# Thread: Related Problems helps (need tips)

1. ## Related Problems helps (need tips)

A spotlight on the ground shines on a wall 200 meters away. A giant 20 meters tall walks from the spotlight to the wall, at a speed of 4 m./sec. ; his path is perpendicular to the wall. Let x be the distance from his feet to the spotlight and let h be the height of the shadow on the wall. Also let θ be the angle of elevation at the spotlight from the horizontal to the top of his head. (a) Draw a sketch of the problem, and find a formula relating h and x. (b) When the giant is 40 meters from the wall, find the height of the shadow and the rate of change of the height of the shadow. (c) What is the rate of change of θ at the time the giant is 40 meters from the wall?

So far, i think i got a good equation going on for this one. For part b particularly.

The equation i got involved similar triangles : h/(200-x) = 20/x, so i differentiated implicitly. However, im not stuck trying to find an h? When i substitute in the variables and have variables that are missing such as height and x, do i find their value at the instant that the giant is 40 meters from the wall (x=40)?

Overall , i'm kinda unsure if im using the right equations and the right numbers to substitute in my implicit derivative.

2. ## Re: Related Problems helps (need tips)

alright, i think my equation is wrong. Would it be h / 200 = 20/x? im kinda confused why i wouldnt subtract 200 by x.. Would the formula be based on the exact value equivalency (if you understand what i mean) or would it be taking into account how the distance from man to the wall changes (so i subtract 200 by x)

3. ## Re: Related Problems helps (need tips)

similar triangles ...

(shadow height)/(large triangle base) = (giant height)/(small triangle base)

$\dfrac{h}{200} = \dfrac{20}{x}$

4. ## Re: Related Problems helps (need tips)

How would i set up the equation for the angle?

How do i know which triangle to use?

I thought of using tantheta = h/200

5. ## Re: Related Problems helps (need tips)

Originally Posted by skeeter
similar triangles ...

(shadow height)/(large triangle base) = (giant height)/(small triangle base)

$\dfrac{h}{200} = \dfrac{20}{x}$
the answer for the rate of angle is suppose to be 1/325 radians / sec, but i dont think i can figure that out with a calculator... is there a way to find it without a calculator? I'm trying to study for my final and they gave us this review practice exam!

6. ## Re: Related Problems helps (need tips)

$\dfrac{h}{200} = \dfrac{20}{x} \implies hx = 4000$

$\dfrac{d}{dt} \left(hx=4000\right)$

$h \cdot \dfrac{dx}{dt} + x \cdot \dfrac{dh}{dt} = 0$

(b) When the giant is 40 meters from the wall, find the height of the shadow and the rate of change of the height of the shadow.
$h(160) = 4000 \implies h = 25 \, m$

$25 \cdot 4 + 160 \cdot \dfrac{dh}{dt} = 0 \implies \dfrac{dh}{dt} = -\dfrac{5}{8} \, m/s$, i.e. the shadow's height is decreasing.

(c) What is the rate of change of θ at the time the giant is 40 meters from the wall?
Note that $\theta$ decreases as the giant moves away from the light $\implies \dfrac{d\theta}{dt} < 0$

using the small triangle ...

$\theta = \text{arccot}\left(\dfrac{x}{20}\right)$

$\dfrac{d\theta}{dt} = - \dfrac{\frac{1}{20}}{1 + \frac{x^2}{400}} \cdot \dfrac{dx}{dt} = - \dfrac{20}{400+x^2} \cdot 4 = - \dfrac{80}{400 + 160^2} = -\dfrac{80}{26000} = - \dfrac{1}{325} \, rad/sec$

using the large triangle ...

$\theta = \arctan\left(\dfrac{h}{200}\right)$

$\dfrac{d\theta}{dt} = \dfrac{\frac{1}{200}}{1+\frac{h^2}{200^2}} \cdot \dfrac{dh}{dt}$

$\dfrac{d\theta}{dt} = \dfrac{200}{200^2+25^2} \cdot \left(-\dfrac{5}{8}\right)$

$\dfrac{d\theta}{dt} = -\dfrac{1000}{8(40625)} = -\dfrac{125}{40625} = -\dfrac{1}{325} \, rad/sec$

7. ## Re: Related Problems helps (need tips)

Oh i see, you used the inverse derivatives for trig. Does this always work for these angle problems in relate rates? I see people use tantheta = a/b to relate .

8. ## Re: Related Problems helps (need tips)

both techniques work