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Tju [1.3M]
3 years ago
8

A man standing on a lighthouse at a height of 124 feet sights two boats directly in front of him. One is at an angle of depressi

on of 62°, and the other is at an angle of depression of 33°. Identify the distance between the two boats. Round your answer to the nearest foot.

Mathematics
1 answer:
Ksivusya [100]3 years ago
5 0

Answer:

125\ ft

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

In the right triangle ABC find the length side BC

we know that

tan(62\°)=\frac{124}{BC}

BC=\frac{124}{tan(62\°)}

step 2

In the right triangle ABD find the length side BD

we know that

tan(33\°)=\frac{124}{BD}

BD=\frac{124}{tan(33\°)}

step 3

we know that

The distance between the two boats is the length side CD

CD=BD-BC

substitute the values  

CD=\frac{124}{tan(33\°)}-\frac{124}{tan(62\°)}=125\ ft

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This is a "water tank" calculus problem that I've been working on and I would really appreciate it if someone could look at my w
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Part A

Everything looks good but line 4. You need to put all of the "2h" in parenthesis so the teacher will know you are squaring all of 2h. As you have it right now, you are saying "only square the h, not the 2". Be careful as silly mistakes like this will often cost you points. 

============================================================

Part B

It looks like you have the right answer. Though you'll need to use parenthesis to ensure that all of "75t/(2pi)" is under the cube root. I'm assuming you made a typo or forgot to put the parenthesis. 

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Answer:
h(t) = [ (75t)/(2pi) ]^(1/3)

============================================================

Part C

Your answer is correct. 
Below is an alternative way to find the same answer

--------------------------------------

Plug in the given height; solve for t
h(t) = [ (75t)/(2pi) ]^(1/3)
8 = [ (75t)/(2pi) ]^(1/3)
8^3 = (75t)/(2pi)
512 = (75t)/(2pi)
(75t)/(2pi) = 512
75t = 512*2pi
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t = 1024pi/75
At this time value, the height of the water is 8 feet

Set up the radius r(t) function 
r = 2*h
r = 2*h(t)
r = 2*[ (75t)/(2pi) ]^(1/3) .... using the answer from part B

Differentiate that r(t) function with respect to t
r = 2*[ (75t)/(2pi) ]^(1/3)
dr/dt = 2*(1/3)*[ (75t)/(2pi) ]^(1/3-1)*d/dt[(75t)/(2pi)] 
dr/dt = (2/3)*[ (75t)/(2pi) ]^(-2/3)*(75/(2pi))
dr/dt = (2/3)*(75/(2pi))*[ (75t)/(2pi) ]^(-2/3)
dr/dt = (25/pi)*[ (75t)/(2pi) ]^(-2/3)

Plug in t = 1024pi/75 found earlier above
dr/dt = (25/pi)*[ (75t)/(2pi) ]^(-2/3)
dr/dt = (25/pi)*[ (75(1024pi/75))/(2pi) ]^(-2/3)
dr/dt = (25/pi)*[ (1024pi)/(2pi) ]^(-2/3)
dr/dt = (25/pi)*(1/64)
dr/dt = 25/(64pi)
getting the same answer as before

----------------------------

Thinking back as I finish up, your method is definitely shorter and more efficient. So I prefer your method, which is effectively this:
r = 2h, dr/dh = 2
dh/dt = (25)/(2pi*h^2) ... from part A
dr/dt = dr/dh*dh/dt ... chain rule
dr/dt = 2*((25)/(2pi*h^2))
dr/dt = ((25)/(pi*h^2))
dr/dt = ((25)/(pi*8^2)) ... plugging in h = 8
dr/dt = (25)/(64pi)
which is what you stated in your screenshot (though I added on the line dr/dt = dr/dh*dh/dt to show the chain rule in action)
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