Solveeit Logo

Question

Question: The rate of increase of length of the shadow of man 2 metres in height, due to a lamp at 10 metres h...

The rate of increase of length of the shadow of man 2 metres in height, due to a lamp at 10 metres height, when he is moving away from it at the rate of 2 m/sec, is
(a) 12\dfrac{1}{2} m/sec
(b) 25\dfrac{2}{5} m/sec
(c) 13\dfrac{1}{3} m/sec
(d) 5 m/sec

Explanation

Solution

We will draw a rough figure to understand the given information. We will be able to show that we see similar triangles in the figure. Using the ratio of similar sides of these triangles, we will form an equation that relates the length of the shadow with the distance between the man and the lamp. Then we will differentiate this equation to obtain the required answer.

Complete step by step answer:
Let us draw a rough figure representing the given information. It looks like the following figure,

The point C is the lamp and segment ED is the man with height 2 meters. The length of the shadow is the length of segment DB. Let us denote this length of the shadow by xx. Let the distance between the man and the lamp be yy metres.
Now, we can see that we have two right angled triangles, ΔCAB\Delta \text{CAB} and ΔEDB\Delta \text{EDB}. In these triangles, we can see that CAB = EDB = 90\angle \text{CAB = }\angle \text{EDB = 90}{}^\circ and CBA = EBD\angle \text{CBA = }\angle \text{EBD}because it is a common angle. Hence, by AA test, we conclude that ΔCAB\Delta \text{CAB} and ΔEDB\Delta \text{EDB} are similar triangles. Now, we will take the ratio of the similar sides in the following manner,
CAAB=EDDB\dfrac{\text{CA}}{\text{AB}}=\dfrac{\text{ED}}{\text{DB}}
We know that CA = 10\text{CA = 10}, AB = AD + DB = y+x\text{AB = AD + DB = }y+x, ED = 2\text{ED = 2} and DB = x\text{DB = }x. Substituting these values in the above equation, we get
10y+x=2x\dfrac{\text{10}}{y+x}=\dfrac{\text{2}}{\text{x}}
Simplifying this equation, we get
10x=2y+2x 10x2x=2y 8x=2y \begin{aligned} & 10x=2y+2x \\\ & \Rightarrow 10x-2x=2y \\\ & \therefore 8x=2y \\\ \end{aligned}
Differentiating the above equation with respect to time, we get
8dxdt=2dydt8\dfrac{dx}{dt}=2\dfrac{dy}{dt}
We are given that the rate of man moving away is 2 m/sec, that is dydt=2 m/sec\dfrac{dy}{dt}=2\text{ m/sec}. Substituting this value in the above equation, we get
8dxdt=2×2 dxdt=48 dxdt=12 \begin{aligned} & 8\dfrac{dx}{dt}=2\times 2 \\\ & \Rightarrow \dfrac{dx}{dt}=\dfrac{4}{8} \\\ & \therefore \dfrac{dx}{dt}=\dfrac{1}{2} \\\ \end{aligned}
Hence, the rate of increase in the length of shadow of man is 12 m/sec\dfrac{1}{2}\text{ m/sec}.

So, the correct answer is “Option a”.

Note: It is useful to draw rough diagrams to understand the given information in this type of question. We see that the word ‘rate’ is associated with a differentiation. It is always a rate of some element with respect to another element. Therefore, the differentiation tells us about a particular change in an element with respect to the other element. It is important that we interpret this correctly so that we can substitute the proper values in the differential equation obtained.