2-2:50pm
Fall 2012
100 points total
Midterm Exam 1
Math 0240
Solutions
1. (10 points) Find an equations of the plane that contains the line x = 3 + 2t, y = t,
z = 8 − t and is parallel to the plane 2x + 4y + 8z = 17.
Solution: The normal vector of both planes is h2, 4, 8i or n̄ = h1, 2, 4i. Put t = 0 to get
the point (3, 0, 8) on the line and hence on the plane that we have to find. The equation
of the plane is
(x − 3) + 2(y − 0) + 4(z − 8) = 0, x − 3 + 2y + 4z − 32 = 0, x + 2y + 4z − 35 = 0.
2. (15 points) Use differentials to estimate the amount of paint needed to apply a coat of
paint 0.04 cm thick to sides of a rectangular building with the base of 10 × 15 meters and
the hight of 8 meters.
Note: 1 cm = 0.01 m, the roof will not be painted, sides only.
Hint: Use V (x, y, z) = xyz, ∆z = 0, where z is the hight.
Solution: Let x be the width, y be the length, and z be the hight. Then the volume is
V = xyz, x = 10 m, y = 15 m, z = 8 m, ∆x = ∆y = 2 · 0.04 cm = 8 · 10−4 m.
∆V ≈ dV = Vx dx + Vy dy + Vz dz, where dx ≈ ∆x, dy ≈ ∆y, dz ≈ ∆z = 0.
Hence ∆V ≈ Vx (10, 15, 8)∆x + Vy (10, 15, 8)∆y.
Vx = yz, Vx (10, 15, 8) = 120, Vy = xz, Vy (10, 15, 8) = 80.
∆V ≈ 120 · 8 · 10−4 + 80 · 8 · 10−4 = 200 · 8 · 10−4 = 16 · 10−2 = 0.16 m3 . The amount of
paint needed to apply a coat is 0.16 m3 .
3. (15 points) Find the length of the curve
Solution:
r̄(t) = t2 ī + (sin t − t cos t) j̄ + (cos t + t sin t) k̄,
Z π
The length of the curve is L =
|r̄ 0 (t)| dt.
0≤t≤π
0
0
r̄ (t) = h2t, cos t − cos t + t sin t, − sin t + sin t + t cos ti = t h2, sin t, cos ti,
√
|r̄ 0 (t)| = t 5.
√
Z π √
√ t2 π
5 2
t 5 dt = 5 =
L=
π .
2 0
2
0
1
4. (15 points) Find and sketch the domain of the function
f (x, y) =
ln(9 − x2 − 9y 2 )
x−y
Domain D = {(x, y) | 9 − x2 − 9y 2 > 0, x − y 6= 0}. Here x2 + 9y 2 < 9 is the
x2
inner part of the ellipse 2 + y 2 = 1 which passes through the points (3, 0), (0, 1), (−3, 0),
3
(0, −1). Use the dashed line to draw the ellipse. The dashed line y = −x is exluded from
the inner part of the ellipse.
Solution:
5. (15 points) A projectile is fired with muzzle speed 80 m/s at angle of elevation 75◦ from
a position 20 m above ground level. Find its hight (to the ground level) after 6 seconds.
Solution:
ā = h0, −gi ⇒ v̄(t) = hv0x , −gt+v0y i, where v0x = 80 cos 75◦ , v0y = 80 sin 75◦ .
Then v̄(t) = h80 cos 75◦ , −gt + 80 sin 75◦ i ⇒ r̄(t) = h0 + 80 cos 75◦ , 20 − gt2 /2 + 80t sin 75◦ i.
The hight of the projectile after 6 sec is y-coordinate of the vector r̄(t), which is
20 − 18g + 480 sin 75◦ m.
(Note, that this number is greater than zero which means that the projectile has not hit
the ground during the first 6 sec.)
6. (15 points) Find
dw
if w = xy + yz 2 , x = et , y = et sin t, z = et cos t.
dt
dw
= wx xt + wy yt + wz zt , where xt , yt , and zt are regular derivatives w.r.t. t.
dt
wx = y, wy = x + z 2 , wz = 2yz, xt = et , yt = et (cos t + sin t), zt = et (cos t − sin t).
Solution:
dw
= et sin t · et + (et + e2t cos2 t)et (cos t + sin t) + 2e2t sin t cos t · et (cos t − sin t),
dt
dw
= e2t (2 sin t + cos t) + e3t (3 cos2 t sin t + cos3 t − 2 cos t sin2 t).
dt
7. (15 points) Find the local maximum and minimum values and saddle point(s) of the
function f (x, y) = x3 y + 12x2 − 8y.
Solution:
CPs: fx (x, y) = 3x2 y + 24x = 3x(xy + 8) = 0, fy (x, y) = x3 − 8 = 0 ⇒
x = 2. If x = 2 then fx (x, y) = 6(2y + 8) = 0 ⇒ y = −4. Both partial derivatives exist
everywhere. Hence the only CP is (2, −4).
2
Second Derivative Test: fxx = 6xy + 24, fyy = 0, fxy = 3x2 , D(x, y) = −fxy
.
D(2, −4) = −(3 · 4)2 < 0. Hence (2, −4) is a saddle point of f (x, y).
There is no local maximum or minimum values.
2
bonus problem [10 points extra] Find the points on the surface xy 2 z 3 = 2 that are closest to
the origin.
Denote f = d2 = x2 + y 2 + z 2 . It is clear that x 6= 0, y 6= 0, z 6= 0.
2
2
From the surface equation y 2 = 3 . Then f (x, z) = x2 + 3 + z 2 .
xz
xz
3 3
2
2(x z − 1)
CPs: fx = 2x − 2 3 =
= 0 ⇒ x3 z 3 = 1 ⇒ xz = 1.
xz
x2 z 3
6
2(xz 5 − 3)
fz = − 4 + 2z =
= 0 ⇒ xz 5 = 3, xz 5 = xz · z 4 = 1 · z 4 = z 4 = 3 ⇒ z = ±31/4 .
xz
xz 4
x = 1/z = ±3−1/4 .
√
√
√
√
√
1
Also, z 2 = 3, x2 = √ , xz 3 = x2 z 6 = xz 5 · xz = 3 · 1 = 3.
3
Solution:
24
+ 2.
xz 5
36
36
36
24
= 4 8 = 3 3
=
= 12, fzz =
+ 2 = 10.
5
xz
(x z )(xz )
1·3
3
Second Derivative Test: fxx = 2 +
4
2
= 6 > 0, fxz
1
D = 6 · 10 − 12 = 48 > 0.
At CPs fxx = 2 +
4
x3 z 3
, fxz =
6
x2 z 4
, fzz =
√
1
1
5
Hence all CPs are the points of local minimum. At CPs f = √ + √ + 3 = √ . All the
3
3
3
values are equal. Local minimum is absolute minimum.
r
2
y = ± √ = ±2 · 3−1/4 .
3
The four points are (±3−1/4 , −2 · 3−1/4 , ±31/4 ), (±3−1/4 , 2 · 3−1/4 , ±31/4 ).
3