Guy pushing girl on a swing models h.displacement with function

[aeromat]

1,2,3. Homework Statement , and work done.
After Lee gives his little sister Kara a big push on a swing, her horizontal position as a function of time is given by the equation $$x(t) = 3cost(t)*e^{-0.05t}$$ , where x(t) is her horizontal displacement, in metres, from the lowest point of her swing, as a function of time, t, in seconds.

a) From what horizontal distance from the bottom of Kara's swing did Lee push his sister?
I said 3m.

b) Determine the highest speed Kara will reach and when this occurs.
$$x'(t) = -3sin(t)*e^{-0.05t} + 3cos(t)*0.05*e^{-0.05t}$$
$$x'(t) = e^{0.05t}(-3sin(t) - 0.15cos(t))$$
$$x'(t) = -e^{0.05t}(3sin(t) + 0.15cos(t))$$

So,
$$0 = -e^{0.05t}$$ or $$0 = 3sin(t) + 0.15cos(t)$$
$$0 = -e^{0.05t}$$ DNE
$$0 = 3sin(t) + 0.15cos(t)$$
$$-3sin(t) = 0.15cos(t)$$
$$-3tan(t) = 0.15$$
$$tan(t) = 0.15/-3$$
$$t = arctan(0.15/-3)$$

But this gives t = -2.8624; a negative time value.

What did I do wrong?

Part C) How long did it take for Kara's maximum horizontal displacement at the top of her swing arc to diminish to 1m? After how many swings will this occur?

Part D) Sketch the graph <--- I am unsure as to what scale I should use.

 

[lanedance]

you seem to have missed some minus signs in your initial derivative, first in the multiplier after differentiating the exponential (though i think you put that in next step) and then in the exponential itself

 

[lanedance]

now following you working through
$$x(t) = 3 cos(t) e^{-\frac{t}{20}}$$
differentiating
$$x'(t) = -3sin(t) e^{-\frac{t}{20}} + \frac{-1}{20}3cos(t) e^{-\frac{t}{20}}$$
$$x'(t) = -3e^{-\frac{t}{20}}(sin(t) + \frac{1}{20}cos(t))$$

then
$$x'(t) = 0 \ \ \to \ \ sin(t) + \frac{1}{20}cos(t) = 0$$

giving
$$tan(t) = \frac{sin(t)}{cos(t)} = -\frac{1}{20}$$

which is what you have, however setting the first derivative equal to zero will maximise the horizontal distance, not the speed

 

[lanedance]

http://www.wolframalpha.com/input/?i=tan(t)
also note if you drew a line y= -1/20 it would intesect tan(x) infinite times, the first positive value is close to pi, eahc one of these corresponds to local maxima in position

 

[Ray Vickson]

1,2,3. Homework Statement , and work done.
After Lee gives his little sister Kara a big push on a swing, her horizontal position as a function of time is given by the equation $$x(t) = 3cost(t)*e^{-0.05t}$$ , where x(t) is her horizontal displacement, in metres, from the lowest point of her swing, as a function of time, t, in seconds.

a) From what horizontal distance from the bottom of Kara's swing did Lee push his sister?
I said 3m.

b) Determine the highest speed Kara will reach and when this occurs.
$$x'(t) = -3sin(t)*e^{-0.05t} + 3cos(t)*0.05*e^{-0.05t}$$
$$x'(t) = e^{0.05t}(-3sin(t) - 0.15cos(t))$$
$$x'(t) = -e^{0.05t}(3sin(t) + 0.15cos(t))$$

So,
$$0 = -e^{0.05t}$$ or $$0 = 3sin(t) + 0.15cos(t)$$
$$0 = -e^{0.05t}$$ DNE
$$0 = 3sin(t) + 0.15cos(t)$$
$$-3sin(t) = 0.15cos(t)$$
$$-3tan(t) = 0.15$$
$$tan(t) = 0.15/-3$$
$$t = arctan(0.15/-3)$$

But this gives t = -2.8624; a negative time value.

What did I do wrong?

Part C) How long did it take for Kara's maximum horizontal displacement at the top of her swing arc to diminish to 1m? After how many swings will this occur?

Part D) Sketch the graph <--- I am unsure as to what scale I should use.

You want to find where x'(t) is largest, so you need to look at its *derivative*. When you set its derivative (d/dt) x'(t) to zero, you are finding either a local minimum or local maximum---both have derivative = 0. You need other tests to determine which points are minima and which are maxima; there will be infinitely many points of either type (that is, *local* optima), but among all the local maxima, one will stand out at the absolute maximum. This will all be clear if you plot x'(t) for a large enough range of t.

In your work above you found times where the horizontal speed = 0, and these occur at times t where tan(t) = -1/20. There are infinitely many such t, all differing by multiples of pi (because tan(w+n*pi) = tan(w) for n = 1,2,3,4,... ).

RGV