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What is the speed of the plane if the distance between the person and the plane is increasing at the rate of 300 ft/sec ?

500 ft/sec

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We now return to the problem involving the rocket launch from the beginning of the chapter.

Chapter opener: a rocket launch

A photo of a rocket lifting off.
(credit: modification of work by Steve Jurvetson, Wikimedia Commons)

A rocket is launched so that it rises vertically. A camera is positioned 5000 ft from the launch pad. When the rocket is 1000 ft above the launch pad, its velocity is 600 ft/sec . Find the necessary rate of change of the camera’s angle as a function of time so that it stays focused on the rocket.

Step 1. Draw a picture introducing the variables.

A right triangle is formed with a camera at one of the nonright angles and a rocket at the other nonright angle. The angle with the camera has measure θ. The distance from the rocket to the ground is h; note that this is the side opposite the angle with measure θ. The side adjacent to the angle with measure θ is 5000 ft.
A camera is positioned 5000 ft from the launch pad of the rocket. The height of the rocket and the angle of the camera are changing with respect to time. We denote those quantities with the variables h and θ , respectively.

Let h denote the height of the rocket above the launch pad and θ be the angle between the camera lens and the ground.

Step 2. We are trying to find the rate of change in the angle of the camera with respect to time when the rocket is 1000 ft off the ground. That is, we need to find d θ d t when h = 1000 ft . At that time, we know the velocity of the rocket is d h d t = 600 ft/sec .

Step 3. Now we need to find an equation relating the two quantities that are changing with respect to time: h and θ . How can we create such an equation? Using the fact that we have drawn a right triangle, it is natural to think about trigonometric functions. Recall that tan θ is the ratio of the length of the opposite side of the triangle to the length of the adjacent side. Thus, we have

tan θ = h 5000 .

This gives us the equation

h = 5000 tan θ .

Step 4. Differentiating this equation with respect to time t , we obtain

d h d t = 5000 sec 2 θ d θ d t .

Step 5. We want to find d θ d t when h = 1000 ft . At this time, we know that d h d t = 600 ft/sec . We need to determine sec 2 θ . Recall that sec θ is the ratio of the length of the hypotenuse to the length of the adjacent side. We know the length of the adjacent side is 5000 ft . To determine the length of the hypotenuse, we use the Pythagorean theorem, where the length of one leg is 5000 ft , the length of the other leg is h = 1000 ft , and the length of the hypotenuse is c feet as shown in the following figure.

A right triangle has one angle with measure θ. The hypotenuse is c, the side length opposite the angle with measure θ is 1000, and the side adjacent to the angle with measure θ is 5000.

We see that

1000 2 + 5000 2 = c 2

and we conclude that the hypotenuse is

c = 1000 26 ft .

Therefore, when h = 1000 , we have

sec 2 θ = ( 1000 26 5000 ) 2 = 26 25 .

Recall from step 4 that the equation relating d θ d t to our known values is

d h d t = 5000 sec 2 θ d θ d t .

When h = 1000 ft , we know that d h d t = 600 ft/sec and sec 2 θ = 26 25 . Substituting these values into the previous equation, we arrive at the equation

600 = 5000 ( 26 25 ) d θ d t .

Therefore, d θ d t = 3 26 rad/sec .

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What rate of change is necessary for the elevation angle of the camera if the camera is placed on the ground at a distance of 4000 ft from the launch pad and the velocity of the rocket is 500 ft/sec when the rocket is 2000 ft off the ground?

1 10 rad/sec

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In the next example, we consider water draining from a cone-shaped funnel. We compare the rate at which the level of water in the cone is decreasing with the rate at which the volume of water is decreasing.

Questions & Answers

what is function?
Ryan Reply
A set of points in which every x value (domain) corresponds to exactly one y value (range)
what is lim (x,y)~(0,0) (x/y)
NIKI Reply
limited of x,y at 0,0 is nt defined
But using L'Hopitals rule is x=1 is defined
Could U explain better boss?
value of (x/y) as (x,y) tends to (0,0) also whats the value of (x+y)/(x^2+y^2) as (x,y) tends to (0,0)
can we apply l hospitals rule for function of two variables
why n does not equal -1
K.kupar Reply
ask a complete question if you want a complete answer.
I agree with Andrew
f (x) = a is a function. It's a constant function.
Darnell Reply
proof the formula integration of udv=uv-integration of vdu.?
Bg Reply
Find derivative (2x^3+6xy-4y^2)^2
Rasheed Reply
no x=2 is not a function, as there is nothing that's changing.
Vivek Reply
are you sure sir? please make it sure and reply please. thanks a lot sir I'm grateful.
i mean can we replace the roles of x and y and call x=2 as function
if x =y and x = 800 what is y
Joys Reply
how do u factor the numerator?
Drew Reply
Nonsense, you factor numbers
You can factorize the numerator of an expression. What's the problem there? here's an example. f(x)=((x^2)-(y^2))/2 Then numerator is x squared minus y squared. It's factorized as (x+y)(x-y). so the overall function becomes : ((x+y)(x-y))/2
The problem is the question, is not a problem where it is, but what it is
I think you should first know the basics man: PS
Yes, what factorization is
Antonio bro is x=2 a function?
Yes, and no.... Its a function if for every x, y=2.... If not is a single value constant
you could define it as a constant function if you wanted where a function of "y" defines x f(y) = 2 no real use to doing that though
Why y, if domain its usually defined as x, bro, so you creates confusion
Its f(x) =y=2 for every x
Yes but he said could you put x = 2 as a function you put y = 2 as a function
F(y) in this case is not a function since for every value of y you have not a single point but many ones, so there is not f(y)
x = 2 defined as a function of f(y) = 2 says for every y x will equal 2 this silly creates a vertical line and is equivalent to saying x = 2 just in a function notation as the user above asked. you put f(x) = 2 this means for every x y is 2 this creates a horizontal line and is not equivalent
The said x=2 and that 2 is y
that 2 is not y, y is a variable 2 is a constant
So 2 is defined as f(x) =2
No y its constant =2
what variable does that function define
the function f(x) =2 takes every input of x within it's domain and gives 2 if for instance f:x -> y then for every x, y =2 giving a horizontal line this is NOT equivalent to the expression x = 2
Yes true, y=2 its a constant, so a line parallel to y axix as function of y
Sorry x=2
And you are right, but os not a function of x, its a function of y
As function of x is meaningless, is not a finction
yeah you mean what I said in my first post, smh
I mean (0xY) +x = 2 so y can be as you want, the result its 2 every time
OK you can call this "function" on a set {2}, but its a single value function, a constant
well as long as you got there eventually
2x^3+6xy-4y^2)^2 solve this
follow algebraic method. look under factoring numerator from Khan academy
volume between cone z=√(x^2+y^2) and plane z=2
Kranthi Reply
answer please?
It's an integral easy
V=1/3 h π (R^2+r2+ r*R(
How do we find the horizontal asymptote of a function using limits?
Lerato Reply
Easy lim f(x) x-->~ =c
solutions for combining functions
Amna Reply
what is a function? f(x)
Jeremy Reply
one that is one to one, one that passes the vertical line test
It's a law f() that to every point (x) on the Domain gives a single point in the codomain f(x)=y
is x=2 a function?
restate the problem. and I will look. ty
jon Reply
is x=2 a function?
What is limit
MaHeSh Reply
it's the value a function will take while approaching a particular value
don ger it
what is a limit?
it is the value the function approaches as the input approaches that value.
Its' complex a limit It's a metrical and topological natural question... approaching means nothing in math
is x=2 a function?
Practice Key Terms 1

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