# 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.

find the domain and range of f(x)= 4x-7/x²-6x+8
find the range of f(x)=(x+1)(x+4)
-1, -4
Marcia
That's domain. The range is [-9/4,+infinity)
Jacob
If you're using calculus to find the range, you have to find the extrema through the first derivative test and then substitute the x-value for the extrema back into the original equation.
Jacob
Good morning,,, how are you
d/dx{1/y - lny + X^3.Y^5}
How to identify domain and range
hello
Akpevwe
He,,
Harrieta
hi
Dr
hello
velocity
I only talk to girls
Dr
women are smart then guys
Dr
Smarter
sorry
Dr
Dr
:(
Shun
was up
Dr
hello
is it chatting app?.. I do not see any calculus here. lol
Find the arc length of the graph of f(x) = In (sinx) on the interval [Π/4, Π/2].
Sand falling freely from a lorry form a conical shape whose height is always equal to one-third the radius of the base. a. How fast is the volume increasing when the radius of the base is (1m) and increasing at the rate of 1/4cm/sec Pls help me solve
show that lim f(x) + lim g(x)=m+l
list the basic elementary differentials
Differentiation and integration
yes
Damien
proper definition of derivative
the maximum rate of change of one variable with respect to another variable
terms of an AP is 1/v and the vth term is 1/u show that the sum of uv terms is 1/2(uv+1)
what is calculus?
calculus is math that studies the change in math, such as the rate and distance,
Tamarcus
what are the topics in calculus
Augustine
what is limit of a function?
what is x and how x=9.1 take?
what is f(x)
the function at x
Marc
also known as the y value so I could say y=2x or f(x)= 2x same thing just using functional notation your next question is what is dependent and independent variables. I am Dyslexic but know math and which is which confuses me. but one can vary the x value while y depends on which x you use. also
Marc
up domain and range
Marc
enjoy your work and good luck
Marc
I actually wanted to ask another questions on sets if u dont mind please?
Inembo
I have so many questions on set and I really love dis app I never believed u would reply
Inembo
Hmm go ahead and ask you got me curious too much conversation here
am sorry for disturbing I really want to know math that's why *I want to know the meaning of those symbols in sets* e.g n,U,A', etc pls I want to know it and how to solve its problems
Inembo
and how can i solve a question like dis *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
next questions what do dy mean by (A' n B^c)^c'
Inembo
The sets help you to define the function. The function is like a magic box where you put inside stuff(numbers or sets) and you get out the stuff but in different shapes (forms).
I dont understand what you wanna say by (A' n B^c)^c'
(A' n B (rise to the power of c)) all rise to the power of c
Inembo
Aaaahh
Ok so the set is formed by vectors and not numbers
A vector of length n
But you can make a set out of matrixes as well
I I don't even understand sets I wat to know d meaning of all d symbolsnon sets
Inembo
High-school?
yes
Inembo
am having big problem understanding sets more than other math topics
Inembo
So f:R->R means that the function takes real numbers and provides real numer. For ex. If f(x) =2x this means if you give to your function a real number like 2,it gives you also a real number 2times2=4
pls answer this question *in a group of 40 students, 32 offer maths and 24 offer physics and 4 offer neither maths nor physics , how many offer both maths and physics*
Inembo
If you have f:R^n->R^n you give to your function a vector of length n like (a1,a2,...an) where all a1,.. an are reals and gives you also a vector of length n... I don't know if i answering your question. Otherwise on YouTube you havr many videos where they explain it in a simple way
I would say 24
Offer both
Sorry 20
Actually you have 40 - 4 =36 who offer maths or physics or both.
I know its 20 but how to prove it
Inembo
You have 32+24=56who offer courses
56-36=20 who give both courses... I would say that
solution: In a question involving sets and Venn diagram, the sum of the members of set A + set B - the joint members of both set A and B + the members that are not in sets A or B = the total members of the set. In symbolic form n(A U B) = n(A) + n (B) - n (A and B) + n (A U B)'.
Mckenzie
In the case of sets A and B use the letters m and p to represent the sets and we have: n (M U P) = 40; n (M) = 24; n (P) = 32; n (M and P) = unknown; n (M U P)' = 4
Mckenzie
Now substitute the numerical values for the symbolic representation 40 = 24 + 32 - n(M and P) + 4 Now solve for the unknown using algebra: 40 = 24 + 32+ 4 - n(M and P) 40 = 60 - n(M and P) Add n(M and P), as well, subtract 40 from both sides of the equation to find the answer.
Mckenzie
40 - 40 + n(M and P) = 60 - 40 - n(M and P) + n(M and P) Solution: n(M and P) = 20
Mckenzie
thanks
Inembo
Simpler form: Add the sums of set M, set P and the complement of the union of sets M and P then subtract the number of students from the total.
Mckenzie
n(M and P) = (32 + 24 + 4) - 40 = 60 - 40 = 20
Mckenzie