# Thread: triangles and rates of change (and some other stuff)

1. ## triangles and rates of change (and some other stuff)

1)A street light is at the top of a 16 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 4 ft/sec along a straight path. How fast is the tip of her shadow moving along the ground when she is 30 ft from the base of the pole?

Okay, so here's what I did

dx/dt=5 x=30

y=16

s/6=(s+x)/16

16s=6s+6s

s=(3/5)x thus (3/5)=ds/dt

That's what my teacher told me to do, but I have no idea why I did that and I STILL don't know how to get the answers. I understand that I probably have to make a smaller triangle where the y is 6 ft, but I just don't know where to go from there.

2)Linear approximation

Use linear approximation, i.e. the tangent line, to approximate as follows: cubed root 63.6
The equation of the tangent line to at f(x) x=64 can be written in the form y=
I don't even understand what the equation is, but shouldn't the answer be close to 4?

I'll probably have some more questions as I work through this homework, but I really think that I'll start to understand it if someone helps me with these few problems. Thanks guys!

2. Originally Posted by nobodygirl
1)A street light is at the top of a 16 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 4 ft/sec along a straight path. How fast is the tip of her shadow moving along the ground when she is 30 ft from the base of the pole?

Okay, so here's what I did

dx/dt=5 x=30

y=16

s/6=(s+x)/16

16s=6s+6s

s=(3/5)x thus (3/5)=ds/dt

That's what my teacher told me to do, but I have no idea why I did that and I STILL don't know how to get the answers. I understand that I probably have to make a smaller triangle where the y is 6 ft, but I just don't know where to go from there.
See the diagram below, ALWAYS DRAW A DIAGRAM WHEN DOING A RELATED RATES PROBLEM.

you were right about forming a smaller triangle within the larger one with a height of 6 ft. where do you go from there? well, you use the concept of similar triangles. remember similar triangles are triangles whose sides are propotional, so the ration of any two sides of one triangle is equal to the ratio of the corresponding sides of a triangle similar to it.

so the height of the big triangle is the 16ft pole with the light on top of it. the 6 ft line x meters away from the pole is the woman (she's on the Atkin's diet which is why she's so thin).

Obviously, we will use similar triangles here.

Now 16/(x + y) = 6/y ……the ratio of height/base of the large triangle = ratio of height/base of small triangle
=> 16y = 6x + 6y
=> 10y = 6x …………….now differentiate implicitly
=> 10 dy/dt = 6 dx/dt = 6(4) = 24
=> dy/dt = 24/10 = 2.4 ft/sec

But the rate at which the tip of her shadow is moving away from the pole is equal to the rate at which she is moving plus the rate at which her shadow is extending in front of her. So then,

ds/dt = dx/dt + dy/dt = 4 + 2.4 = 6.4 ft/sec

so this is a slightly different way from how your professor did the problem, well not really, it is essentially the same, except he solved for y where i just left 10y.

i'm kind of in a rush now, i'll do the second problem later

3. Originally Posted by nobodygirl

2)Linear approximation

Use linear approximation, i.e. the tangent line, to approximate as follows: cubed root 63.6
The equation of the tangent line to at f(x) x=64 can be written in the form y=
I don't even understand what the equation is, but shouldn't the answer be close to 4?
Ok, so linear approximation is not that hard, all you have to do is memorize one formula. the formula is:

f(x) ~= f(a) + f ' (a)(x - a)

x is the value that you want to find f(x) for but obviously can't, while a is a value close to x that you do know the value of f(a) for and you will use to approximate f(x). let's see how it works.

you want to find (63.6)^(1/3), (this is the same as the cube root of 63.6, i just wrote it in terms of powers). so we don't know what (63.6)^(1/3) is, however, 64 is very close to 63.6 and we do know what (64)^(1/3) is, that's just 4. so we want x = 63.6 and a = 64, now let's do the problem.

Consider the function f(x) = x^(1/3), we want to approximate f(63.6), that is, (63.6)^(1/3).

now f ' (x) = (1/3)x^(-2/3)

choose x = 63.6 and a = 64

=> f(a) = 64^(1/3) = 4
=> f ' (a) = (1/3)64^(-2/3) = (1/3)(1/64^(2/3)) = (1/3)(1/16) = 1/48
=> (x - a) = 63.6 - 64 = -0.4

so using f(x)~= f(a) + f '(a)(x - a), we have

(63.6)^(1/3) = f(63.6) ~= f(64) + f ' (64)(63.6 - 64) = 4 - (1/48)(4) = 47/12 = 3.9167

the actual value is 3.991649245, pretty darn close if you ask me

4. Originally Posted by Jhevon
...

so using f(x)~= f(a) + f '(a)(x - a), we have

(63.6)^(1/3) = f(63.6) ~= f(64) + f ' (64)(63.6 - 64) = 4 - (1/48)(4) = 47/12 = 3.9167

the actual value is 3.991649245, pretty darn close if you ask me
Hello, Jhevon,

I don't want to pick at you but you have made a small but nevertheless effective mistake:

(63.6)^(1/3) = f(63.6) ~= f(64) + f ' (64)(63.6 - 64) = 4 - (1/48)(0.4) = 479/120 = 3.99167

And now your last remark makes more sense

EB

5. Originally Posted by earboth
Hello, Jhevon,

I don't want to pick at you but you have made a small but nevertheless effective mistake:

(63.6)^(1/3) = f(63.6) ~= f(64) + f ' (64)(63.6 - 64) = 4 - (1/48)(0.4) = 479/120 = 3.99167

And now your last remark makes more sense

EB