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Math 11-Worksheet 04 Sol

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Calculus I (MATH 011)

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1 Section 3.

At time t in seconds, a particle’s distance s(t), in micrometers (μm), from a point is given by

s(t) = 3 4

t 2 − 1

What is the average velocity of the particle from t = 2 to t = 4?

vavg =

s(4) − s(2)

4 − 2

= 9

2

μm/s

Consider the graph below of the position of a model rocket, h(t). Here h is in meters and t is measured in seconds.

(a) Estimate the average velocity of the rocket in the first 3 seconds of flight.

v = 35

− 0

3 − 0

= 35

3

m/s (≈ 11 .67 m/s)

(b) What is the average velocity of the rocket in the last 3 seconds of flight?

v = 0

− 35

6 − 3

= −

35

3

m/s

v = 35

− 20

3 − 1

= 15

2

or 7 m/s

(d) Estimate the average velocity of the rocket in the last 5 seconds of flight.

v = 0

− 20

6 − 1

= −4 m/s

(e) In general, under what conditions is the average velocity for the rocket zero? When the two times are equidistant from t = 3, i. t = 1 and t = 5, or t = 2 and t = 3, etc.

For each of the following situations, sketch a graph of the distance the car has traveled as a function

of time.

(a) A car is driven at a constant speed.

(b) A car is driven at an increasing speed.

Write out and simplify the difference quotient

f (x + h) − f (x)

for each of the following functions.

(a) f (x) = 2x 2 + 3

f (x + h) − f (x)

= 2(

x + h) 2 + 3 − (2x 2 + 3)

= 4

xh + 2h 2

h = 4x + 2h

(b) g(x) = 4 x

g(x + h) − g(x)

=

4

x + h

−

4

x h

=

4

x + h

−

4

x h

·

x(x + h)

= 4

x − 4(x + h)

xh(x + h)

= −

4 h

xh(x + h)

= −

4

x(x + h)

(b) B(x) =

1

x 2

at x = − 2

B

′ (−2) = lim h→ 0

1

(−2 + h) 2

−

1 (−2) 2

= lim h→ 0

4 − (−2 + h) 2

4 h(−2 + h) 2

= lim h→ 0

4 h − h 2

4 h(−2 + h) 2

= lim h→ 0

4 − h

4(−2 + h) 2

= 1

4 √

x at x = 4 Hint: Rationalize the numerator.

C

′ (4) = lim h→ 0

√

4 + h − 2

= lim h→ 0

√

4 + h − 2

· √

4 + h + 2 √ 4 + h + 2

= lim h→ 0

4 + h − 4

√

4 + h + 2)

= lim h→ 0

1 √

4 + h + 2

= 1

4

Label whether

is positive, negative, or neither at the indicated points for the function, f (x),

given below. A - negative, B - positive, C - positive, D - neither (zero), E - positive, F - negative

Use the definition of the derivative

f ′ (x) = lim h→ 0

f (x + h) − f (x)

on the given functions to produce the derivative function for each. Be sure to write “ lim h→ 0

” on every

step until the limit is actually evaluated.

(a) f (x) = 20x 2 + 15x

′ (x) = lim h→ 0

20(x + h) 2 + 15(x + h) − 20 x 2 − 15 x

= lim h→ 0

40 xh + 20h 2 + 15h

= lim h→ 0

(40x + 20h + 15)

= 40x + 15

(b) f (x) =

20

x + 15

f ′ (x) = lim h→ 0

20

x + h + 15

−

20

x + 15 h

= lim h→ 0

20(x + 15) − 20(x + h + 15)

h(x + 15)(x + h + 15)

= lim h→ 0

−

20

(x + 15)(x + h + 15)

= −

20

(x + 15) 2

√

20 x + 15

f ′ (x) = lim h→ 0

√

20(x + h) + 15 − √ 20 x + 15

= lim h→ 0

20(x + h) + 15 − (20x + 15)

√

20(x + h) + 15 +

√

20 x + 15)

= lim h→ 0

20 √

20(x + h) + 15 +

√

20 x + 15

=

10 √

20 x + 15

A aircraft approaching a runway and landing follows the flightpath given (in meters) by y(t) =

50

t + 1 , with t in seconds. Find the instantaneous velocity for the plane, and use it to calculate the velocity after 3 seconds.

lim h→ 0

y(t + h) − y(t)

= lim h→ 0

50

t + h + 1

−

50

t + 1 h

= lim h→ 0

−

50 h

h(t + 1)(t + h + 1)

= −

50

(t + 1) 2

Evaluating this at t = 3, we find the velocity is − 25 /8 m/s.

A chemical reaction combines two products, x (in milligrams), producing a current (in milliamps)

according to the function f (x) = 2 +

1

x 2

. Find the instantaneous rate of change for the current

produced when using 2 milligrams of the product.

lim h→ 0

f (x + h) − f (x)

= lim h→ 0

2 +

1

(x + h) 2

−

(

2 +

1

x 2

)

= lim h→ 0

x 2 − (x + h) 2

hx 2 (x + h) 2

= lim h→ 0

−h(2x + h)

hx 2 (x + h) 2

= −

2 x

x 4

= −

2

x 3

Evaluating this at x = 2, we find the instantaneous rate of change for the current produced to be − 1 /4 milliamp/mg.

2 Section 3

Find the intervals where f (x), given in the figure below, is (a) increasing and (b) decreasing. Increasing on (−∞, −3) ∪ (0, 2) and decreasing on (− 3 , 0) ∪ (2, ∞).
Sketch a graph of the derivative of the given functions given in the figures below.

Solutions are plotted below.

Find the values of a and b that make f (x) continuous and differentiable at x = −2.

f (x) =











ax + b, x ≤ − 2

2 x 3 , x > − 2

We require for continuity that

lim x→− 2 −

f (x) = lim x→− 2 +

f (x) = f (−2).

This yields the equation −16 = − 2 a + b. For differentiability, we simply require that f ′ (x) exists at x = −2. This gives the equation, 24 = a, which we can use in the previous equation to compute that b = 32.

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Math 11-Worksheet 04 Sol

Course: Calculus I (MATH 011)

81 Documents

Students shared 81 documents in this course

University: University of California, Merced

Was this document helpful?

Math 11 Worksheet 4 - Sections 3.1 and 3.2a (Solutions) Spring 2023

1 Section 3.1

1. At time tin seconds, a particle’s distance s(t), in micrometers (µm), from a point is given by

s(t) = 3

4t2−1

What is the average velocity of the particle from t= 2 to t= 4?

vavg =s(4) −s(2)

4−2=9

2µm/s

2. Consider the graph below of the position of a model rocket, h(t). Here his in meters and tis measured

in seconds.

(a) Estimate the average velocity of the rocket in the first 3 seconds of flight.

v=35 −0

3−0=35

3m/s (≈11.67 m/s)

(b) What is the average velocity of the rocket in the last 3 seconds of flight?

v=0−35

6−3=−35

3m/s

v=35 −20

3−1=15

2or 7.5 m/s

(d) Estimate the average velocity of the rocket in the last 5 seconds of flight.

v=0−20

6−1=−4 m/s

Math 11-Worksheet 04 Sol

Calculus I (MATH 011)

University of California, Merced

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1 Section 3.

= 9

2

− 0

3 − 0

= 35

3

− 35

6 − 3

= −

35

3

− 20

3 − 1

= 15

2

− 20

6 − 1

= 2(

= 4

=

4

−

4

=

4

−

4

·

= 4

= −

= −

4

1

B

1

−

1

(−2) 2

= 1

4

√

C

√

√

·

√

√

1

√

= 1

4

20

20

−

20

−

20

= −

20

√

√

√

√

20

√

√

=

10

√

50

50

−