Answer:
Camel's shadow is growing at a rate of 8 ft/sec.
Step-by-step explanation:
Given:
Height of camel = 8 ft
Height of streetlight = 20 ft
Speed of camel = 12 ft / sec
To find:
How fast is the camel's shadow growing = ?
Solution:
First of all, have a look at the image attached for the situation as given in the question statement.
PQ be the streetlight.
AB be the camel.
Camel is loping away from the streetlight.
BT will be the shadow of the camel.
We are given that speed of camel = 12 ft / sec
And we know that Speed is the Rate of change of distance w.r. to time.
i.e. as per the image, we are given:
[tex]\dfrac{dx}{dt} = 12 ft /sec[/tex]
And we have to find the value of:
[tex]\dfrac{ds}{dt} = ?[/tex]
(Because s is the shadow length and we have to find the rate of change of shadow's length w.r.to time.)
Here, let us consider the two similar triangles [tex]\triangle TAB\ and\ \triangle TPQ[/tex]as per the image attached:
Ratio of corresponding sides will be same.
[tex]\dfrac{AB}{PQ}=\dfrac{TB}{TQ}\\\Rightarrow \dfrac{8}{20}=\dfrac{s}{s+x}\\\Rightarrow 8s+8x=20s\\\Rightarrow 12s = 8x\\\Rightarrow s = \dfrac{2}{3}x[/tex]
Now, differentiating w.r.to t:
[tex]\dfrac{ds}{dt}=\dfrac{2}{3}\dfrac{dx}{dt}\\\Rightarrow \dfrac{ds}{dt}=\dfrac{2}{3}\times 12\\\Rightarrow \dfrac{ds}{dt}=\bold{8\ ft/sec}[/tex]
Camel's shadow is growing at a rate of 8 ft/sec.
