微甲

991微甲^08-12班期中考解答和評分標準 1. (10%) Given g(2) = 4, f (2) = 2 and g⁰(x) =√

x²+ 5, f⁰(x) =√

x³+ 1 for all x > 0, find the derivative of g(f (x)) at x = 2.

Sol:

d

dxg(f (x))¯¯¯

x=2 = g⁰(f (2))f⁰(2) = 9. (10%) 2. (10%) Find f⁰(x), where f (x) =

Z x² 2−x

dt t³+ 1. Sol:

Let u = x², v = 2− x, then f⁰(x) = d

dx Z u

v

dt t³+ 1

= d du

Z _u

0

dt

t³+ 1 ∗ du dx + d

dv Z ₀

v

dt

t³+ 1 ∗ dv dx

= 2x

x⁶+ 1 + 1

(2− x)³+ 1. (10%)

3. (10%) Find dy

dx and d²y

dx² at the point (1, 1) of the curve x³+ x²y + 4y² = 6.

Sol:

Let y=y(x), then take derivative of x³+ x²y + 4y² = 6 on both side over x variable , we get

3x²+ 2xy + x²y⁰+ 8yy⁰ = 0 (a)

.Then substitute x=1,y=1 into this equation we obtain 3 + 2 + y⁰+ 8y⁰ = 0,i.e,y⁰ =−5/9.

Taking derivative of 3x²+ 2xy + x²y⁰+ 8yy⁰ = 0 again we have second derivative of y, it become

6x + 2y + 2xy⁰ + 2xy⁰+ x²y⁰⁰+ 8(y⁰)²+ 8yy⁰⁰ = 0 (b) . When evaluating at (1, 1) point, and using y⁰(1) =−5/9 from above, we have

6 + 2 + 2(−5

9 ) + 2(−5

9 ) + y⁰⁰+ 8(−5

9 )²+ 8y⁰⁰= 0. That is, y⁰⁰(1) =−668 729. Correction standard for this problem:

(1) Correct answer got whole points.Equations for (a) and (b) cost 4 points.

(2) If you just lost one to two terms and you knew the calculation of chain rule along your calculation, it would cost you one to two points.

(3) Methods that are different from above are OK. And your calculation will be graded by the same principles as (1) and (2).

4. (12%) A particle is moving along the curve y = sin⁻¹(x

2). As the particle passes the point (√

2,π

4), its y-coordinate increases at a rate of 2 cm/s. How fast is the distance from the particle to the origin changing at this moment?

Sol:

Let x(t) and y(t) be the x-coordinate and y-coordinate of the particle at time t, respec- tively. And t₀ be the time point such that x(t₀) =√

2 and y(t₀) = π

4. Then the distance from the particle to the origin at time t is z(t) = p

x²(t) + y²(t). By differentiating z with respect to t, we have

z⁰(t₀) = 1

2(x²(t₀) + y²(t₀))⁻¹² (2x(t₀)x⁰(t₀) + 2y(t₀)y⁰(t₀)) where y⁰(t0) = 2 and

x⁰(t₀) = d dtx(t)¯¯¯

t=t0

= d

dt2 sin y(t)¯¯¯

t=t0

= (2 cos y(t))y⁰(t)¯¯¯

t=t0

= (2)(

√2

2 )(2) = 2√ 2.

With all the information above, one can clearly get the solution to this problem z⁰(t₀) = 4 + ^π₂

q 2 + ^π₁₆²

.

Grading Policy:

(1) Two points for successfully specifying the objective function z(t).

(2) An extra of eight points for correctly computing z⁰(t).

(3) You get the final two points for getting the above z⁰(t₀).

5. (10%) Find g⁰(e⁻¹), where g(x) is the inverse function of f (x) = e

√−1

x2−1, 1 < x <∞.

Sol:

step 1 g(x) = p

1 + (log x)² (6%) step 2 g⁰(x) = −h

1 + (log x)² i⁻¹

2 ³ log x

´₋₃³1 x

´

(2%) step 3 g⁰(e⁻¹) = 1

√2e (2%)

6. (12%) Find the maximal volume of a cylindrical can (with top and bottom) with a fixed surface area A.

Sol:

V = πr²h A = 2πrh + 2πr² (2%)

(0≤ r ≤ rA

2π) (2%)

⇒ h = A− 2πr² 2πr

⇒ V = πr²A− 2πr²

2πr = A

2r− πr³ (3%)

⇒ V⁰ = A

2 − 3πr²

critical number: V⁰ = 0⇒ 3πr² = A

2 ⇒ r = rA

6π (∵ r ≥ 0) (3%)

∵ V (0) = 0, V ( rA

6π) = A 3

rA 6π, V (

rA 2π) = 0

⇒ Vmax = A 3

rA

6π (2%)

7. (26%) Given f (x) = (2x²+ 3x)e^−x, Answer the following and show all your work. Each blank is worth 3%.

(a) The function is increasing on the interval(s)

and decreasing on the interval(s) .

(b) The function has local maxima at

and local minima at .

(c) The function is concave upward on the interval(s)

and concave downward on the interval(s) .

(d) The function has inflections at .

(e) The function has asymptotes .

(f) Sketch the graph of f (x). (2%) Sol:

f (x) = (2x²+ 3x)e^−x then

f⁰(x) = (4x + 3)e^−x− (2x²+ 3x)e^−x =−(2x²− x − 3)e^−x=−(x + 1)(2x − 3) e^−x and f⁰⁰(x) =−(4x − 1)e^−x+ (2x²− x − 3)e^−x = (2x²− 5x − 2)e^−x.

(a) f⁰(x) = 0 as x =−1 or 3

2, f⁰(x) > 0 on (−1,3 2) and f⁰(x) < 0 on (−∞, −1), (3

2,∞)

⇒ f is increasing on (−1,3

2) and decreasing on (−∞, −1) and (3 2,∞).

(b) f has local maxima 9e⁻³²as x = 3

2 and local minima −e as x = −1.

(c) f⁰⁰(x) = 0 as x = 5−√ 41

4 or 5 +√ 41

4 ,

f⁰⁰(x) > 0 on (−∞,5−√ 41

4 ), (5 +√ 41

4 ,∞) and f⁰⁰(x) < 0 on (5−√ 41

4 ,5 +√ 41

4 )

⇒ f is concave upward on (−∞,5−√ 41

4 ), (5 +√ 41 4 ,∞) and concave downward on (5−√

41

4 ,5 +√ 41 4 ).

(d) f⁰⁰(x) = 0 or f⁰⁰(x) doesn’t exist and f⁰⁰ changes sign at x it follows that such x = 5 +√

41

4 and x = 5−√ 41

4 which become the inflection points of f .

(e) lim

x→∞f (x) = 0 which implies f has y = 0 as an asymptote.

(f)

.3 .2 2 3 4 5 6

.3 3 5 7 9

評分標準^:

(a)∼(c) : 每格³分。

(d)∼(f) : If there exists any mistakes in your answer, you get zero credit for that blank.

8. (10%) Find the following limits.

(a) lim

x→∞

x⁷ e^x. (b) lim

x→∞

³ 1 + 3

x + 5 x²

´x²

.

Sol:

(a) Using L’Hospital rule seven times, (2pts) we get lim

x→∞

x⁷

e^x = lim

x→∞

7!

e^x = 0 (3pts) (b)

xlim→∞

³ 1 + 3

x + 5 x²

´x²

= exp( lim

x→∞x²ln(1 + 3 x + 5

x²))

= exp( lim

x→∞

ln(1 +_x³ + _x⁵2)

3 x +_x⁵2

x²(3 x + 5

x²))

= exp( lim

x→∞

ln(1 +_x³ + _x⁵2)

3 x +_x⁵2

xlim→∞x²(3 x+ 5

x²)) (3pts)

= exp(1· ∞) (2pts)

=∞

Using L’Hospital rule is ok.