Derivatives in real life (Mooculus)

Question

A light on the ground is 30 feet away from a building. A 4 foot tall man is walking from the light to the building at a rate of 3 feet per second.

He is casting a shadow on the side of the building. At what rate is his shadow shrinking when he is 5 feet from the building?

Comments

Is the answer is -0.5 ft/s

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-0.576 ft/s is the answer

Answer 1

From the above pictorial illustration we can seen that,

$\color{maroon}{\bigtriangleup{ABC}} \text{ & } \color{maroon}{\bigtriangleup{DEC}}$ are similar triangle,

As, $\color{blue}{\angle{ACB}} = \color{blue}{\angle{DCE}}$

 & $\color{gold}{\measuredangle{ABC}} =$ $\color{gold}{90^\circ}$ , $\color{gold}{\measuredangle{DEC}} =$ $\color{gold}{90^\circ}$

$∴\hspace{0.1cm} \color{maroon}{\bigtriangleup{ABC}} \equiv\hspace{0.1cm} \color{maroon}{\bigtriangleup{DEC}}$

Now,

$\dfrac{AB}{BC} = \dfrac{DE}{EC}$

Taking $EC = x$ and $AB = y$

$\dfrac{y}{30} = \dfrac{4}{x}$

Or, $y = \dfrac{120}{x}$ 

Differentiate both side w.r.t  $t$

$\dfrac{dy}{dt} = \dfrac{-120\dfrac{dx}{dt}}{x^{2}}$ $\hspace{2cm}\color{orange}{[\dfrac{d}{dx}(\dfrac{u}{v}) = \dfrac{v\dfrac{du}{dx} -u\dfrac{dv}{dx}}{v^{2}}]}$

As, $x = 30-5 = 25$

Or, $\dfrac{dy}{dt} = \dfrac{-120(3)}{25^{2}}$

Or, $\color{lightblue}{\dfrac{dy}{dt} = -0.576 ft/s \approx -0.5ft/s}$

$∴ \color{green}{\text{ The shadow of man is shrinking at the rate of 0.5 ft/s when he is 5 ft. from the building.}}$

$\color{red}{\text{If we get,}}$ $\color{red}{\dfrac{dy}{dt} = + 0.576 \hspace{0.1cm}ft/s}$,

$\color{red}{\text{then that would be mean that, the man is walking towards the light}}$

$\color{red}{\text{and hence, his shadow keeps on increasing.}}$

Comments

this answer is clearly very clear to understand... :) a big Thumbs up

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@Sukanya, where do you draw these diagrams? They are just beautiful.... :)

Answer 2

i

if you do math you wil get 0.576
references for similar Triangles : http://academic.sun.ac.za/mathed/shoma/MATUNIT25_04.htm