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Math 11-Worksheet 08 Sol

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Course

Calculus I (MATH 011)

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University of California, Merced

Academic year: 2022/2023

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Note that each of these worksheets are designed to offer plenty of practice material. You might not be able to finish the worksheet during the discussion section time. The strategy should be to look over all of the questions, discuss the basics of problem solving for each, and work in depth on those exercises that appear unfamiliar. You do not have to turn in your work for these worksheets, but one of the questions will be on the quiz during your discussion section.

1 Section 3.

Compute the derivative

dy dx

for each of the following settings.

(a)

√

xy = cos− 1 (xy)

1 2 ( xy)− 1 / 2 (y + xy′) = −

1 √

1 − (xy) 2

(y + xy′)

⇒

1 2 (

xy)− 1 / 2 xy′ +

x √ 1 − (xy) 2

y′ = −

1 2 (

xy)− 1 / 2 y −

y √ 1 − (xy) 2

⇒ y′

[

1

2 √

x y

+

x √ 1 − (xy) 2

]

= −

1

2 √

y x

−

y √ 1 − (xy) 2

⇒ y′ =

−

1

2 √

y x

−

y √ 1 − (xy) 2 1 2

√

x y

−

x √ 1 − (xy) 2

(b) x 5 y 7 = tan(x) tan− 1 (y)

5 x 4 y 7 + 7x 5 y 6 y′ = sec 2 x tan− 1 y + tan x

(

y′ 1 + y 2

)

⇒

(

7 x 5 y 6 −

tan x 1 + y 2

)

y′ = sec 2 x tan− 1 y − 5 x 4 y 7

⇒ y′ = sec

2 x tan− 1 y − 5 x 4 y 7

7 x 5 y 6 −

tan x 1 + y 2

(c)

sin x cos y

=

x y

cos x cos y + (sin x sin y)y′ cos 2 x

=

y − xy′ y 2 ⇒ y 2 [cos x cos y + (sin x sin y)y′] = (y − xy′) cos 2 y ⇒

(

y 2 sin x sin y + x cos 2 y

)

y′ = y cos 2 y − y 2 cos x cos y

⇒ y′ =

y cos 2 y − y 2 cos x cos y y 2 sin x sin y + x cos 2 y

(d) y = sin− 1 (y 2 ) +

√

1 − x

y′ = 2 y √ 1 − y 4

y′ −

1 2 (

− x)− 1 / 2

⇒ y′ −

2 y √ 1 − y 4

y′ = −

1 2 (

− x)− 1 / 2

⇒ y′

(

1 −

2 y √ 1 − y 4

)

= −

1 2 (

− x)− 1 / 2

⇒ y′ = −

1 2 (1 − x)

− 1 / 2

1 − √ 2 y 1 −y 4

The part of the graph of sin(x 2 + y) = x that contains the point (0, π) defines y as a function of x implicitly.

(a) Verify that (0, π) is a solution to the equation. sin(0 2 + π) = sin(π) = 0 (b) Is this graph increasing or decreasing near (0, π)? Implicitly differentiating the equation gives

y′ =

1

cos(x 2 + y)

− 2 x,

and evaluating this at (0, π) gives y′ = −1. Hence, the graph is decreasing here.

Find all points where the tangent line to the curve described by

y 3 = xy − 6

is either horizontal or vertical. We find that y′ =

y 3 y 2 − x . Horizontal tangent lines occur when y′ = 0, in other words, when y = 0. However, seeking the x-coordinate, by substituting back into the original equation, of this point results in the false statement, 0 = −6. Hence, there are no such points where the tangent line is horizontal. Vertical tangents occur when the denominator is zero, i. when 3y 2 − x = 0, which yields x = 3y 2. If we substitute this back into the equation, we find that y = 3 1 / 3 = 3

√

3 and x = 3 5 / 3 = 3 3

√

9, which is the only point this occurs at.

Find an equation of the tangent line to the ellipse x 2 + 2y 2 = 1 at the point

(

1 √

2 ,

1

2 )

.

Here, we find that y′ = −

x 2 y ,

which yields a slope of − 1 /

√

With this, we find that y = −

1 √

2

x + 1.

A voltage V across a resistance R generates a current

I =

V

R

If a constant voltage of 9 volts is put across a resistance that is increasing at a rate of 0 ohms per second when the resistance is 5 ohms, at what rate is the current changing? Implicitly differentiating, we get dI dt

=

dV dt R − V

dR dt R 2

.

The given quantities for the moment of interest are V = 9, dV /dt = 0 (V is constant), R = 5, and dR/dt = 0. This gives the decreasing rate of dI/dt = − 9 /125 amp/sec.

A water trough in the shape of an inverted triangle, as in the image below, is being filled with water. The width, W , is 1/3 the overall depth of the trough. Find the rate of change of the volume of water in the trough when the height of the water is 3 feet and is increasing at a rate of 2 ft/s for a 10 foot long trough. Note: the volume of a triangular prism such as this is V = 12 LW H.

Here, we find the objective equation to be V = (L/2)xh and given constraint of x = h/3. This gives V = (15/2)x 2 , which we can differentiate implicitly, yielding dV /dt = 15x dx/dt. Since the given values for the moment of interest are h = 3 and dh/dt = 2, we can find that x = 1 and dx/dt = 2/3. This gives dV /dt = 20 ft 3 /sec.

In an adiabatic process—one in which no heat transfer takes place—the pressure P and volume V of an ideal gas such as oxygen satisfy the equation

P 5 V 7 = C

where C is a constant. Suppose that at a certain instant of time, the volume of the gas is 4 L, the pressure is 100 kPa, and the pressure is decreasing at the rate of 5 kPa/sec. Find the rate at which the volume is changing. 123 Implicitly differentiating, we get

5 P 4 V 7

dP dt

+ 7V 6 P 5

dV dt

= 0.

The moment of interest quantities are V = 4, P = 100, and dP/dt = −5, giving dV /dt = 1/7 L/sec.

A plumber is inspecting the pipes for the We Arc Heap electric company’s main building. He reports that the main line is becoming clogged (refer to diagram below). He estimates the flow of water has reduced from its maximum, unimpeded flow of 30π in 3 /s to 16π in 3 /s over 1 inch of pipe length. If the pipes were installed 28 months ago, how fast is the radius (per month) of the open space in the pipe shrinking?

Here, we use the area formula V = πr 2 , which we differentiate implicitly to get

dV dt

= 2πr dr dt.

We may use the information given to us to estimate the value of dV /dt to be

dV dt

= 16

π − 30 π 28

= −

π 2. We also know that since V = 16π presently, we can deduce that r = 4 at the moment of interest. This allows us to find that dr/dt = − 1 /16 in/mth.

1 This was an exam question during Fall 2014. 2 Hint: isolate dV 3 dt after taking the derivatives with respect to t This is a variant of the ideal gas law; the exponents are indeed part of the math problem.

❼ Since s(t) is continuous on [1, 3], we can use MVT. ❼ s′(t) = 4t.

❼ Slope between endpoints is m = s(3) − s(1) 3 − 1

= 16

2 = 8.

❼ Hence, 4t = 8, which indicates that t = c = 2.

The student never confirmed that s(t) was differentiable, which is one of the hypothesis of the Mean Value Threorem. It is simply not enough to find the derivative to justify differentiability. Lastly, finding a location (particularly without satisfying the hypothesis of the theorem) does not justify usage of the theorem.

Verify that the function satisfies the two hypotheses of the Mean Value Theorem on the given interval, then find all numbers c that satisfy the conclusion of the Mean Value Theorem.

(a) f (x) = 2x 2 − 3 x + 1, [0, 2] f (x) is a polynomial and, hence, continuous and differentiable everywhere. Hence, f (x) has some value, x = c, in [0, 2] where f ′(c) = 1. Here, c = 1. (b) f (x) = x 3 + x − 1 , [0, 2] f (x) is a polynomial and, hence, continuous and differentiable everywhere. Hence, f (x) has some value, x = c, in [0, 2] where f ′(c) = 5. Here, c = +

√

3 /3 (Notice the negative version, − 2

√

3 /3,

is not in the interval!).

x x + 2

, [1, 4]

f (x) is a rational function, which is continuous and differentiable everywhere except its vertical asymptotes. Since f (x) has no vertical asymptotes inside of the interval (the only one is x = −2), the function is continuous and differentiable on [1, 4]. Hence, f (x) has some value, x = c, in [1, 4] where f ′(c) = 1/9. Here, c = −2 + 3

√

2 (Notice the negative version, − 2 − 3

√

2, is not in the interval!).

Consider the function f (x) = 3

√

x over the interval [− 1 , 1].

(a) Show that this function does not satisfy the hypothesis of the Mean Value Theorem. Since f ′(x) = (1/3)x− 2 / 3 , we see that f ′(x) is not defined at x = 0, which lay in the interval. Hence, f (x) fails the second hypothesis of the Mean Value Theorem. (b) Compute the slope of the secant line between the two endpoints. m = 1 (c) Show that, despite MVT not being applicable, there is exists a point x = c as in the conclusion of MVT.

x =

(

1

3 ) 3 / 2

=

√

3

9

Note: This is one justification that you cannot draw any conclusions, one way or the other, when the hypothesis of a theorem is not satisfied!

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Math 11-Worksheet 08 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 8 for Sections 3.8, 4.1, 4.2, and 4.4 (Solutions) Spring 2023

Note that each of these worksheets are designed to offer plenty of practice material. You might not be

able to finish the worksheet during the discussion section time. The strategy should be to look over all of the

questions, discuss the basics of problem solving for each, and work in depth on those exercises that appear

unfamiliar. You do not have to turn in your work for these worksheets, but one of the questions will be on

the quiz during your discussion section.

1 Section 3.8

1. Compute the derivative dy

dx for each of the following settings.

(a) √xy = cos−1(xy)

2(xy)−1/2(y+xy′) = −1

p1−(xy)2(y+xy′)

⇒1

2(xy)−1/2xy′+x

p1−(xy)2y′=−1

2(xy)−1/2y−y

p1−(xy)2

⇒y′"1

2rx

y+x

p1−(xy)2#=−1

2ry

x−y

p1−(xy)2

⇒y′=−1

2ry

x−y

p1−(xy)2

2rx

y−x

p1−(xy)2

(b) x5y7= tan(x) tan−1(y)

5x4y7+ 7x5y6y′= sec2xtan−1y+ tan xy′

1 + y2

⇒7x5y6−tan x

1 + y2y′= sec2xtan−1y−5x4y7

⇒y′=sec2xtan−1y−5x4y7

7x5y6−tan x

1 + y2

cos y=x

cos xcos y+ (sin xsin y)y′

cos2x=y−xy′

⇒y2[cos xcos y+ (sin xsin y)y′] = (y−xy′) cos2y

⇒y2sin xsin y+xcos2yy′=ycos2y−y2cos xcos y

⇒y′=ycos2y−y2cos xcos y

y2sin xsin y+xcos2y

Math 11-Worksheet 08 Sol

Calculus I (MATH 011)

University of California, Merced

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1 Section 3.

√

1

√

⇒

1

2 (

1

2 (

[

1

2

√

+

]

= −

1

2

√

−

−

1

2

√

−

√

−

(

)

⇒

(

)

=

=

(

)

√

1

2 (

1

2 (

(

1 −

)

= −

1

2 (

1

√

√

(

1

√

2

,

1

2

)

.

√

1

√

2

I =

V

R

=

.

5 P 4 V 7

+ 7V 6 P 5

= 0.

= 16

= −