Question

A person 6ft tall stands 10ft from point P directly beneath a lantern hanging 30 ft above the ground. The lantern start to fall, thus causing the person shadow to lengthen(L). Given that the lantern falls 16t2 feet in t seconds, how fast will the shadow be lengthening when t=1

Answer 1

Answer:

$\frac{dL}{dt}=30 \frac{ft}{s}$

Step-by-step explanation:

In order to solve this it is always a good idea to start by drawing a diagram of the situation (See attached picture).

From the diagram we can see that we are dealing with similar triangles. We can use similar triangles to build an equation that relates the length of the shadow with the height of the lamp, so we get:

$\frac{L}{6}=\frac{10+L}{h}$

the height of the lamp can be found by subtracting the $16t^{2}$ distance the lamp falls in a given time t from the original 30ft the lamp was located at.

So the equation will now lok like this:

$\frac{L}{6}=\frac{10+L}{30-16t^{2}}$

So now we can solve the equation for L, we can start by multiplying by the LCD SO WE GET:

$L(30-16t^{2})=6(10+L)$

next, we can distribute the right side of the equation so we get:

$L(30-16t^{2})=60+6L$

and subtract 6L from both sides so we get:

$L(30-16t^{2})-6L=60$

and factor L, so we get:

$L(30-16t^{2}-6)=60$

and solve for L:

$L=\frac{60}{24-16t^{2}}$

now, we can differentiate this equation by using the chain rule, so we get:

$dL=-\frac{60}{(24-16t^{2})^{2}}(-32t)dt$

which can be simplified to:

$\frac{dL}{dt}=\frac{1920t}{(24-16t^{2})^{2}}$

and now we can substitute t for 1s so we get:

$\frac{dL}{dt}=\frac{1920(1)}{(24-16(1)^{2})^{2}}$

$\frac{dL}{dt}=30 \frac{ft}{s}$