A Rate of change and a differentials problem

[aaroffl]

First off i'd like to say Hi to the forums, hehe. I didn't really see a new member area but I suppose this will do. Right now I'm kind of struggling with these two problems that I recently took a quiz on and didn't do so well. I've been trying to figure out how to work them out but i just don't really get it. if anyone could help out it'd be awesome.

(1)

The position (in cm) of a particle along the x-axis at time t(in s) is given by X(t) = t^3-6t^2+8t, t is greater than or equal to 0.

the question:
When is the particle moving to the left? the right?
Acceleration of the particle when the speed is zero.
Total distance traveled from t=0 to t=2.


(2)

( and the one that REALLY gets me)
A person standing 30ft from the base of a building measures the angle of elevation to the top of the building as being 75 degrees with an error of +or- 1.2 degrees.

What is the max error in estimating the height of the building, and the percentage error?

This one has taken me quite a while, I am thinking it has something to do with the law of sines, but when i try it with that method I'm getting a completely unreasonable answer. I kind of have the right idea, i hope, but for some reason I'm just not seeing it.

Thank you in advance for those who wish to take on these.

 

[ozymandias]

(1.) Let x' be the first derivative with respect to t, x'' the second.
a.) Right when x'>0, left when x'<0.
b.) Find at what time t the velocity x'=0, and then substitute this into x''.
c.) compute x(2)-x(0)

(2.) If you know the angle of elevation, a, the height of the building is given by
h = 30ft * tg(a)
Now the error in your case is given by Delta a = +-1.2 degrees = +- 0.02 radians, and the resulting error in h is:
$$\Delta h = (1+\tan(a)^2) \Delta a$$
(If you need further explanations on that one, ask.)
Now you merely need to substitute.
Good luck!



--------
Assaf
http://www.physicallyincorrect.com/"

 

[tacman]

Hi there! Welcome to the forums. Let's see if we can help you out with these problems.

(1) To determine when the particle is moving to the left or right, we need to look at the velocity of the particle. Velocity is the rate of change of position with respect to time, so we can find it by taking the derivative of the position function X(t):

V(t) = dX/dt = 3t^2 - 12t + 8

Now, to find when the particle is moving to the left or right, we need to find the values of t where V(t) is positive and negative. If V(t) is positive, the particle is moving to the right, and if it is negative, the particle is moving to the left.

So, we need to solve the inequality 3t^2 - 12t + 8 > 0. This can be factored as (3t - 4)(t - 2) > 0. The solutions are t < 4/3 and t > 2. Therefore, the particle is moving to the left when t < 4/3 and to the right when t > 2.

To find the acceleration of the particle when the speed is zero, we need to find the values of t where the speed (magnitude of velocity) is zero. This occurs when V(t) = 0, which can be solved using the quadratic formula. We get t = 1 and t = 2 as solutions. Therefore, the acceleration of the particle when the speed is zero is given by the second derivative of the position function at t = 1 and t = 2. This can be found by taking the derivative of V(t), which gives us the acceleration function A(t) = 6t - 12. Plugging in t = 1 and t = 2, we get A(1) = -6 and A(2) = 6.

To find the total distance traveled from t = 0 to t = 2, we need to find the definite integral of the speed function from t = 0 to t = 2. This represents the area under the speed curve, which gives us the total distance traveled. So, we need to evaluate the integral of V(t) from 0 to 2, which gives us 8 cm.

(2) This problem involves using trigonometry and the