A man of height 180 centimetre is moving away from a lamp post at a rate of 1.2 m/s. If the height of the lamp is 4.5m , find the rate at which
(I) It's shadow is lengthening
(II) Tip of the shadow is moving
Detailed explanation needed !!!
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Answers
Answer :
I) 0.8 m/s
II) 2 m/s
Solution :
[ Situation described in the attachment . ]
Given :
• Height of the lamp , AB = 4.5 m
• Height of the man , DE = 180 cm = 1.8 m
• Rate at which the man is moving from the lamp , d(BD)/dt = 1.2 m/s
To find :
• The rate at which the man's shadow is lengthening , d(CD)/dt = ?
• The rate at which the tip of the man's shadow is moving , d(BC)/dt = ?
Now ,
In ∆CED , we have
=> tanθ = DE/CD
=> tanθ = 1.8/CD ------(1)
Also ,
In ∆ABC , we have
=> tanθ = AB/BC
=> tanθ = 4.5/(BD + CD) -------(2)
Now ,
From eq-(1) and (2) , we have ;
=> 1.8/CD = 4.5/(BD + CD)
=> 1.8•(BD + CD) = 4.5•CD
=> 1.8•BD + 1.8•CD = 4.5•CD
=> 1.8•BD = 4.5•CD - 1.8•CD
=> 1.8•BD = 2.7•CD
=> CD = (1.8/2.7)•BD
=> CD = ⅔•BD
Now ,
Differentiating both the sides with respect to time t , we get ;
=> d(CD)/dt = d(⅔•BD)/dt
=> d(CD)/dt = ⅔•d(BD)/dt
=> d(CD)/dt = ⅔•1.2
=> d(CD)/dt = 0.8
Hence ,
The man's shadow is lengthening at the rate of 0.8 m/s .
Also ,
BC = BD + CD
Now ,
Differentiating both the sides with respect to time t , we get ;
=> d(BC)/dt = d(BD + CD)/dt
=> d(BC)/dt = d(BD)/dt + d(CD)/dt
=> d(BC)/dt = 1.2 + 0.8
=> d(BC)/dt = 2
Hence ,
The tip of the man's shadow is moving at the rate of 2 m/s .
Answer:
Here is your answer!!
Step-by-step explanation:
In order to solve this question we will try to make the sketch of the situation and then see the lengths which we need to find and then we can apply the differentiation method to get the speed as we know that the differentiation of the distance with respect to the time gives us speed .
For example ; if we get y = 3x then differentiating both these sides with respect to time will give us derivation shift between their speeds.
» complete process by process answer
Here we are given that the height of the man is =
180 cm = 1.8 m
Now we also know that is moving away from the
lamp post at the rate of 1.2 metre per second
and height of the lamp is also given as 4.5 m
Now let us visualise the situation the situation and draw a roughly figure. [ please refer the attachment]
here we can say ;
» length AB = lamp post = 4.5 m
» CD = Man = 180 CM = 1.8 m
» DE = Shadow
and distance of the tip E at the shadow from the pole is BE = X + Y
Now in ;
∆ ABE AND ∆ CDE
< E = < E { angles common in both}
<ABE = < CDE = 90°
< BAE = < DCE [ CORRESPONDING ANGLES]
so we can say ∆ ABE ~ ∆ CDE
so we can write.
AB = BE
CD DE
»» 4.5 = X+ Y
1.8 Y
»» 4.5 Y = 1 .8 X + 1.8 Y
Y = 2 X
3
Now differentiating both the sides with respect to the time" t" we get
»» dy = 2 dx
dt 3 dt
we know that according to the question we are given
dx = 1.2 meter/sec
dt
Hence we can say that!!
dy = 2 dx = 2 ( 1.2) = 0.8 m/ sec
dt 3 dt 3
as the length of the shadow is by hence we can say that the shadow is lengthening with the speed of of 0.8 metre per second.
also we know that BE = X + Y
so d ( BE) = dx + dy
dt dt dy
= 1.2 + 0.8 = 2m/sec
hence the rate at which the tip of the shadows moving away is 2 metre per second.
hopefully this might help you! !!
