You don't have to do the problem, but I just want to know how you would at least set these problems up.
1) A particle moves along a line with a velocity function v(t) = (v^2) - t, where v is measured in meters per second. Find (a) the displacement and (b) the distance traveled by the particle during the time interval [0,5].
2) Find the area of the region bounded by the curves: x + y = 0, and x = (y^2) + 3y.
3) Find the volume of the solid obtained by rotating the region bounded by:
x = 1 + (y^2) and y = x - 3
1) A particle moves along a line with a velocity function v(t) = (v^2) - t, where v is measured in meters per second. Find (a) the displacement and (b) the distance traveled by the particle during the time interval [0,5].
2) Find the area of the region bounded by the curves: x + y = 0, and x = (y^2) + 3y.
3) Find the volume of the solid obtained by rotating the region bounded by:
x = 1 + (y^2) and y = x - 3
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1) Huh? What is the v in "v^2 - t"? If it's the same v as on the left, you have a screwy problem. If its a constant, you have an easy problem, but the units are screwy....how do you subtract some number f seconds from another number of meters^2 per second^2?
Assuming v on the right is a constant, then you get x(t) by integrating from 0 to t:
x(t) = x(0) + ∫[0,t]v(t) = x(0) + v²t - t∫²/2
The displacement is x(5) - x(0), and the total distance depends on whether there was a reversal in velocity in the first 5 seconds. If not, the distance traveled is |x(5) - x(0)|. If there was a reversal at time t1, then the distance traveled is |x(t1) - x(0)| + |x(5) - x(t1)|. The reversal happens when v(t) = 0, so that v² - t = 0. So, if the constant v² is in (0,5) then t1=v² is also in (0,5) and you have to do the two-part version.
2) Solve for the intersections to find your limits of integration. The first eq. gives x = -y. Substituting that into the second equation gives
y² + 3y = -y
y² + 4y = 0
y(y + 4) = 0
So the two points of intersection are when y = 0 or -4 and the corresponding x=-y values are x=0 or 4. The intersections are at (0,0) and (4,-4).
Because the second equation is in the form x=f(y), I'd do the same for the first, giving:
x = -y
and integrate |(y² + 3y) - (-y)| dy from y=-4 to 0.
3). Again, start by finding the intersections. Rewrite the 2nd eq. as x = y+3 and then the 1st eq. becomes:
y + 3 = 1 + y²
y² - y - 2 = 0
(y - 2)(y + 1) = 0
So y is 2 or -1 and the intersection points are (5,2) and (2,-1).
That's as far as you can go for now, since you didn't specify an axis of rotation.
Assuming v on the right is a constant, then you get x(t) by integrating from 0 to t:
x(t) = x(0) + ∫[0,t]v(t) = x(0) + v²t - t∫²/2
The displacement is x(5) - x(0), and the total distance depends on whether there was a reversal in velocity in the first 5 seconds. If not, the distance traveled is |x(5) - x(0)|. If there was a reversal at time t1, then the distance traveled is |x(t1) - x(0)| + |x(5) - x(t1)|. The reversal happens when v(t) = 0, so that v² - t = 0. So, if the constant v² is in (0,5) then t1=v² is also in (0,5) and you have to do the two-part version.
2) Solve for the intersections to find your limits of integration. The first eq. gives x = -y. Substituting that into the second equation gives
y² + 3y = -y
y² + 4y = 0
y(y + 4) = 0
So the two points of intersection are when y = 0 or -4 and the corresponding x=-y values are x=0 or 4. The intersections are at (0,0) and (4,-4).
Because the second equation is in the form x=f(y), I'd do the same for the first, giving:
x = -y
and integrate |(y² + 3y) - (-y)| dy from y=-4 to 0.
3). Again, start by finding the intersections. Rewrite the 2nd eq. as x = y+3 and then the 1st eq. becomes:
y + 3 = 1 + y²
y² - y - 2 = 0
(y - 2)(y + 1) = 0
So y is 2 or -1 and the intersection points are (5,2) and (2,-1).
That's as far as you can go for now, since you didn't specify an axis of rotation.