Stumped by these problems

[Guest BaNZai]

Hello, I am completely stumped by some of these problems, and have no where else to get help, can anyone help me with these problems? Any help would be greatly appreciated. Thank you very much.

 

1. The function v(t)=12t(squared) - 16t is the velocity in m/sec of a particle moving along the x-axis, where t is measured in seconds. Use analytic methods to find the particles displacement for 0<(greater than or equal to)t<(less than or equal to)5. (round to nearest 10m). (Note: just t is squared in the equation, and it's 0 to 5 including both 0 and 5) (I got 120m but it doesn't seem right)

 

2. A certain spring obeys Hooke's Law and requires a force of 10N to stretch it 8cm beyond its natural length. How much work would be done in stretching it to 12cm beyond its natural length? (Note: I got 90N*m but I'm not sure if it's right)

 

3. Find the area of the region enclosed by y=sin2x and y=cosx for -(negative)PI/2 greater than or equal to X less than or equal to PI/6. (I got -4.5 which seems completely wrong :frown: )

 

These next two I had no idea how to do

 

4. The base of a solid is the region between the line y=4 and the parabola y=x(squared). THe cross sections perpendicular to the x-axis are semi-circles. Find the volume of the solid.

 

5. A region is bounded by the lines: y=(square root of)X , y= X-2, and y=0. Find the volume of the solid generated by rotating this region about the x-axis.

 

THanks again for any help, sorry if it is too much.

[Dave]

Hello' date=' I am completely stumped by some of these problems, and have no where else to get help, can anyone help me with these problems? Any help would be greatly appreciated. Thank you very much.

 

1. The function v(t)=12t(squared) - 16t is the velocity in m/sec of a particle moving along the x-axis, where t is measured in seconds. Use analytic methods to find the particles displacement for 0<(greater than or equal to)t<(less than or equal to)5. (round to nearest 10m). (Note: just t is squared in the equation, and it's 0 to 5 including both 0 and 5) [b'](I got 120m but it doesn't seem right)[/b]

 

[math]s = \int_0^5 v(t)\, dt = \int_0^5 (12t^2-16t) \, dt[/math]

So we have [math] s = \left[ 4t^3 - 8t^2 \right]_0^5[/math]

 

And so s = 4*125 - 8*25 = 300 metres.

 

Unless I've made a big mistake (which is quite plausible considering the time).

 

2. A certain spring obeys Hooke's Law and requires a force of 10N to stretch it 8cm beyond its natural length. How much work would be done in stretching it to 12cm beyond its natural length? ([b']Note: I got 90N*m but I'm not sure if it's right)[/b]

 

Haven't done springs for about a year now so can't remember anything about them, sorry

 

3. Find the area of the region enclosed by y=sin2x and y=cosx for -(negative)PI/2 greater than or equal to X less than or equal to PI/6. (I got -4.5 which seems completely wrong :frown: )

 

If you draw a graph, things will become much clearer. First off, a negative value for an integral means that the area lies below the x-axis; so the area is the modulus (absolute value) of the integral.

 

Also realise that when you calculate an integral on a function that moves from being negative to positive (for example, sin(2x) in the region you've given), it's value will not be the area under the curve.

 

I think the integral you'll want to evaluate will be this:

 

[math]\left| \int_{-\tfrac{\pi}{2}}^{0} \sin(2x) \, dx \right| + \int_{0}^{\tfrac{\pi}{6}} \sin(2x) \, dx + \int_{-\tfrac{\pi}{2}}^{\frac{\pi}{6}} \cos(x) \, dx[/math].

 

As I said, draw a graph.

 

These next two I had no idea how to do

 

4. The base of a solid is the region between the line y=4 and the parabola y=x(squared). THe cross sections perpendicular to the x-axis are semi-circles. Find the volume of the solid.

 

I'm not quite sure of the question there to be honest. Perhaps someone else can shed some light on it.

 

5. A region is bounded by the lines: y=(square root of)X , y= X-2, and y=0. Find the volume of the solid generated by rotating this region about the x-axis.

 

Just use the standard formula for rotation of a solid about the x-axis, it's not too hard. You're going to rotate the region bounded by sqrt(x) and the x-axis, then take a conical chunk out of it with y = x-2.

 

Hope this helps

[Guest BaNZai]

AH! I see, I took the derivative for the first one instead of doing the anti-derivative. Thank you very much, I should have noticed that error. I think I got some of the others too. Thank you very much.

[bloodhound]

i looked at question 4 and it didnt make any sense to me as well