Unformatted text preview: METU
Department of Mathematics
Group Code
Acad. Year
Semester
Instructor : Math 120
: 20042005
: Spring
: Date
Time
Duration Calculus II
MidTerm 2
Last Name :
:
Name
Department :
:
Signature : May 7th,2005
: 09:30
: 120 minutes 1 2 3 4 List No. Student No. :
:
Section 6 QUESTIONS ON 5 PAGES
TOTAL 60 POINTS 5 6 SOLUTION KEY
Question 1 (10 points) Show that for any tangent plane to the surface √ √ √ x+ √ y+ z = a , (a > 0) the sum of x, y, and z axis intercepts is constant.
( x0 is called x intercept if the graph intersects x axis at the point (x0 , 0, 0). )
√ √ √ √ Solution: Let f (x, y, z) = x+ y+ z− a then the surface , S, is given by f (x, y, z) = 0.
Let the point P0 (x0 , y0 , z0 ) ∈ S be the point of tangency. Then
1
1
1
f (x, y, z) = √ i + √ j + √ k
2 y
2 x
2 z
and hence
n= 1
1
1
f (x0 , y0 , z0 ) = √ i + √ j + √ k
2 x0
2 y0
2 z0 is a normal vector to the tangent plane. Therefore, the equation of the tangent plane is
obtained as
1
1
1
√ (x − x0 ) + √ (y − y0 ) + √ (z − z0 ) = 0.
2 x0
2 y0
2 z0
Now,
√ √
√
√
x−intercept is obtained, for y = z = 0, as: ( x0 + y0 + z0 ) x0
√
√ √
√
y−intercept is obtained, for x = z = 0, as: ( x0 + y0 + z0 ) y0
√
√ √
√
z−intercept is obtained, for x = y = 0, as: ( x0 + y0 + z0 ) z0 . Therefore the sum of
intercepts is:
√
√
√
√
( x0 + y0 + z0 )2 = ( a)2 = a
which is a constant. Question 2 ( 12 points)
Consider the function f (x, y) = x3
if (x, y) = (0, 0), f (0, 0) = 0. Show that;
x2 + y 2 a)f is continuous everywhere Solution: We need to check only continuity at the point (0, 0).
0≤ x2 x2 .x
x3
= 2
≤ x
2
+y
x + y2 so by the squeeze theorem, we have
lim f (x, y) = (x,y)→(0,0) lim (x,y)→(0,0) x2 x3
= 0 = f (0, 0)
+ y2 meaning that f is continuous at (0, 0). b)all directional derivatives of f exist at (0, 0), using limit deﬁnition of directional derivative Solution: Let u = ai + bj with a2 + b2 = 1. Then
a3 h3 /[h2 (a2 + b2 )] − 0
f (ah, bh) − f (0, 0)
= lim
= a3
h→0
h→0
h
h Du f (0, 0) = lim which completes the proof. c)f is not diﬀerentiable (0, 0)
f (h, k) − f (0, 0) − hfx (0, 0) − kfy (0, 0)
√
(h,k)→(0,0)
h2 + k 2 = f (h, k) − h
√
(h,k)→(0,0)
h2 + k 2 = lim h3
h2 +k2 − h
√
lim
(h,k)→(0,0)
h2 + k 2 lim along the line k = h we have
−h/2
lim √
2h h→0 which does not exist. Hence, f is not diﬀerentiable at (0, 0). d)ﬁnd the set of all vectors v for which the equation Dv f (0, 0) = f (0, 0) · v is satisﬁed. 3
Solution: Let v = v1 i + v2 j. Then, by (b) we have Dv f (0, 0) = v1 and f (0, 0) = i.
which gives us v1 =
Therefore, f (0, 0) · v = v1 . Now, the required condition is v1 =
0, v1 = ±1. For, v1 = 0 we have v2 = ±1 and for v1 = ±1 we have v2 = 0. Thus, the
directions are ±i and ±j.
3
v1 Question 3 (8 points) Let z = f (x, y) = 20 − x2 − 7y 2 . Use Tangent Plane Ap proximation to approximate f (1.95, 1.08). Solution: We have
−x fx (x, y) = 20 − x2 − 7y 2 , fy (x, y) = −7y
20 − x2 − 7y 2 . So fx (2, 1) = −2/3 and fy (2, 1) = −7/3. Therefore
f (1.95, 1.08) ≈ f (2, 1) + fx (2, 1)(1.95 − 2) + fy (2, 1)(1.08 − 1)
=
= 3 + (−2/3)(−0.05) + (−7/3)(0.08)
0.1 − 0.56
3+
.
3 x y
y x
derivatives in the neighborhood of the point (x, y) = (1, −1) with
f1 (−1, −1) = 2 = f2 (1, −1),
f1 (1, −1) = −2 = f2 (−1, −1),
f11 (−1, −1) = 6 = f12 (1, −1),
f11 (1, −1) = 8 = f21 (1, −1),
f12 (−1, −1) = −5 = f21 (−1, −1),
f22 (1, −1) = 1 = f22 (−1, −1).
∂2z
Find z12 =
at the point (x, y) = (1, −1).
∂y∂x Question 4 (10 points) Given that the function z = f ( , ) has continuous partial Solution: Let s = x/y, t = y/x then (x, y) = (1, −1) gives (s, t) = (−1, −1) and z =
f (x/y, y/x) = f (s, t). Now,
z1 = ∂z
∂z ∂s
∂z ∂t
1
y
=
+
= f1 (s, t) − 2 f2 (s, t).
∂x
∂s ∂x
∂t ∂x
y
x Similarly,
z12 ∂ 1
y
f1 (s, t) − 2 f2 (s, t)
∂y y
x
−1
1
∂s
∂t
=
f (s, t) +
f11 (s, t)
+ f12 (s, t)
2 1
y
y
∂y
∂y
1
∂t
y
∂s
− 2 f2 (s, t) − 2 f21 (s, t)
+ f22 (s, t)
x
x
∂y
∂y
−1
1
−x
1
=
f1 (s, t) +
f11 (s, t) 2 + f12 (s, t)
y2
y
y
x
1
y
−x
1
− 2 f2 (s, t) − 2 f21 (s, t) 2 + f22 (s, t)
x
x
y
x
= Put x = 1, y = −1 to get
z12 (1, −1) = −f1 (−1, −1) − [−f11 (−1, −1) + f12 (−1, −1)]
−f2 (−1, −1) + [−f21 (−1, −1) + f22 (−1, −1)]
= 17. Question 5 (10 points) Use the Method of Lagrange Multipliers to ﬁnd the least and
greatest distance between the origin and the ellipse 17x2 + 12xy + 8y 2 − 100 = 0. Solution: Let the point be (x, y) then the distance is d = x2 + y 2 . Now, we deﬁne L(x, y, λ) = x2 + y 2 + λ(17x2 + 12xy + 8y 2 − 100)
∂L
∂x
∂L
∂y
∂L
∂λ = 2x + λ(34x + 12y) = 0 (1) = 2y + λ(12x + 16y) = 0 (2) = 17x2 + 12xy + 8y 2 − 100 = 0 (3) −y
−x
and λ =
, respectively.
17x + 6y
6x + 8y
Equating them and performing cross multiplication we get From the equations (1) and (2) we write λ = 2x2 − 3xy − 2y 2 = 0. (4) Multipying, (4) by 4, and adding with (3) we obtain, 25x2 − 100 = 0 or x = ±2.
For x = 2, (4) becomes 8 − 6y − 2y 2 = 0 which has the roots y = 1 and y = −4.
For x = −2, (4) becomes 8 + 6y − 2y 2 = 0 which has the roots y = −1 and y = 4. Therefore
we have P1 (2, 1), P2 (2, −4), P3 (−2, −1), P4 (−2, 4). Therefore, the least distance is √ √
5 and the greatest distance is 2 5. Surname, Name: ............................, ............................ Student Id:.......................
Question 6 (10 points) Find and classify all the critical points of the function
1
f (x, y) = x3 + xy 2 − x.
3 Solution: We ﬁrst ﬁnd the critical points:
f1 (x, y) = x2 + y 2 − 1 = 0 (5) f2 (x, y) = 2xy = 0 (6) The equation in (6) gives us x = 0 or y = 0. If x = 0, (5) implies that y = ±1 and if y = 0,
(5) implies that x = ±1. Therefore, we have P1 (0, 1), P2 (0, −1), P3 (1, 0), P4 (−1, 0) as critical
points.
f11 (x, y) = 2x, f12 (x, y) = f21 (x, y) = 2y, f22 (x, y) = 2x
∆= f11
f21 f12
f22 = 2x 2y
2y 2x = 4(x2 − y 2 ) and
∆(P1 ) = −4 < 0 so P1 is a SADDLE point,
∆(P2 ) = −4 < 0 so P2 is a SADDLE point,
∆(P3 ) = 4 > 0 and f11 (P3 ) = 2 > 0 so P3 is a LOCAL MIN point,
∆(P4 ) = 4 > 0 and f11 (P4 ) = −2 < 0 so P4 is a LOCAL MAX point. ...
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This note was uploaded on 05/09/2010 for the course MATHEMATIC 120 taught by Professor A during the Spring '10 term at Middle East Technical University.
 Spring '10
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 Calculus

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