# 4.1 Related rates  (Page 3/7)

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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 $300\phantom{\rule{0.2em}{0ex}}\text{ft/sec}?$

$500\phantom{\rule{0.2em}{0ex}}\text{ft/sec}$

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

## Chapter opener: a rocket launch

A rocket is launched so that it rises vertically. A camera is positioned $5000\phantom{\rule{0.2em}{0ex}}\text{ft}$ from the launch pad. When the rocket is $1000\phantom{\rule{0.2em}{0ex}}\text{ft}$ above the launch pad, its velocity is $600\phantom{\rule{0.2em}{0ex}}\text{ft/sec}.$ 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.

Let $h$ denote the height of the rocket above the launch pad and $\theta$ 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 $\frac{d\theta }{dt}$ when $h=1000\phantom{\rule{0.2em}{0ex}}\text{ft}.$ At that time, we know the velocity of the rocket is $\frac{dh}{dt}=600\phantom{\rule{0.2em}{0ex}}\text{ft/sec}.$

Step 3. Now we need to find an equation relating the two quantities that are changing with respect to time: $h$ and $\theta .$ 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 $\text{tan}\phantom{\rule{0.1em}{0ex}}\theta$ is the ratio of the length of the opposite side of the triangle to the length of the adjacent side. Thus, we have

$\text{tan}\phantom{\rule{0.1em}{0ex}}\theta =\frac{h}{5000}.$

This gives us the equation

$h=5000\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}\theta .$

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

$\frac{dh}{dt}=5000\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{2}\theta \phantom{\rule{0.1em}{0ex}}\frac{d\theta }{dt}.$

Step 5. We want to find $\frac{d\theta }{dt}$ when $h=1000\phantom{\rule{0.2em}{0ex}}\text{ft}.$ At this time, we know that $\frac{dh}{dt}=600\phantom{\rule{0.2em}{0ex}}\text{ft/sec}.$ We need to determine ${\text{sec}}^{2}\theta .$ Recall that $\text{sec}\phantom{\rule{0.1em}{0ex}}\theta$ 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 $5000\phantom{\rule{0.2em}{0ex}}\text{ft}.$ To determine the length of the hypotenuse, we use the Pythagorean theorem, where the length of one leg is $5000\phantom{\rule{0.2em}{0ex}}\text{ft},$ the length of the other leg is $h=1000\phantom{\rule{0.2em}{0ex}}\text{ft},$ and the length of the hypotenuse is $c$ feet as shown in the following figure.

We see that

${1000}^{2}+{5000}^{2}={c}^{2}$

and we conclude that the hypotenuse is

$c=1000\sqrt{26}\phantom{\rule{0.2em}{0ex}}\text{ft}.$

Therefore, when $h=1000,$ we have

${\text{sec}}^{2}\theta ={\left(\frac{1000\sqrt{26}}{5000}\right)}^{2}=\frac{26}{25}.$

Recall from step 4 that the equation relating $\frac{d\theta }{dt}$ to our known values is

$\frac{dh}{dt}=5000\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{2}\theta \phantom{\rule{0.1em}{0ex}}\frac{d\theta }{dt}.$

When $h=1000\phantom{\rule{0.2em}{0ex}}\text{ft},$ we know that $\frac{dh}{dt}=600\phantom{\rule{0.2em}{0ex}}\text{ft/sec}$ and ${\text{sec}}^{2}\theta =\frac{26}{25}.$ Substituting these values into the previous equation, we arrive at the equation

$600=5000\left(\frac{26}{25}\right)\frac{d\theta }{dt}\text{.}$

Therefore, $\frac{d\theta }{dt}=\frac{3}{26}\phantom{\rule{0.2em}{0ex}}\text{rad/sec}.$

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 $4000\phantom{\rule{0.2em}{0ex}}\text{ft}$ from the launch pad and the velocity of the rocket is 500 ft/sec when the rocket is $2000\phantom{\rule{0.2em}{0ex}}\text{ft}$ off the ground?

$\frac{1}{10}\phantom{\rule{0.2em}{0ex}}\text{rad/sec}$

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.

what is function?
A set of points in which every x value (domain) corresponds to exactly one y value (range)
Tim
what is lim (x,y)~(0,0) (x/y)
limited of x,y at 0,0 is nt defined
Alswell
But using L'Hopitals rule is x=1 is defined
Alswell
Could U explain better boss?
emmanuel
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)
NIKI
can we apply l hospitals rule for function of two variables
NIKI
why n does not equal -1
Andrew
I agree with Andrew
Bg
f (x) = a is a function. It's a constant function.
proof the formula integration of udv=uv-integration of vdu.?
Find derivative (2x^3+6xy-4y^2)^2
no x=2 is not a function, as there is nothing that's changing.
are you sure sir? please make it sure and reply please. thanks a lot sir I'm grateful.
The
i mean can we replace the roles of x and y and call x=2 as function
The
if x =y and x = 800 what is y
y=800
800
Bg
how do u factor the numerator?
Nonsense, you factor numbers
Antonio
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
The problem is the question, is not a problem where it is, but what it is
Antonio
I think you should first know the basics man: PS
Vishal
Yes, what factorization is
Antonio
Antonio bro is x=2 a function?
The
Yes, and no.... Its a function if for every x, y=2.... If not is a single value constant
Antonio
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
zach
Why y, if domain its usually defined as x, bro, so you creates confusion
Antonio
Its f(x) =y=2 for every x
Antonio
Yes but he said could you put x = 2 as a function you put y = 2 as a function
zach
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)
Antonio
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
zach
The said x=2 and that 2 is y
Antonio
that 2 is not y, y is a variable 2 is a constant
zach
So 2 is defined as f(x) =2
Antonio
No y its constant =2
Antonio
what variable does that function define
zach
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
zach
Yes true, y=2 its a constant, so a line parallel to y axix as function of y
Antonio
Sorry x=2
Antonio
And you are right, but os not a function of x, its a function of y
Antonio
As function of x is meaningless, is not a finction
Antonio
yeah you mean what I said in my first post, smh
zach
I mean (0xY) +x = 2 so y can be as you want, the result its 2 every time
Antonio
OK you can call this "function" on a set {2}, but its a single value function, a constant
Antonio
well as long as you got there eventually
zach
2x^3+6xy-4y^2)^2 solve this
femi
moe
volume between cone z=√(x^2+y^2) and plane z=2
Fatima
It's an integral easy
Antonio
V=1/3 h π (R^2+r2+ r*R(
Antonio
How do we find the horizontal asymptote of a function using limits?
Easy lim f(x) x-->~ =c
Antonio
solutions for combining functions
what is a function? f(x)
one that is one to one, one that passes the vertical line test
Andrew
It's a law f() that to every point (x) on the Domain gives a single point in the codomain f(x)=y
Antonio
is x=2 a function?
The
restate the problem. and I will look. ty
is x=2 a function?
The
What is limit
it's the value a function will take while approaching a particular value
Dan
don ger it
Jeremy
what is a limit?
Dlamini
it is the value the function approaches as the input approaches that value.
Andrew
Thanx
Dlamini
Its' complex a limit It's a metrical and topological natural question... approaching means nothing in math
Antonio
is x=2 a function?
The