homework

MAT 1214 – Brucks – Spring 2014 Name: ____________________________________________

Sample Exam 2 Part 1

Instructions: Answer all of the following on your own paper. Show all necessary work neatly, using proper notation, and

box your answers. When necessary, solutions may be expressed as decimals rounded to 3 places.

Find the second derivative, 𝑑2𝑦

𝑑𝑥2 .

1. 4√𝑦 = 𝑥 − 𝑦

Find the lines that are a) tangent and b) normal to the given curve at the given point.

2. 𝑥2 + 𝑥𝑦 − 𝑦2 = 9, (3, 6)

Use logarithmic differentiation to find the derivative of 𝑦 with respect to 𝑥.

3. 𝑦 = 𝑥+8

𝑥 cos 𝑥

4. 𝑦 = (9𝑥 + 2)𝑥

Find the derivative of 𝑦 with respect to 𝑥.

5. 𝑦 = csc−1(𝑒3𝑥)

Solve the related rates problem.

6. A 13-ft ladder is leaning against a house when its base starts to slide away. By

the time the base is 5 ft from the house, the base is moving away at the rate of

12 ft/sec. At what rate is the angle between the ladder and the ground

changing at that time?

Find the linearization 𝐿(𝑥) of the given function at the given point, 𝑥0. Then use 𝐿(𝑥) to approximate 𝑓(𝑎).

7. 𝑓(𝑥) = 𝑥 + 1

𝑥 , 𝑥0 = 1, 𝑎 = 1.1

Find 𝑑𝑦.

8. 9𝑦8 3⁄ + 4𝑥𝑦 − 5𝑥 = 0

MAT 1214 – Brucks – Spring 2014 Sample Exam 2 – Part 1

1. 𝑑

𝑑𝑥 [4𝑦1 2⁄ ] =

𝑑

𝑑𝑥 [𝑥 − 𝑦]

2𝑦−1 2⁄ 𝑑𝑦

𝑑𝑥 = 1 −

𝑑𝑦

𝑑𝑥

2𝑦−1 2⁄ 𝑑𝑦

𝑑𝑥 +

𝑑𝑦

𝑑𝑥 = 1

𝑑𝑦

𝑑𝑥 (2𝑦−1 2⁄ + 1) = 1

𝑑𝑦

𝑑𝑥 =

1

2𝑦−1 2⁄ +1

𝑑𝑦

𝑑𝑥 = (2𝑦−1 2⁄ + 1)

−1

𝑑2𝑦

𝑑𝑥2 =

𝑑

𝑑𝑥 [(2𝑦−1 2⁄ + 1)

−1 ] = −1(2𝑦−1 2⁄ + 1)

−2 ⋅

𝑑

𝑑𝑥 [2𝑦−1 2⁄ + 1]

= −(2𝑦−1 2⁄ + 1) −2

(−𝑦−3 2⁄ ) = 1

𝑦3 2⁄ (2𝑦−1 2⁄ +1) 2

2. 𝑑

𝑑𝑥 [𝑥2 + 𝑥𝑦 − 𝑦2] =

𝑑

𝑑𝑥 [−9]

2𝑥 + 𝑥 𝑑𝑦

𝑑𝑥 + 𝑦 − 2𝑦

𝑑𝑦

𝑑𝑥 = 0

(𝑥 − 2𝑦) 𝑑𝑦

𝑑𝑥 = −2𝑥 − 𝑦

𝑑𝑦

𝑑𝑥 =

2𝑥+𝑦

2𝑦−𝑥

𝑑𝑦

𝑑𝑥 |

(𝑥,𝑦)=(3,6) =

2(3)+(6)

2(6)−(3) =

4

3

a) Tangent Slope = 4

3 → Tangent Line: 𝑦 − 6 =

4

3 (𝑥 − 3) → 𝑦 =

4

3 𝑥 + 2

b) Normal Slope = − 3

4 → Normal Line: 𝑦 − 6 = −

3

4 (𝑥 − 3) → 𝑦 = −

3

4 𝑥 +

33

4

3. ln 𝑦 = ln ( 𝑥+8

𝑥 cos 𝑥 )

ln 𝑦 = ln(𝑥 + 8) − ln 𝑥 − ln(cos 𝑥)

𝑑

𝑑𝑥 [ln 𝑦] =

𝑑

𝑑𝑥 [ln(𝑥 + 8) − ln 𝑥 − ln(cos 𝑥)]

𝑦′

𝑦 =

1

𝑥+8 −

1

𝑥 −

− sin 𝑥

cos 𝑥

𝑦′ = 𝑦 ( 1

𝑥+8 −

1

𝑥 + tan 𝑥) =

𝑥+8

𝑥 cos 𝑥 (

1

𝑥+8 −

1

𝑥 + tan 𝑥)

4. ln 𝑦 = ln[(9𝑥 + 2)𝑥]

ln 𝑦 = 𝑥 ln(9𝑥 + 2)

𝑑

𝑑𝑥 [ln 𝑦] =

𝑑

𝑑𝑥 [𝑥 ln(9𝑥 + 2)]

𝑦′

𝑦 = 𝑥 ⋅ [ln(9𝑥 + 2)]′ + [𝑥]′ ⋅ ln(9𝑥 + 2)

𝑦′

𝑦 = 𝑥 ⋅

9

9𝑥+2 + 1 ⋅ ln(9𝑥 + 2)

𝑦′ = 𝑦 ( 9𝑥

9𝑥+2 + ln(9𝑥 + 2)) = (9𝑥 + 2)𝑥 (

9𝑥

9𝑥+2 + ln(9𝑥 + 2))

MAT 1214 – Brucks – Spring 2014 Sample Exam 2 – Part 1

5. 𝑑𝑦

𝑑𝑥 = −

[𝑒3𝑥] ′

|𝑒3𝑥|√(𝑒3𝑥)2−1 = −

3𝑒3𝑥

𝑒3𝑥√𝑒6𝑥−1 = −

3

√𝑒6𝑥−1

6. Given 𝑥 = 5 ft and 𝑑𝑥

𝑑𝑡 = 12 ft/sec, the question asks us to find

𝑑𝜃

𝑑𝑡 .

An equation that relates the given quantities is cos 𝜃 = 𝑥

13 , or 𝑥 = 13 cos 𝜃.

Differentiating this equation with respect to time gives

𝑑

𝑑𝑡 [𝑥] =

𝑑

𝑑𝑡 [13 cos 𝜃] →

𝑑𝑥

𝑑𝑡 = −13 sin 𝜃 ⋅

𝑑𝜃

𝑑𝑡

To solve for 𝑑𝜃

𝑑𝑡 we find 𝜃 when 𝑥 = 5:

cos 𝜃 = 5

13 → cos 𝜃 =

5

13 → 𝜃 = cos−1 (

5

13 ) ≈ 1.176

Thus, we have

𝑑𝑥

𝑑𝑡 = −13 sin 𝜃 ⋅

𝑑𝜃

𝑑𝑡

12 = −13 sin(1.176) ⋅ 𝑑𝜃

𝑑𝑡

12 = −13 ( 12

13 ) ⋅

𝑑𝜃

𝑑𝑡

𝑑𝜃

𝑑𝑡 = −1 rad/sec

7. 𝑓′(𝑥) = 1 − 1

𝑥2

𝐿(𝑥) = 𝑓(1) + 𝑓′(1)(𝑥 − 1)

= (1 + 1

1 ) + [1 −

1

12 ] (𝑥 − 1)

= 2 + 0(𝑥 − 1)

𝐿(𝑥) = 2

𝑓(1.1) ≈ 𝐿(1.1) = 2

8. 𝑑(9𝑦8 3⁄ + 4𝑥𝑦 − 5𝑥) = 𝑑(0)

24𝑦5 3⁄ 𝑑𝑦 + 4𝑥𝑑𝑦 + 4𝑦𝑑𝑥 − 5𝑑𝑥 = 0

24𝑦5 3⁄ 𝑑𝑦 + 4𝑥𝑑𝑦 = 5𝑑𝑥 − 4𝑦𝑑𝑥

(24𝑦5 3⁄ + 4𝑥)𝑑𝑦 = (5 − 4𝑦)𝑑𝑥

𝑑𝑦 = ( 5−4𝑦

24𝑦5 3⁄ +4𝑥 ) 𝑑𝑥