A girl of height $120 $cm is walking towards a light on the ground at a speed of $0.6$ m/s. Her shadow is being cast on a wall behind her that is $5 $m from the light. How is the size of her shadow changing when she is $1.5$ m from the light?
Not sure what to do here, if someone could help me out that would be fantasic.
Let $l(t)$ be the distance between the girl and the light changing with time and $h(t)$ be the hight of her shadow.
You can get that $$\dfrac {l(t)}{5}=\dfrac{1.2}{h(t)} \implies 6=l.h$$
By taking derivative with respect to $t$
$$0=l_th+lh_t$$ and we know that $l_t=-0.6$ and $l=1.5 \implies h=4$
$$0=-0.6\times4+1.5\times h_t \implies h_t=\dfrac{8}{5} $$
Note: The first equation can be got by the properties of similiar triangles and I used the notation $x_t$ as $\dfrac{dx}{dt}$. We take $l_t=-0.6$ but not $0.6$ since the girl is approaching the light so $l(t)$ is decreasing.
Assume she is $x$ m from the light. Find the height of your shadow in terms of $x$ using similar triangles.
Differentiate it with respect to $t$ where $t$ is time. Use $dx/dt=\text{speed of girl}$
