(6)
dy/dt = 10 ft/s
B) From the diagram, length of shadow is;
BD = y - x
Rate at which the shadow is changing will be gotten by differentiation of this length with respect to time.
Thus;
d(BD)/dt = dy/dt - dx/dt
Plugging in the relevant values gives;
d(BD)/dt = 10 - 6
d(BD)/dt = 4 ft/s
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