MATH 2250 - 83220 MIDTERM EXAM 2 MODEL SOLUTION Write ...

MATH 2250 - 83220 MIDTERM EXAM 2 MODEL SOLUTION SPRING 2012 - MOON

Write your answer neatly and show intermediate steps. Calculator, computer and any electronic devices are not allowed. (1) (1pt) Write any suggestion for improving this class. (For instance, give written homework twice in a week, give more examples in class, the tempo of this class is too fast, ...) dy of dx y = (x2 + 2)(x − 3).

(2) (4pts) Compute the derivative

Solution: dy d d = (x2 + 2) (x − 3) + (x − 3) (x2 + 2) dx dx dx 2 = (x + 2) · 1 + (x − 3) · 2x = 3x2 − 6x + 2 d d • Writing the step (x2 + 2) (x − 3) + (x − 3) (x2 + 2): 2 pts. dx dx • Computing the answer (x2 + 2) · 1 + (x − 3) · 2x: 3pts. • Simplifying and getting answer 3x2 − 6x + 2: 4 pts.

Date: March 1, 2012. 1

MATH 2250 - 83220 MIDTERM EXAM 2 MODEL SOLUTION


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dy of dx x−3 y= . 2x + 5

Solution: d d (2x + 5) (x − 3) − (x − 3) (2x + 5) dy dx dx = dx (2x + 5)2 (2x + 5) · 1 − (x − 3) · 2 11 = = (2x + 5)2 (2x + 5)2 d d (2x + 5) (x − 3) − (x − 3) (2x + 5) dx dx • Writing the step : 3pts. (2x + 5)2 (2x + 5) · 1 − (x − 3) · 2 : 4pts. • Computing the answer (2x + 5)2 11 • Simplifying and getting : 5pts. (2x + 5)2 (4) (5pts) Compute the derivative

Solution: y=

dy of dx √ 5 y = ln x. √ 5

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ln x = (ln x) 5 . 1

u = ln x ⇒ y = u 5 . 4 1 4 dy du 1 41 1 1 dy = · = u− 5 = (ln x)− 5 = (ln x)− 5 . dx du dx 5 x 5 x 5x

• Finding appropriate intermediate variable u = ln x and express y as a func1 tion with respect u, y = u 5 : 2pts. 1 41 • Applying chain rule and obtaining u− 5 : 4pts. 5 x 1 − 45 • Getting correct answer (ln x) : 5pts. 5x


(5) (5pts) By using logarithmic differentiation, compute the derivative y=

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dy of dx

(x2 + 1)6 (5x − 1) . (x − 5)10

Solution: (x2 + 1)6 (5x − 1) = ln(x2 + 1)6 + ln(5x − 1) − ln(x − 5)10 ln y = ln (x − 5)10 = 6 ln(x2 + 1) + ln(5x − 1) − 10 ln(x − 5). d d ln y = 6 ln(x2 + 1) + ln(5x − 1) − 10 ln(x − 5) . dx dx 1 dy 2x 5 1 12x 5 10 =6· 2 + − 10 · = 2 + − . y dx x + 1 5x − 1 x−5 x + 1 5x − 1 x − 5 dy 12x 5 10 5 10 (x2 + 1)6 (5x − 1) 12x =y + − + − = . dx x2 + 1 5x − 1 x − 5 (x − 5)10 x2 + 1 5x − 1 x − 5

• Computing logarithm of right side 6 ln(x2 + 1) + ln(5x − 1) − 10 ln(x − 5): 2pts. 5 10 (x2 + 1)6 (5x − 1) 12x + − • Getting correct answer by com(x − 5)10 x2 + 1 5x − 1 x − 5 12x 5 10 1 dy = 2 + − : 5pts. puting y dx x + 1 5x − 1 x − 5 dy of dx y = cos(ln(x2 − 1)).


Solution: u = ln(x2 − 1) ⇒ y = cos u. dy dy du du du = · = −(sin u) = − sin(ln(x2 − 1)) . dx du dx dx dx v = x2 − 1 ⇒ u = ln v. du du dv 1 1 2x = · = · 2x = 2 · 2x = 2 . dx dv dx v x −1 x −1 dy du 2x = − sin(ln(x2 − 1)) = − sin(ln(x2 − 1)) 2 . dx dx x −1 dy du • Computing = − sin(ln(x2 − 1)) : 3pts. dx dx du 2x • Computing = 2 : 2pts. dx x −1


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(7) (5pts) Find the tangent line of the curve x2 + y + y 2 = 11 at (3, 1). Solution:

dy dy + 2y = 0. dx dx dy dy −2x (1 + 2y) = −2x ⇒ = . dx dx 1 + 2y dy −2 · 3 = −2. x=3 = dx y=1 1 + 2 · 1 The equation of tangent line: y = −2(x − 3) + 1 = −2x + 7. dy −2x = : 2pts. • Computing dx 1 + 2y dy • Computing = −2: 3pts. dx x=3 y=1 • Finding the equation of tangent line: 5pts. x2 + y + y 2 = 11 ⇒ 2x +

(8) (5pts) For y = f (x) = ex + x, let f −1 (y) be the inverse function of f (x). Compute 0

f −1 (e + 1). Solution: f −1 (e + 1) = x ⇔ f (x) = e + 1 ⇒ ex + x = e + 1 ⇒ x = 1. f 0 (x) = ex + 1 ⇒ f 0 (1) = e + 1. By the inverse function theorem, 1 1 0 f −1 (e + 1) = 0 = . f (1) e+1 • Writing precise statement of inverse function theorem: + 2pts. • Finding x such that f (x) = e + 1: + 2pts. 1 0 • Finding f 0 (1) = e + 1 and computing f −1 (e + 1) = : + 1pt. e+1


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(9) The following figure is the graph of velocity v(t) of a car moving along a straight road. For the following questions, you don’t need to write the reason.

(a) (2pts) When does the car move forward? (0, 2) and (7, 9).

(b) (2pts) When does the car reverse direction? At 2 and 7.

(c) (2pts) When is the car moving at a constant speed? (3, 5).

(d) (2pts) When does the car move at its greatest speed? At 6.

(e) (2pts) When is the car’s acceleration negative? (1, 3), (5, 6) and (8, 9).


(10) (5pts) By using linearization, compute an approximation value of e0.03 . Solution: f (x) = ex ⇒ e0.03 = f (0.03). f (0) = e0 = 1, f 0 (x) = ex ⇒ f 0 (0) = e0 = 1. So the linearization L(x) at 0 is L(x) = f (0) + f 0 (0)(x − 0) = 1 + x. Therefore the linear approximation of f (0.03) is L(0.03) = 1 + 0.03 = 1.03. • • • •

Introducing appropriate function f (x): + 1pt. Knowing the formula of linearization: + 1pt. Computing linearization L(x) at 0: + 2pts. Calculating approximation value 1.03: + 1pt.

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(11) (10pts) You are recording a race from a stand 100 ft from the track, following a car that is moving at 200 ft/sec. How fast the distance between you and the car π is increasing when θ = ? 4 1 1 π π π Note that sin = √ , cos = √ , tan = 1. 4 4 4 2 2

Solution: s: distance between you and the car. x: the length of the base of above right triangle. ds Goal: π =? dt θ= 4 Known facts: • s2 = x2 + 1002 . dx = 200. • dt √ √ π x θ= ⇒1= ⇒ x = 100 ⇒ s = 1002 + 1002 = 100 2. 4 100 ds dx ds x dx 2 2 s = x + 1002 ⇒ 2s = 2x ⇒ = . dt dt dt s dt ds 100 200 √ 200 = √ (ft/sec). π = dt θ= 4 100 2 2 ds • Writing precise goal π : + 2pts. dt θ= 4 dx • Writing related equations s2 = x2 + 1002 and = 200: + 2pts for each. dt x dx ds • Computing relation = : + 2pts. dt s dt 200 • Getting correct answer √ : + 2pts. 2