Calculus question involving rates and derivatives.?

Question

A rocket is rising vertically according to the position equation s = 60t^2. If a camera 4000ft away is filming the launch, find the rate of change of the angle of elevation after 3 seconds. Use radians.

Answer 1

Let tan theta = (60 t^2) / 4000

sec^2 theta * dtheta / dt = 60/4000 * 2t - Eq. (1)

But, we know theta = arctan (540/4000)

sec theta = 1 / cos(arctan (540/4000))

You can find sec^2 theta with your calculator and solve for dtheta/dt which will be in radians - the right hand side of Eq (1). simplifies to 1/9 and you can easily find d theta / dt. AbeLincolnsParty gave you the answer but he did not explain his work and I wouldn't trust if he's right - he may have pressed the wrong keys on his calculator (which happens sometimes).

Hope this helps and I hope you understand how to solve these problems - they can be tricky at first but with practice - they become easier.

Answer 2

The angle of elevation is arctan ((60t^2)/4000). If you differentiate this with respect to t using the chain rule you get the answer.

Answer 3

tan(theta) = oppsite/adjacent = height/4000 = s(t)/4000

s(t) = 60t^2 ---------------> ds/dt = 120t = velocity in ft/sec

since t=3 seconds, s(3) = 60*9 = 540ft, velocity(3sec) = 360ft/s

d/dt tan(theata) =[sec(theata)]^2 * dtheata/dt = 120t/4000

remember that sec(theata)*cos(theata) = 1, so we multiply by [cos(theata)]^2 on each side of the above equation to get every thing on the right side of the equation except dtheata/dt

since at t=3, tan(theata) = 540/4000 = 0.135

Inverse Tangent(0.135) which is about 7.68 degrees, but don't need to know this.

dtheata/dt = [120(3)/4000]*[cos(0.135)]^2 = 0.0899 rads/sec

round to 0.09 rads/sec

please check for errors.

Answer 4

enable c(t) = automobile enable b(t) = bus enable r(t) = distance between them c(t) = one hundred twenty + 80t - automobile reaches intersection a million/2 hour earlier bus. b(t) = ninety + 60t - bus travels a million a million/2 hour after automobile reaches intersection At t=0, bus is one hour previous the intersection and the automobile is one hundred twenty miles previous the intersection. they are one hundred fifty miles aside as obtrusive by skill of the pythagorean theorem. Use the pythagorean theorem to stumble on r as a function of c and b. r(t) = sqrt((c^2(t)+b^2(t)) r(t) = sqrt((one hundred twenty + 80t)^2 + (ninety + 60t)^2)) dr(t) / d(t) = a million/2((one hundred twenty + 80t)^2 + (ninety + 60t)^2)^-0.5 * (2*80*(one hundred twenty+80t) + 2*60*(ninety+60t)) enable t=0 a million/2(one hundred twenty^2+ninety^2)^-0.5 * ((100 and sixty*one hundred twenty)+(one hundred twenty*ninety)) inspite of the above expression simplifies to is the speed of separation.

Answer 5

take first derivative and substitute for t.