# 4.1 Related rates  (Page 3/7)

 Page 3 / 7

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 derivative of antilog x dx ?
what's the meaning of removable discontinuity
what's continuous
Brian
an area under a curve is continuous because you are looking at an area that covers a range of numbers, it is over an interval, such as 0 to 4
Lauren
using product rule x^3,x^5
Kabiru
may god be with you
Sunny
Luke 17:21 nor will they say, See here or See there For indeed, the kingdom of God is within you. You've never 'touched' anything. The e-energy field created by your body has pushed other electricfields. even our religions tell us we're the gods. We live in energies connecting us all. Doa/higgsfield
Scott
if you have any calculus questions many of us would be happy to help and you can always learn or even invent your own theories and proofs. math is the laws of logic and reality. its rules are permanent and absolute. you can absolutely learn calculus and through it better understand our existence.
Scott
ya doubtless
Bilal
help the integral of x^2/lnxdx
Levis
also find the value of "X" from the equation that follow (x-1/x)^4 +4(x^2-1/x^2) -6=0 please guy help
Levis
Use integration by parts. Let u=lnx and dv=x2dx Then du=1xdx and v=13x3. ∫x2lnxdx=13x3lnx−∫(13x3⋅1x)dx ∫x2lnxdx=13x3lnx−∫13x2dx ∫x2lnxdx=13x3lnx−19x3+C
Bilal
itz 1/3 and 1/9
Bilal
now you can find the value of X from the above equation easily
Bilal
Pls i need more explanation on this calculus
usman
usman from where do you need help?
Levis
thanks Bilal
Levis
Do we ask only math question? or ANY of the question?
yh
Gbesemete
How do i differentiate between substitution method, partial fraction and algebraic function in integration?
usman
you just have to recognize the problem. there can be multiple ways to solve 1 problem. that's the hardest part about integration
Lauren
test
we asking the question cause only the question will tell us the right answer
Sunny
find integral of sin8xcos12xdx
don't share these childish questions
Bilal
well find the integral of x^x
Levis
bilal kumhar you are so biased if you are an expert what are you doing here lol😎😎😂😂 we are here to learn and beside there are many questions on this chat which you didn't attempt we are helping each other stop being naive and arrogance so give me the integral of x^x
Levis
Levis I am sorry
Bilal
Bilal it okay buddy honestly i am pleasured to meet you
Levis
x^x ... no anti derivative for this function... but we can find definte integral numerically.
Bilal
thank you Bilal Kumhar then how we may find definite integral let say x^x,3,5?
Levis
evaluate 5-×square divided by x+2 find x as limit approaches infinity
i have not understood
Leo
Michael
welcome
Sunny
I just dont get it at all...not understanding
Michagaye
0 baby
Sunny
The denominator is the aggressive one
Sunny
wouldn't be any prime number for x instead ?
Harold
or should I say any prime number greater then 11 ?
Harold
just wondering
Harold
I think as limit Approach infinity then X=0
Levis
ha hakdog hahhahahaha
ha hamburger
Leonito
Fond the value of the six trigonometric function of an angle theta, which terminal side passes through the points(2x½-y)²,4
What's f(x) ^x^x
What's F(x) =x^x^x
Emeka
are you asking for the derivative
Leo
that's means more power for all points
rd
Levis
iam sorry f(x)=x^x it means the output(range ) depends to input(domain) value of x by the power of x that is to say if x=2 then x^x would be 2^2=4 f(x) is the product of X to the power of X its derivatives is found by using product rule y=x^x introduce ln each side we have lny=lnx^x =lny=xlnx
Levis
the derivatives of f(x)=x^x IS (1+lnx)*x^x
Levis
what is a maximax
A maxima in a curve refers to the maximum point said curve. The maxima is a point where the gradient of the curve is equal to 0 (dy/dx = 0) and its second derivative value is a negative (d²y/dx² = -ve).
Viewer
what is the limit of x^2+x when x approaches 0
it is 0 because 0 squared Is 0
Leo
0+0=0
Leo
simply put the value of 0 in places of x.....
Tonu
the limit is 2x + 1
Nicholas
the limit is 0
Muzamil
limit s x
Bilal
The limit is 3
Levis
Leo we don't just do like that buddy!!! use first principle y+∆y=x+∆x ∆y=x+∆x-y ∆y=(x+∆x)^2+(x+∆x)-x^2+x on solving it become ∆y=3∆x+∆x^2 as ∆x_>0 limit=3 if you do by calculator say plugging any value of x=0.000005 which approach 0 you get 3
Levis
find derivatives 3√x²+√3x²
3 + 3=6
mujahid
How to do basic integrals
the formula is simple x^n+1/n+1 where n IS NOT EQUAL TO 1 And n stands for power eg integral of x^2 x^2+1/2+1 =X^3/3
Levis
write something lmit
find the integral of tan tanxdx
-ln|cosx| + C
Jug
lnSecx+c
Levis