Please answer all questions for full points and show your work. Calc 3 Problems
ID: 2831572 • Letter: P
Question
Please answer all questions for full points and show your work. Calc 3 Problems
An athlete throws a ball at an angle of 45 degree to the horizontal at an initial speed of 43 ft/s. The ball leaves the athlete's hand 7 ft above the ground. Where is the shot 2 seconds later? How high does the shot go? Where does the shot land? Find the local maximum, local minimum and saddle point given f(x, y) = y3 + 3x2y - 6x2 - 6y2 + 2 Find the volume of the solid above the paraboloid z = x2 + y2 and below the half-cone z = . You do not need to sketch the region of integration, but here is a grid in case you need it. Find the volume under the surface z = x + 2 that lies above the triangle with vertices (1, 1), (0, 2) and (3, 3). Sketch the region of integration.Explanation / Answer
1) As sin(45 degrees) = 1/sqrt(2) and the cos(45 degrees) = 1/sqrt(2), letting g = -32 ft/sec^2, we get, after 2 seconds,
x = 43 cos(45 degrees) t = 43 * 1/sqrt(2) * 2 = 43 sqrt(2) = 60.8111831820431
y = y0 + vy t + 1/2at^2 = 7 + 43/sqrt(2) * 2 - 16 * 2^2 = 7 + 60.8111831820431 - 64 = 3.8111831820431
b) max. height is when t = 0. This is when v0 = - a t, or t = -v0/a. Then, as the height at this point is
7 -1/2a t^2 = 7 + 16 * (43/sqrt(2)/32)^2 = 21.4453125
We solve for t = 0
0 = -16t^2 + 43/sqrt(2) t + 7 or 16t^2 - 43/sqrt(2) t - 7 = 0
t = 43/32 sqrt(2) +- 1/32 sqrt(43^2/2 + 4*16*7)
We only need the positive solution. It is 2.10790183427209
Then, x = vx t = 43/sqrt(2)*2.10790183427209 = 64.0920022868423
2. f(x,y) = y^3 + 3x^2y -6x^2 - 6y^2 + 2
The gradient is (-12x + 6xy, 3y^2 + 3x^2 - 12y)
First, we find the stationary points where the gradient is the 0 vector
Then, -12x + 6xy = 0
3y^2 + 3x^2 - 12y = 0
Factoring the first expression, we get
6x(y - 2) = 0
Thus, either x = 0 or y = 2
If x = 0, we have
3y^2 + 3x^2 - 12y = 0
3y^2 - 12y = 0
3y(y - 4) = 0
y = 0, 4
(0, 0) or (0, 4)
If y = 2, we have
3y^2 + 3x^2 - 12y = 0
3*2^2 + 3x^2 -12*2 = 0
3x^2 = 12
x = +- 2
(2, 2), (-2, 2)
Thus, our critical points are
The Hessian is
(0, 0) or (0, 4)
(2, 2), (-2, 2)
The Hessian is
(-12x + 6xy, 3y^2 + 3x^2 - 12y)
-12 + 6y 6x
6x 6y - 12
(0, 0), (0, 4), (2, 2), (-2, 2)
(0, 0)
-12 0
0 -12
This is negative def., with 2 -12 eigenvalues, so this is a maximum
(0, 4)
12 0
0 12
This is positive definite, with 2 positive eigenvalues of 12, so this is a minimum.
(2, 2)
This is
0 12
12 0
This has eigenvalues of 12 and -12, so this is a saddlepoint.
For (-2, 2), this is still
0 12
12 0
which is again a saddlepoint.
f(0, 0) = y^3 + 3x^2y -6x^2 - 6y^2 + 2 = 0+0+0+0+2 is a maximum
f(0, 4) = y^3 + 3x^2y -6x^2 - 6y^2 + 2 = 64+0+0-96+2 = -30 is a minimum
f(-2, 2) = y^3 + 3x^2y -6x^2 - 6y^2 + 2 = 8+24-24-24+2 = -14 is a saddlepoint
f(2, 2) = y^3 + 3x^2y -6x^2 - 6y^2 + 2 = 8+24-24-24+2 = -14 is a saddlepoint
4. x^2+y^2 = sqrt(x^2+y^2) r^2 = r r^2 - r = 0 r(r-1) = 0. r = 0, 1
The radius goes from 0 to 1.
We use cylindrical coordinates. r dz dr dtheta= dxdydz
Then, we integrate, using I[a, b] fpr the integral from a to b and E[a, b] for the evaluationfrom a to b,
I[0, 2 pi] I[0, 1] I[r^2, r] r dz dr dtheta =
I[0, 2pi] I[0, 1] rz E[r^2, r] dr dtheta =
I[0, 2pi] I[0,1] r(r-r^2) dr dtheta =
I[0, 2pi] I[0, 1] r^2 - r^3 dr dtheta =
I[0, 2pi] r^3/3 - r^4/4 E[0, 1] dtheta =
I[0, 2pi] 1/12 dtheta =
1/12 theta E[0, 2pi] = 1/12 * (2 pi-0) = 1/6 pi
5. You can draw the region of integration just by plotting the 3 points and making the triangle.
The line connecting (0, 2) and (1, 1) is y = 2-x
The line connecting (0, 2) and (3, 3) is y = 1/3x + 2
The line connecting (1, 1) and (3, 3) is y = x
We have I[0, 1]I[2-x,1/3x+2]x+2 dy dx +
I[1, 3]I[x,1/3x+2]x+2 dy dx =
I[0, 1](x+2) 4/3x dx +
I[1, 3](x+2)(2-2/3x) dy dx =
I[0, 1] 4/3x^2 + 8/3x dx + I[1, 3] -2/3x^2 +2/3x + 4 dx =
4/9x^3 + 4/3x^2 E[0, 1] + -2/9x^3 + 1/3x^2 + 4 x E[1, 3] =
4/9*1+4/3*1 - 0 + (-2/9*3^3 + 1/3*3^2+4*3)-(-2/9*1 + 1/3*1 +4*1) =
20/3 = 6.6666666667