IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

kane

a man 2 metres high walks at a uniform speed of 5km/hr away from a lamp post 6 metres high.find the rate at which the length of his shadow increases?

answer is 5/2 km/hr

mukundmadhav

Let the distance of the man from the post be x and length of shadow be y..
6/(x+y) = 2/y (From geometry)
Simplifying, you get
2y = x
Differentiate with respect to t
dy/dt = dx/ dt x (1/2)

dx/dt is his walking speed
dy/dt is the rate of change of length of shadow with time

feynmann

let the height of the post = H , distance of the man from post = x , heightof the man = h , length of his shadow = y

so from simple geometry we have

h/y = H/ ( x +y )

solvinf for y gives y = hx / ( H- h )

so dy/dt = h/ ( H- h ) dx/dt = 2/(4 ) * 5 kmph = 5/2 kmph

feynmann

OOpps !!

Did not notice ur post !!

allamraju

Let the distance between the pole and man be x and the length of the shadow be y.Draw a fig.u will understand,now given that dx/dt=5kmph.

using the idea of similar triangles,6/x+y=2/y $\Rightarrow$ x=2y $\Rightarrow$ dx/dt=5=2.dy/dt.Hence the ans is 5/2kmph.

kane

thnxs guys!!!!!!!!!!!

allamraju

Oh...I am very slow.When I started,there was no ans at all,But now I see that there are already 2.

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Let the distance between the pole and man be x and the length of the shadow be y.Draw a fig.u will understand,now given that dx/dt=5kmph.

using the idea of similar triangles,6/x+y=2/yx=2ydx/dt=5=2.dy/dt.Hence the ans is 5/2kmph.

using the idea of similar triangles,6/x+y=2/y $\Rightarrow$ x=2y $\Rightarrow$ dx/dt=5=2.dy/dt.Hence the ans is 5/2kmph.