1062微微微甲甲甲01-04、、、06-10班班班期期期中中中考考考解解解答答答和和和評評評分分分標標標準準準
1. (15 points) Determine whether the series is absolutely convergent, conditionally convergent, or divergent. Please state the tests which you use.
(a) (5 points)
∞
∑
n=2
(−1)nln(n!) n3ln n (b) (5 points)
∞
∑
n=1(−1)n( 1
√3
n −sin ( 1
√3
n)) (c) (5 points)
∞
∑
n=1
(−1)n3−n2 (n + 1 n )
n2
Solution:
(a) Question
∞
∑
n=2
(−1)nln(n!) n3ln n Solution Consider that
ln(n!)
n3ln n ≤ n ln n n3ln n = 1
n2. And series
∞
∑
n=1
1
n2 is convergent by p-series for p = 2 > 1.
Therefore we can apply Limit Comparison Test to determine
∞
∑
n=2
(−1)nln(n!)
n3ln n is abso- lutely convergent.
2pts Those who thought
∞
∑
n=1
ln(n!)
n3ln n is divergent for any reason and then prove that
∞
∑
n=1(−1)nln(n!) n3ln n is convergent by Alternating Series Test with a correct process get 2 points.
Correct (a)
∞
∑
n=1
1
n2 is going to be convergent by Integral Test.
(b) ln(n!) = ln 1 + ln 2 + ⋯ + ln n < ln n + ln n + ⋯ + ln n = n ln n (c) ln(n!) ≤ ln(nn) =n ln n
(d) n ln n − n + 1 =∫
n
1
ln x dx < ln(n!) <∫
n+1 1
ln x dx = (n + 1) ln(n + 1) − n (e) Stirling Formula: n! ∼
√
2πn (n e)
n
Ô⇒ ln(n!) ∼ n ln n − n + ln
√ 2πn Incorrect (a) Ratio Test: lim
n→∞∣ an+1
an ∣ =1 leads no conclusion.
(b) L’Hˆospital Rule: Differentiate ln(n!) leads mistakes.
(c) Test for Divergence: The limit of an as n → ∞ is zero. So we can not use it to conclude the series is divergent.
(d) Limit Comparison with
∞
∑
n=1
1
n3 leads ∞ and no conclusion.
(e)
∞
∑
n=1
1 n2 =
π2
6 ≠1 =∫
∞ 1
1 x2dx (f) lim
n→∞
ln 1 ln n+
ln 2 ln n+ ⋯ +
ln n
ln n =0 + 0 + ⋯ + 1 is wrong.
(b) Let an= 1
√3
n −sin 1
√3
n, for any positive integer n. Then, an≥0.
(For x ≥ 0, sin(x) = sin(x) − sin(0) = x ⋅ cos(ξ) ≤ x, for some ξ ∈ (0, x), by Mean Value
Page 1 of 19
Theorem. Hence, 1
√3
n −sin 1
√3
n ≥0.)
This problem can be decomposed into two parts 1. Convergence of
∞
∑
n=1
(−1)nan (2 points) 2. Divergence of
∞
∑
n=1an (3 points)
Convergence of
∞
∑
n=1
(−1) n a n
There are two kinds of grading, depending on what kind of method one used.
1. Directly applying Alternating Series Test:
(1 point) an is decreasing : Let f (x) = 1
√3
x −sin 1
√3
x. f′(x) = −1
3 x−4/3(1 − cos(x)) ≤ 0, for x > 0. Hence, f (x) is decreasing as x increases (when x > 0). Since an = f (n) for all n, an is decreasing as n increases.
(1 point) lim
n→∞an=0 :
n→∞lim an=n→∞lim( 1
√3
n−sin( 1
√3
n)) =n→∞lim 1
√3
n −n→∞lim sin( 1
√3
n) =0 − 0 = 0, since sin(x) is continuous with respect to x.
Hence, by Alternating Series Test,
∞
∑
n=1(−1)nan=
∞
∑
n=1(−1)n( 1
√3
n−sin 1
√3
n)is convergent.
2. Considering convergence of
∞
∑
n=1
(−1)n
√3
n and
∞
∑
n=1
(−1)nsin( 1
√3
n):
(1 point) Convergence of ∑∞
n=1
(−1)n
√3
n : 1
√3
n ≥0, lim
n→∞
1
√3
n =0 and 1
√3
n is decreasing as n increases. Therefore, by Alter- nating Series Test,
∞
∑
n=1
(−1)n
√3
n is convergent.
(1 point) Convergence of ∑∞
n=1
(−1)nsin( 1
√3
n):
sin( 1
√3
n) ≥0, lim
n→∞sin( 1
√3
n) =0 and sin( 1
√3
n)is decreasing as n increases. There- fore, by Alternating Series Test,
∞
∑
n=1(−1)nsin( 1
√3
n) is convergent.
Divergence of
∞
∑
n=1
a n
We will use Comparison Test to demonstrate that
∞
∑
n=1an is divergent. Note that, if one only use upper bound or negative lower bound of anto get the divergence of
∞
∑
n=1
an, he/she will get 0 point in this part.
(2 points) Compare ∑∞
n=1an with some appropriate series:
x→∞lim
x − sin(x)
x3 = x→∞lim 1 − cos(x)
3x2 = x→∞lim sin(x) 6x = 1
6 ⇒ n→∞lim an
1/n = n→∞lim
1
√3
n−sin3√1 n
(1/√3
n)3 = 1
6 >0.
Therefore,
∞
∑
n=1an and
∞
∑
n=1
1
n both converges or both diverges.
(1 point) Divergence of the ”appropriate” series:
By p-series test or integral test or comparison test, we know that
∞
∑
n=1
1
n is divergent, and hence, so is
∞
∑
n=1an. That is,
∞
∑
n=1
( 1
√3
n −sin 1
√3
n) is divergent.
(c) Let an= (−1)n3−nn ( n + 1
n )n
2, for all positive integer n.
(2 points) lim
n→∞
n
√
∣an∣ = lim
n→∞( n + 1
n )n=e
(2 points) e > 1 (strictly larger than 1)
(1 point) By Root Test,
∞
∑
n=1an=
∞
∑
n=1
(−1)n3−nn (n + 1 n )n
2 is divergent.
Page 3 of 19
(01-02班) Determine whether the series is absolutely convergent, conditionally convergent, or di- vergent. Please state the tests which you use.
(a) (5 points)
∞
∑
n=2(−1)nln(n!) n3ln n (b) (5 points)
∞
∑
n=2(−1)n
√n
2 − 1 ln n (c) (5 points)
∞
∑
n=1
n5 4n−n3 (−5)n+3n Solution:
(a)
(b) Observe that 21/n−1 is decreasing to zero, and ln(n) is increasing to infinity. So 21/n−1
ln(n)
is descreasing to zero. Hence by alternating series test,
∞
∑
n=2
(−1)n21/n−1 ln(n) is convergent. (2 %)
On the other hand, observe that
21/n−1 ln(n) =
eln(2)/n−1 ln(n) ≥
(1 +ln(2) n ) −1
ln(n) =
ln(2) n ln(n). Since
∞
∑
n=2
1
n ln(n) is divergent by integral test, the series
∞
∑
n=2
21/n−1 ln(n) is also divergent by comparison test. (3 %) Thereofore,
∞
∑
n=2
(−1)n21/n−1
ln(n) is conditionally convergent.
(c) Let an=n5 4n−n3
(−5)n+3n. Observe that
n→∞lim∣ an+1
an ∣ = lim
n→∞(
(n + 1)5 n5 ×
5n+ (−3)n 5n+1+ (−3)n+1 ×
4n+1− (n + 1)3 4n−n3 )
=n→∞lim (
(n + 1)5 n5 ×
1 + (−3/5)n 5 + (−3) × (−3/5)n ×
4 − (n + 1)3/4n 1 − n3/4n )
=1 ×1 5×4
= 4
5. (5 %) Hence by ratio test,
∞
∑
n=1an is absolutely convergent.
2. (10 points) Find the radius of convergence and the interval of convergence of the power series
∞
∑
n=2
(2x − 1)n n(ln n)34 .
Solution:
Write an=
(2x − 1)n
n(ln n)3/4, by ratio text, we have
n→∞lim ∣ an+1
an
∣ = lim
n→∞
RR RR RR RR RR R
(2x − 1)n + 1 n (
ln(n + 1) ln n )
3/4RRRRRRRRRRR
= ∣2x − 1∣
Hence, we have ∣2x − 1∣ < 1, or ∣x − 1 2∣ <
1 2(4%).
Now we check the convergence of endpoints:
When x = 0:
Now an =
(−1)n
n(ln n)3/4. Note that lim
n→∞
1
n(ln n)3/4 = 0(1%) and 1
n(ln n)3/4 is obviously decreasing.(1%) By Leibnitz test, it is convergent.(1%)
When x = 1:
Now an = 1
n(ln n)3/4. Write f (x) = 1
x(ln x)3/4, then f is obviously positive (for x > 1), decreasing(1%), and continuous. By integral test, we have
∫
∞ 2
dx
x(ln x)3/4 =4(ln x)1/4∣
∞
2 = ∞(1%) Therefore, it is divergent.(1%)
Hence, the radius of convergence is 1
2 and the convergence interval is [0, 1)
Page 5 of 19
(01-02班)
(a) (5 points) Find the constant p such that lim
n→∞
√1 1+√1
2 + ⋯ +√1 n
np is a finite nonzero constant.
(b) (5 points) Find the interval of convergence of the power series
∞
∑
n=1
(1 − 3x)n
√1 1 +√1
2+ ⋯ +√1
n
.
Solution:
(a) Let L = lim
n→∞
√1 1 +√1
2+ ⋯ + √1 n
np . Observe that
∫
n+1 1
1
√x dx ≤ 1
√ 1+ 1
√
2+ ⋯ + 1
√n ≤1 +∫
n
1
1
√x dx. (3 %) So
2√
n + 1 − 2
np ≤
√1 1+√1
2 + ⋯ +√1 n
np ≤
2√ n − 1 np . Hence by squeeze theorem,
⎧⎪
⎪⎪
⎪
⎨
⎪⎪
⎪⎪
⎩
If p > 1/2, then L = 0, If p = 1/2, then L = 2, If p < 1/2, then L = ∞.
Thus the constant p = 1/2. (2 %) (b) Let fn(x) = 1
√1 1 +√1
2+ ⋯ +√1 n
(1 − 3x)n. Then
n→∞lim ∣
fn+1(x) fn(x) ∣ =
⎛
⎝
n→∞lim
√1 1+√1
2 + ⋯ +√1
n
√1 1+√1
2 + ⋯ +√n+11
⎞
⎠
∣1 − 3x∣
=
⎛
⎝
n→∞lim
√1 1+√1
2 + ⋯ +√1
√ n
n ×
√ n + 1
√1 1 +√1
2+ ⋯ +√1 n+1
×
√n
√ n + 1
⎞
⎠
∣1 − 3x∣
=2 ×1
2×1 × ∣1 − 3x∣
= ∣1 − 3x∣.
So by ratio test, if ∣1−3x∣ < 1, or equiveltnly, 0 < x < 2 3, then
∞
∑
n=1fn(x) is convergent (3 %), and if x < 0, or 2
3 <x, then
∞
∑
n=1fn(x) is divergent. Now we are going to check x = 0 and x = 2
3.
Observe that
∞
∑
n=1fn(0) =
∞
∑
n=1
1
√1 1 +√1
2+ ⋯ +√1
n
. By (a), since
n→∞lim RR RR RR RR RR R
1
√n/
1
√1 1 +√1
2+ ⋯ + √1
n
RR RR RR RR RR R
=2,
∞
∑
n=1fn(0) is divergent by limit comparison test. (1 %)
Observe that
∞
∑
n=1fn(2 3) =
∞
∑
n=1
(−1)n 1
√1 1 +√1
2+ ⋯ +√1 n
. So it is clear that
∞
∑
n=1fn(2
3)is conver- gent by alternating series test. (1 %)
Therefore, the interval of convergence of
∞
∑
n=1fn(x) is (0,2 3].
Page 7 of 19
3. (10 points) Let F (x) =∫
x
0
ln (1 + t2 2)dt.
(a) (6 points) Find the Maclaurin series of F (x) and its radius of convergence.
(b) (4 points) Estimate F (10−1)up to an error within 10−7.
Solution:
(a)
Let F (x) = ∫
x
0
ln(1 + t2
2)dt and by Fundamental Theorem of Calculus we have F′(x) = ln(1 +x2
2 ).
Since ln(1+x) =
∞
∑
n=1
(−1)n−1xn
n , substitute x2
2 with x and we have ln(1+x2 2) =
∞
∑
n=1
(−1)n−1(x
2
2 )n
n =
∞
∑
n=1
(−1)n−1x2n 2nn
F (x) is integrate F′(x) term by term F (x) =
∞
∑
n=1
(−1)n−1x2n+1 2nn(2n + 1) Next, use ratio test to find radius of convergence.
n→∞lim ∣ an+1
an ∣ = lim
n→∞∣
(−1)nx2n+3 2n+1(n + 1)(2n + 3)⋅
2nn(2n + 1) (−1)n−1x2n+1∣ = ∣
x2 2∣ In order to make this alternative series converge, ∣x2
2 ∣need to be less than 1. We have ∣x∣ <
√ 2, hence the radius of convergence is
√ 2
(b)
Suppose bn =
(101 )2n+1
2nn(2n + 1) and Mn =
n
∑
n=1
(−1)n−1(101)2n+1
2nn(2n + 1) , we know that the error of Mn(x) is bounded by bn+1. Note that
b1 = (1
10)−3 2 ⋅ 1 ⋅ 3 =
1
6000 =0.000167 > 10−7 b2 =
(101)−5 4 ⋅ 2 ⋅ 5 =
1
4000000 =2.5 × 10−5>10−7 b3=
(101 )−7
8 ⋅ 3 ⋅ 7 = 1
168 × 107 =5.952 × 10−10 <10−7
Hence the summation of first two term of F (10−1)is sufficient to make the error less than 10−7. F (10−1) ≈
1 6000−
1
4000000 ≈0.0001664
GRADING CRITERIA
(a) Finding the Maclaurin series and radius of convergence are
3
points respectively. Write down the basic formula of Maclaurin series will get1
point, answer correct will get2
points.For radius of convergence, use ratio test to find the answer will get
1
point, answer correct will get2
points.You will loss
1
point for each calculation error.(b) If you try to estimate the error F (10−1), you will get
2
points even if the final answer is wrong. If the answer is correct, you will get another2
points.You will loss
1
point for each calculation error.Page 9 of 19
4. (8 points)
(a) (4 points) Identify the power series
∞
∑
n=0
(−1)n22n+1
2n + 1 x2n+1 as an elementary function.
(b) (4 points) Find the sum 1
√ 3 −
1 9
√ 3 +
1 45
√ 3 −
1 189
√ 3 + ⋯
Solution:
(a)
Method 1.
∵ tan−1x =
∞
∑
n=0
(−1)n x2n+1 2n + 1 (2%)
∴
∞
∑
n=0
(−1)n(2x)2n+1
2n + 1 =tan−12x (2%)
Method 2.
for ∣x∣ < 1 2
∵ (
∞
∑
n=0
(−1)n(2x)2n+1 2n + 1 )
′
=2 ⋅
∞
∑
n=0
(−1)n(2x)2n = 2
1 + 4x2 (2%)
∴
∞
∑
n=0
(−1)n(2x)2n+1 2n + 1 = ∫
2
1 + 4x2dx = tan−12x + C (1%) The series equals 0 when x = 0, so C = 0. (1%)
Therefore,
∞
∑
n=0(−1)n(2x)2n+1
2n + 1 =tan−12x (b)
1
√ 3−
1 9√
3+ 1 45√
3− 1 189√
3+...
= 1
√ 3−
1 3(√
3)3 + 1 5(√
3)5 − 1 7(√
3)7 +...
=
∞
∑
n=0
(−1)n 2n + 1( 1
√
3)2n+1 (2%)
=tan−1 1
√
3 (1%)
=π 6 (1%)
5. (12 points) Let r(t) = (sin t − t cos t)i + (cos t + t sin t)j + t2k, 0 ≤ t ≤ π, be a vector function that parametrizes a curve in space.
(a) (3 points) Find the arc length of the curve.
(b) (6 points) At what point on the curve is the osculating plane parallel to the plane x+
√
3y−z = 0
?
(c) (3 points) Find the curvature of the curve.
Solution:
(a) r′(t) = (cos t − cos t + t sin t)i + (− sin t + sin t + t cos t)j + 2tk = t sin ti + t cos tj + 2tk
⇒ ∣r′(t)∣ =
√
t2sin2t + t2cos2t + (2t)2 =
√
5t (2 points)
∴ Arc length L =∫
π
0
∣r′(t)∣ dt =∫
π
0
√ 5tdt =
√ 5
2 π2 (1 point)
(b) Osculating plane is spanned by the tangent and normal vector of the curve r(t), so we need to find T(t) and N(t).
∵r′(t) = (t sin t, t cos t, 2t) and ∣r′(t)∣ =
√
5t ⇒ T(t) = r′(t)
∣r′(t)∣ = 1
√
5(sin t, cos t, 2)
∴T′(t) = 1
√
5(cos t, − sin t, 0) ⇒ N(t) = (cos t, − sin t, 0) Normal vector of osculating plane ⇀n = (1,
√
3, −1) parallel to T × N = 1
√
5(2 sin t, 2 cos t, −1)
⇒ t = π
6, that is, the osculating plane at (1 2−
√ 3π 12 ,
√ 3 2 +
π 12,π2
36) is parallel to the plane x +√
3y + z = 0. (2 points for each T, T′, 1 point for each N, Point) (c) (Method I)By (b) we have ∣T′(t)∣ = 1
√
5, (1 point) Curvature κ = ∣T′(t)∣
∣r′(t)∣ =
√1
√5
5t = 1
5t (2 points)
(Method II) By (a) we have r′(t) = t sin ti + t cos tj + 2tk
⇒r′′(t) = (sin t + t cos t)i + (cos t − t sin t)j + 2k (1 point)
∴r′(t) × r′′(t) = 2t2sin ti + 2t2cos tj − t2k ⇒ ∣r′(t) × r′′(t)∣ =
√
5t2 (1 point)
⇒κ = ∣r′(t) × r′′(t)∣
∣r′(t)∣3
=
√ 5t2 (
√ 5t)3
= 1
5t (1 point)
Page 11 of 19
(sol II)
(i) r′(t) = t sin ti + t cos tj + 2tk ⇒ ∣r′(t)∣ =
√ 5t (ii) r′′(t) = (sin t + t cos t)i + (cos t − t sin t)j + 2k
(iii) r′(t) × r′′(t) = 2t2sin ti + 2t2cos tj − t2k ⇒ ∣r′(t) × r′′(t)∣ =
√ 5t2 (iv) T(t) = r′(t)
∣r′(t)∣ = 1
√
5(sin t, cos t, 2) (v) T′(t) = 1
√
5(cos t, − sin t, 0) ⇒ ∣T′(t)∣ = 1
√ 5 (vi) N(t) = T′(t)
∣T′(t)∣ = (cos t, − sin t, 0) (vii) B(t) = T(t) × N(t) = 1
√
5(2 sin t, 2 cos t, −1) (a) Arc length L =∫
π
0
∣r′(t)∣ dt =∫
π
0
√ 5tdt =
√ 5 2 π2 (b) Normal vector of osculating plane ⇀n = (1,
√
3, −1) parallel to B(t) = 1
√
5(2 sin t, 2 cos t, −1)
⇒ t = π
6, that is, the osculating plane at (1 2−
√ 3π 12 ,
√ 3 2 +
π 12,π2
36) is parallel to the plane x +
√
3y + z = 0.
(c) κ = ∣T′(t)∣
∣r′(t)∣ =
√1
√5
5t = 1
5t or κ = ∣r′(t) × r′′(t)∣
∣r′(t)∣3 =
√ 5t2 (
√ 5t)3
= 1 5t
6. (10 points) Let surface S be given by S = {(x, y, z) ∈ R3∣sin(xyz) = x + 2y + 3z}.
(a) (4 points) On the surface, compute ∂z
∂x and ∂y
∂x.
(b) (2 points) Find an equation of the tangent plane to the surface S at (2, −1, 0).
(c) (4 points) Suppose, when restricted to the surface S, a differentiable function f attains a local maximum value at the point (2, −1, 0) with f (2, −1, 0) = 10 and fx(2, −1, 0) = 2. Let (x0, y0, z0)be a point which is close to the point (2, −1, 0) and lies on another surface sin(xyz) = z + 2y + 3z + 10−2. Use the linear approximation to estimate f (x0, y0, z0).
Solution:
Define g(x, y, z) = sin(xyz) − x − 2y − 3z.
(a) Treating z implicitly as a function of x and y, by chain rule we can differentiate the equation g(x, y, z) = 0 as follows:
∂g
∂x =
∂g
∂x
∂x
∂x+
∂g
∂y
∂y
∂x+
∂g
∂z
∂z
∂x =
∂g
∂x +
∂g
∂z
∂z
∂x =0.
We obtain
∂z
∂x = − gx gz = −
yz cos(xyz) − 1 xy cos(xyz) − 3. Similarly,
∂y
∂x = − gx gy = −
yz cos(xyz) − 1 xz cos(xyz) − 2. (b) The tangent plane to the surface S at (2, −1, 0) is
∇g(2, −1, 0) ⋅ < x − 2, y − (−1), z − 0 > = −(x − 2) − 2(y + 1) − 5z = 0, or x + 2y + 5z = 0.
(c) Since f (2, −1, 0) is a local maximum value, by the method of Lagrange multiplier, there is a number λ such that ∇f (2, −1, 0) = λ∇g(2, −1, 0).
From the x-exponent of the equation and the fact that fx(x, y, z) = 2 we find that λ = −2 and thus ∇f (2, −1, 0) = −2∇g(2, −1, 0). It follows from the linear approximation of g at the point (2, −1, 0) that
10−2=g(x0, y0, z0) −g(2, −1, 0) ≈ ∇g(2, −1, 0) ⋅ < x0−2, y0+1, z0>
Therefore, the linear approximation of f at (2, −1, 0) yields
f (x0, y0, z0) ≈f (2, −1, 0) + ∇f (2, −1, 0) ⋅ < x0−2, y0+1, z0 >
=10 − 2 ∇g(2, −1, 0) ⋅ < x0−2, y0+1, z0> ≈10 − 2 ⋅ 10−2=9.98.
Marking Scheme
(a) 1 point for each derivation using chain rule or direct use of formula;
1 point for each correct answer.
(b) 1 point for the formula of the tangent plane and 1 point for the correct equation.
(c) 1 point for using Lagrange’s method; 0.5 point for the correct λ.
1 point for each approximation of f and g; 0.5 point for the correct estimate.
Page 13 of 19
7. (13 points) Let f (x, y) =
⎧⎪
⎪⎪
⎨
⎪⎪
⎪⎩
x3+y3
x2+y2 if (x, y) ≠ (0, 0).
0 if (x, y) = (0, 0).
(a) (3 points) Is f (x, y) continuous at (0, 0)? Justify your answer.
(b) (2 points) Find the gradient vector ∇f (0, 0).
(c) (4 points) Is fx(x, y) continuous at (0, 0)? Justify your answer.
(d) (4 points) Find the maximum and minimum directional derivatives of f at the point (0, 0) among the directions of all the unit vectors u.
Solution:
(a) x = r cos θ, y = r sin θ
∣f (x, y) ∣=∣ r3(cos3θ + sin3θ)
r2 ∣
=r ∣ cos3θ + sin3θ ∣
≤r ∣ cos3θ ∣ +r ∣ sin3θ ∣≤ 2r
So, f (x, y) → 0 = f (0, 0) as r → 0 is as (x, y) → (0, 0) Therefore, f is continuous at (0,0).
Grading Policy:
(1) 3 points for correct proof.
(2) No partical points.
(b)fx(0, 0) = lim
h→0
f (h, 0) h =lim
h→0
1 h
h3 h2 =1 fy(0, 0) = lim
h→0
f (0, h) h =lim
h→0
1 h
h3 h2 =1
∇f (0, 0) = ⃗i + ⃗j Grading Policy:
(1)Correct limits for 1 point.
(2)Correct answer for 1 point.
(c)Away from (0, 0),
fx= 3x2(x2+y2) − (x3+y3)(2x) (x2+y2)2
=
x4+3x2y2−2xy3 (x2+y2)2
Let x = r cos θ, y = r sin θ
fx=cos4θ + 3 cos2θ sin2θ − 2 cos θ sin3θ fx is Dθ=0.
θ = 0 → f (x, y) = 1, and θ = π
2 →f (x, y) = 1 fx(x, y) is NOT continuous at (0,0).
Other method:
fx=
x4+3x2y2−2xy3 (x2+y2)2 Let y = mx → fx =
2m2 1 + m2
So, the limit as (x, y) → (0, 0) along the different lines y = mx is different for different m.
fx(x, y) is NOT continuous.
Grading Policy:
(1)Correct fx for 2 points.
(2)Correct limit at (0, 0) for 2 points.
(d) ⃗u = (cos θ, sin θ) Duf (0, 0) = lim
r→0
f (r cos θ, r sin θ) r
=lim
r→0
r3(cos3θ+sin3θ) r2
r
=cos3θ + sin3θ
Let g(θ) = cos3θ + sin3θ
→g′(θ) = −3 sin θ cos2θ + 3 cos θ sin2θ If g′(θ) = 0, then θ = 0 or π
4 or π 2 or 5π
4 and so on.
Maximum is g(0) = g(π 2) =1.
Minimum is g(π) = g(3π
2 ) = −1.
Grading Policy:
(1)Find the directional derivative in direction of ⃗u = (cos θ, sin θ) for 2 points.
(2)Correct maximum and minimum argunents for 1 point.
(3)Correct answer for 1 point.
Page 15 of 19
8. (12 points) Let f (x, y) = 1 + 3x2−2x3+3y − y3.
(a) (6 points) Find the local maximum and minimum values and saddle point(s) of f (x, y).
(b) (6 points) Find the extreme values of f (x, y) on the region D bounded by the triangle with vertices (−2, 2), (2, 2) and (2, −2).
Solution:
(a)
fx=6x − 6x2=6x (1 − x) fy =3 − 3y2=3 (1 − y) (1 + y) solve
⎧⎪
⎪
⎨
⎪⎪
⎩ fx=0
fy=0 imply critical points are :(1, 1) , (1, −1) , (0, 1) , (0, −1) fxy =fyx=0
fxx=6 − 12x
fyy= −6y D = fxxfyy=72xy − 36y2 at point (1, 1)
D (1, 1) = 36 > 0, fxx(1, 1) = −6 < 0,local maximumf (1, 1) = 4 at point (1, −1)
D (1, −1) = −36 < 0,saddle point at point (0, 1)
D (0, 1) = −36 < 0,saddle point at point (0, −1)
D (0, −1) = 36 > 0, fxx(0, −1) = 6 > 0,local minimumf (0, −1) = −1 (b)
for y=2,−2 ≤ x ≤ 2
f (x, 2) = −2x3+3x2−1 denote g1(x) g′1(x) = −6x (x − 1),solve g1′(x) = 0, x = 0, 1
f (−2, 2) = 27, f (0, 2) = −1, f (1, 2) = 0, f (2, 2) = −5 for x=2,−2 ≤ y ≤ 2
f (2, y) = −y3+3y − 3 denote g2(y)
g′2(y) = −3 (y − 1) (y + 1),solve g2′(y) = 0, y = −1, 1 f (2, −2) = −1, f (2, −1) = −5, f (2, 1) = −1, f (2, 2) = −5 for x+y=0,−2 ≤ x ≤ 2
f (x, −x) = −x3+3x2−3x + 1 denote g3(x) g′3(x) = −3 (x − 1)2,solve g3′(x) = 0, x = 1 f (−2, 2) = 27, f (1, −1) = 0, f (2, −2) = −1
Comparing above point and critical points we get ,
⎧⎪
⎪
⎨
⎪⎪
⎩
maximum = 27, at (−2, 2) minimum = −5, at (2, −1) , (2, 2) [Grading]
(a)
(2 points)
fx, fy, fxx, fyyand find 4 critical points (4 points)
correct determine each critical point is location maximum,location minimum,saddle point if you don’t write the local maximum and local minimum values, you will lose 1 point (b)
if you only consider points in the interior of the triangle and extremum on boundary, you will get at most 4 points
if you only consider points in the interior of the triangle and corners of the triangle, you will get at most 3 points
if you consider all points but doesn’t dissusion, you will get at most 5 points
if you only consider points in the interior of the triangle, you will get at most 1 point
Page 17 of 19
9. (10 points) By the Extreme Value Theorem, a continuous function on a sphere attains both absolute maximum and minimum values. Find the extreme values of f (x, y, z) = ln(x+2)+ln(y +2)+ln(z +2) on the sphere x2+y2+z2=3.
Solution:
Step1.
Let g(x, y, z) = x2+y2+z2=3.
According to the method of Lagrange multipliers, we solve the equation ∇f = λ∇g and g(x, y, z) = 3. This gives
1
x + 2 =λ ⋅ 2x 1
y + 2 =λ ⋅ 2y 1
z + 2 =λ ⋅ 2z x2+y2+z2 =3 (3pts)
Step2.
Note that λ ≠ 0 because λ = 0 implies 1
x + 2 =0, which is impossible.
Thus we have
1
λ =2x(x + 2) = 2y(y + 2) = 2z(z + 2) From 2x(x + 2) = 2y(y + 2), we have
0 = x2−y2+2x − 2y = (x − y)(x + y + 2) which gives
y = x or y = −x − 2 Similarly, from 2x(x + 2) = 2z(z + 2), we have
z = x or z = −x − 2
Case1. y = x and z = x
From x2+y2+z2=3, we have 3x2=3 and then x = 1,−1.
Thus we have two points (1, 1, 1), (−1, −1, −1).
Case2. y = x and z = −x − 2
From x2+y2+z2=3, we have 3x2+4x + 4 = 3 and then x = −1 3,−1.
Thus we have two points (−1 3, −1
3, −5
3), (−1, −1, −1).
Case3. y = −x − 2 and z = x
From x2+y2+z2=3, we have 3x2+4x + 4 = 3 and then x = −1 3,−1.
Thus we have two points (−1 3, −5
3, −1
3), (−1, −1, −1).
Case4. y = −x − 2 and z = −x − 2
From x2+y2+z2=3, we have 3x2+8x + 8 = 3 and then x = −5 3,−1.
Thus we have two points (−5 3, −1
3, −1
3), (−1, −1, −1).
Hence f has possible extreme values at the points (1, 1, 1), (−1, −1, −1), (−1 3, −1
3, −5 3), (−1
3, −5 3, −1
3) and (−5
3, −1 3, −1
3). (5pts)
Step3.
We compare the values of f (x, y, z) at these points:
f(1, 1, 1) = ln 27
f(−1, −1, −1) = 0
f(−1 3, −1
3, −5
3) =f (−1 3, −5
3, −1
3) =f (−5 3, −1
3, −1
3) =ln25 27
Therefore the maximum value of f on the sphere x2+y2+z2 =3 is ln 27 and the minimum value is ln25
27. (2pts)
Page 19 of 19
