A spotlight on the ground shines on a wall $12 m$ away. If a man $2m$ tall walks from the spotlight toward the wall at a speed of $1.6 m/sec$, how fast is the length of his shadow decreasing when he is $4m$ from the building.
[Math] Calculus Word Problem Related Rates
calculus
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The $15$ foot tall lightpole and the $6$ foot tall man aren’t sides of the same triangle. You actually have two triangles. Let $A$ be the point at the base of the lightpole, $B$ the point at the top of the lightpole, $C$ the point where the man is standing, $D$ the top of his head, and $E$ the end of his shadow. Then you have a right triangle $ABE$ with a vertical side of $15$ feet and another right triangle $CDE$ with a vertical side of $6$ feet. These triangles are related in a rather special way: they’re _______?
Now let $x$ be the distance of the man from the lightpole, i.e., the length of $\overline{AC}$. You know what $dx/dt$ is. Let $y$ be the length of $\overline{AE}$, the distance from the lightpole to the tip of his shadow. You want to find $dy/dt$. If you can answer the question above, you should be able to find a relationship between $x$ and $y$ that you can use to find $dy/dt$.
This is something that you should draw to understand the details. The 11-ft tall pole, the distance from the bottom of the pole to the tip of the man's shadow, and the distance from the top of the pole to the tip of the man's shadow form a large triangle. Additionally, the man's height, the top of the man's head to the tip of his shadow, and the distance from the feet of the man to the tip of the shadow form a smaller triangle. We can use this information to find the desired answer. Indeed, since these triangles described are similar, there is proportionality at play here - we can glean the following equation from examination of the situation, where $h$ is the height of the man's shadow and $x$ is the distance from the bottom of the pole to the man's feet at any given time (thus, $x+h$ is the distance from the bottom of the pole to the tip of the man's shadow - this is the base of the larger triangle), $$\frac{11}{x+h}=\frac{5.5}{h}.$$ Now, rearrange this equation to find a linear relationship between $h$ and $x$: $\,\,\,h=x$. Taking the derivative of both sides of this with respect to $t$, where $t$ is the time in seconds, to get $$\frac{dh}{dt}=\frac{dx}{dt}=3.5$$ It is important to recognize that the speed of the tip of the shadow does not depend upon the distance of the man from the pole, as we can see from the form of $dh/dt$. In other words, the speed of the tip of the shadow is constant at the rate at which the man walks away from the pole, with these dimensions, regardless of how far away he is from the pole.
Best Answer
Hint: if you draw a picture, there are similar triangles that let you find a relation between the distance between the man and the wall and the height of the shadow.
Added: Let us measure from the spotlight at $x=0$ with the wall at $x=12$. The height of the shadow is then $\frac {12}x \cdot 2$ meters from similar triangles.