Calculus question?

Does anyone know how to find the answer to this problem?

A television camera at ground level is filming the lift-off of a space shuttle that is rising according to the formula s=60t^2, where s is measured in feet and t is measured in seconds. The camera is 2000 feet from the launchpad. Find the rate of change in the angle of elevation of the television camera (d angle/dt) 10 seconds after liftoff.

I'd like to see a step by step answer to this one. I'll admit it. I'm lost.

Any calculus gurus around now?

Loved algebra, trig and geometry, though. :shrug:

There might be a tangent in there or cosecant, its a triangle solution provided
you know the acceleration equation of the lifting rocket.

because the shuttle actually does a roll and has a curved path.

It's a problem that has me stumped. I'm new to calculus and trying to get better at it.

square of the hypotenus is equal to the sum of the square of the right angle at a given point in time.

10 seconds x 60 feet = 600 feet, 600 x 600 =

2000 ft distance x2000 =

ad the two, find the square to find the length of the angle.

from that point I gotta go to a book or draw it and use a protractor to find angle

divide angle difference by 10.

I think.

Yes, a triangle is involved, in fact a right-triangle. One side of this right triangle is formed by the vertical ascent of the space shuttle, s(t) = 60*t^2. The other side of this right triangle is the distance between the camera and the rocket launch pad, which is equal to 2000. We are not too concerned about the hypotenuse. The crucial insight is to notice the right triangle. Anyway, from there you can get the angle, call it angle theta, that the camera makes as it rolls upward in the following way:

tan (theta) = s(t) / 2000 = 60*t^2 / 2000

Then solve for the angle theta:

(theta) = arctan ( 60*t^2 / 2000 )

and then take its derivative:

d/dt (theta) = <1 / ((60*t^2 / 2000)^2 + 1)> * (120*t / 2000)

Now you have the equation for the rate of change for the angle theta.

looks a lot better than my answer. At least it has a derivative in there which is a part of calculus

I just don't get it but I failed general math twice and dropped out at 17, by the way welcome to DU and don't wait for the next math question for your second reply, it may be a while.

:applause:

necessary to find?

Yes, I'm that lost. And thank you for trying to help me out here. :hi:

No, we are not too concerned with an actual numerical value for angle theta. We just want to know, firstly, what angle theta is as a function of time, and, secondly, what the derivative(s) of this function is(are). In this problem we are just looking for the first derivative, that is, what the rate of change of angle theta is with respect to time (d/dt). The derivative of the arctan function is as follows, which might be listed somewhere in your textbook:

d/dt arctan (t) = 1 / (t^2 + 1)

And of course don't forget to apply the chain rule for the argument (t) of the arctan function when taking its derivative.

Anyway, the function of s(t) = 60*t^2, if you graph it out, is a parabola opened upwards, with the very bottom located at (0,0). What the derivative of this function tells you, is what the slope of this function is at whatever point in time you are looking at, say at t = 10 seconds. This value will of course vary, depending on where you are on the curve. The units for the time rate of change for angle theta will be radians / second (or radians per second).

I will have to work on it again later. It's bedtime now. I appreciate all the responses. Sooner or later, maybe I'll start to get it. It's already clearer than it was.

Thanks again.

Very, very sorry. Had the wrong function listed. I should have said "...the function (theta) = arctan( 60*t^2 / 2000), when you graph it out,... ". Still, everything that follows remains the same, i.e. that this function is a parabola opened upwards, etc. Cheers.

And thanks for this VERY astute answer! :hi:

By definition the tangent of an angle is the length of the opposite side of a triangle divided by the adjacent side. So the tangent of the angle you are looking for is 60t^2 / 2000 or more importantly find the inverse tangent of 60t^2 / 2000 and you should get the angle. Divide this angle by 10 seconds and you would get the rate of change. I think this is right. Try it and let us know.

You need to imagine a right triangle whose base is 2000 feet and whose vertical leg is the height of the shuttle at any given moment, which is what you have called "s."

The angle of elevation is therefore the arctangent of s/2000.

So the value you are looking for is d(arctan((60t^2)/2000))/dt.

Then I had to go to this page to find the formulas I need. The formula you need is about 80% of the way down. It looks like d(arctan(ax^2)/dx = 2ax/(1+(ax)^2). So substituting 60/2000 for a and t for x, and knowing that t=10, we arrive at a solution of 2(60/2000)(10)/1+((60/2000)*10)^2.

This is a lot easier when you can put stuff on multiple lines, but it looks like the numerator simplifies to 1200/2000, and the denominator simplifies to 1+(600/2000)^2.

I leave the rest to you. The answer should be expressed in radians.

Fortunately for me, others got here first so I don't have to do all that typing! :)

I'll only comment that the answer should be in radians per second, and that -- unless I'm having a bad algebra moment -- your denominator should be 1+(6000/2000)^2.

My bad -- you're correct that the answer should be expressed in radians per second, not just radians.

Your bad -- I checked the algebra in my denominator and it appears correct: Variable a is 60/2000, which is multiplied by t which is 10 to get 600/2000, then the whole thing is squared.

Still, nice to know the calc I took 20 years ago is still with me!

I will have to study this further before it's totally clear to me, but it is clearer than it was now thanks to all of you.

Again, thanks. Nighty night.

...to get someone else to do one's homework?????

Sorry if i'm wrong, but that just looks too much like: "Joe set's out from...." for more advanced students.

:spank: :spank: all round if I'm right. :D