Question

A man 2m tall, walks at the rate of $1\dfrac{2}{3}m/sec$ towards a street light which is $5\dfrac{1}{3}m$ above the ground. At what rate is the tip of his shadow moving? At what rate is the length of his shadow changing when he is $3\dfrac{1}{3}m$ from the base of the light?

Answer 1

Hint: We are going to solve this problem by proving both the triangles similar and then we will differentiate the rate of change shadow with respect to time.

$\angle ABE$ and$\angle CDE = {90^ \circ }$.
$\angle A = \angle C$ (Angles on the same side of a triangle are equal)
$\left. {AB} \right\|CD$
Hence, $\Delta ABE$ and $\Delta CDE$ are similar triangles.
So, similar triangles have their corresponding sides in proportion are equal $\dfrac{{AB}}{{CD}} = \dfrac{{BE}}{{DE}} = \dfrac{{AE}}{{CE}}$

Formula Used:
For finding the rate of changes in shadow we will differentiate it with respect to time $ = \dfrac{{dx}}{{dt}}$ and $\dfrac{{dy}}{{dt}}$

Complete step by step answer:
Suppose \[AB\] is the height of the street light, and \[CD\] is the height of the man. Suppose at any time \[t\] the man \[CD\] is at a distance \[x\] meter from the street light. \[y\] is the shadow of the man.
So, the rate of change of the distance of man at time \[t\] is $\dfrac{{dx}}{{dt}} = 2\dfrac{1}{3}m/s$
We know that,
$\angle ABE$ and $\angle CDE$ are equal, because both are the right angle.
$\Delta ADE$ and $\Delta CDE$ are similar triangle
Hence, $\dfrac{{AB}}{{CD}} = \dfrac{{BE}}{{DE}} = \dfrac{{AE}}{{CE}}$ as sides of similar triangles are equal.
It is given that \[CD = 2m\] and $AB = 5\dfrac{1}{3} = \dfrac{{16}}{3}m$
From the diagram, we can conclude, \[BD = x,{\text{ }}DE = y,{\text{ and }}BE = x + y\]
So putting the values of \[AB,{\text{ }}BE,{\text{ }}DE,{\text{ }}CD\] in the above equation:
$ \Rightarrow \dfrac{{\dfrac{{16}}{3}}}{{\dfrac{2}{1}}} = \dfrac{{x + y}}{y}$
Simplifying the above equation we get,
$ \Rightarrow \dfrac{{16}}{6} = \dfrac{{x + y}}{y}$
Cross multiply the denominator to the numerator to solve the above equation,
$ \Rightarrow 16y = 6(x + y)$
$ \Rightarrow 16y = 6x + 6y$
Arranging the same variables in the above equation,
$ \Rightarrow 16y - 6y = 6x$
$\Rightarrow 10y = 6x$
Now, we will find the rate of change of shadow. We will differentiate the rate of change of shadow with respect to time \[t\].
$ \Rightarrow 10\dfrac{{dy}}{{dt}} = 6\dfrac{{dx}}{{dt}}$
Value of $\dfrac{{dx}}{{dt}}$ is given = $\dfrac{5}{3}$
$ \Rightarrow 10\dfrac{{dy}}{{dt}} = 6 \times \dfrac{5}{3}$
Simplifying we get,
$ \Rightarrow \dfrac{{dy}}{{dt}} = \dfrac{{6 \times 5}}{{10 \times 3}}$
Multiplying the terms,
$ \Rightarrow \dfrac{{dy}}{{dt}} = \dfrac{{30}}{{30}} = 1$
So, the rate of change of shadow $\dfrac{{dy}}{{dt}}$ is \[1\]
Now, we have to find out the rate at which the tip of the shadow \[CE\] changes.
$AB = \dfrac{{16}}{3}$
We have to find the rate at which $BE = BD + DE$ moves
From similar property = $\dfrac{{AB}}{{CD}} = \dfrac{{BE}}{{DE}}$
Let \[BE = P\]
Now putting the values of \[AB\], \[CD\], \[DE\] in the above equation
$ \Rightarrow \dfrac{{\dfrac{{16}}{3}}}{{\dfrac{2}{1}}} = \dfrac{P}{y}$
Simplifying we get,
$ \Rightarrow \dfrac{8}{3} = \dfrac{P}{y}$
Solving for $P$,
$ \Rightarrow \dfrac{8}{3}y = P$
Now differentiating it with respect to time \[t\],
$ \Rightarrow \dfrac{8}{3} \times \dfrac{{dy}}{{dt}} = \dfrac{{dP}}{{dt}}$
Now putting the value of $\dfrac{{dy}}{{dt}} = 1$ in the above equation
$\Rightarrow \dfrac{{dP}}{{dt}} = \dfrac{8}{3}$ is the rate at which the tip of the shadow changes.

Therefore, the rate of change in shadow length is $\dfrac{8}{3}$.

Note:
From the similar property of the triangle, we will take the sides in proportion in equal and find out the changes in the rate of the shadow and the tip of the shadow. In general, these real-life problems convert the problem into a diagram. It is easy to understand the problem to find and it will make the problem easiest to solve.