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Sample Calculus Problems

• Part 1: Single Variable Functions
• Part 2: Multi-Variable Functions
• Part 3: Sequences and Series
• Part 4: Vector Analysis
  • Last revision: January 12,

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• One must also not use other symbols, which have completely different meanings, in place of “=”. The solution above should not go like:

lim x→ 5

3 x + 1 x^2 − 7 x + 10 ⇒ lim x→ 5

3 x + 1 )( 4 +

3 x + 1 ) (x^2 − 7 x + 10 )( 4 +

3 x + 1 ) → lim x→ 5 15 − 3 x (x − 5 )(x − 2 )( 4 +

3 x + 1 )

 lim x→ 5

3 ( 5 − x) (x − 5 )(x − 2 )( 4 +

3 x + 1 )

• The equal sign always stands between two things, although sometimes one of these things are at the end of the previous line or at the beginning of the next line. The solution above should not start like:

= lim x→ 5

3 x + 1 )( 4 +

3 x + 1 ) (x^2 − 7 x + 10 )( 4 +

3 x + 1 )

This begs the question: What is equal to lim x→ 5

3 x + 1 )( 4 +

3 x + 1 ) (x^2 − 7 x + 10 )( 4 +

3 x + 1 )

• The equal sign can be used between two functions when we deal with identities, like

x^2 − 1 x − 1

= x + 1 for all x ~= 1

or when we deal with equations, like

Find all x such that x^2 = 4. Therefore we can not just drop some of the limit signs in the solution above to make it look like:

lim x→ 5

3 x + 1 x^2 − 7 x + 10

3 x + 1 )( 4 +

3 x + 1 ) (x^2 − 7 x + 10 )( 4 +

3 x + 1 )

(x − 2 )( 4 +

3 x + 1 )

The equalities on the lines marked with 7 are not correct.

(x − 2 )( 4 +

3 x + 1 )

is

not equal to −

because, for instance, if we let x = 1 then

(x − 2 )( 4 +

3 x + 1 )

2. Let m be the slope of the tangent line to the graph of y = x^2 x + 2 at the point (− 3 , − 9 ).

Express m as a limit. (Do not compute m.)

Solution: The slope m of the tangent line to the graph of y = f (x) at the point (x 0 , f (x 0 )) is given by the limit

m = (^) xlim→x 0

f (x) − f (x 0 ) x − x 0 or equivalently by the limit

m = lim h→ 0

f (x 0 + h) − f (x 0 ) h

Therefore two possible answers are

m = (^) xlim→− 3

x^2 x + 2

x − (− 3 )

= lim h→ 0

(− 3 + h)^2 − 3 + h + 2

h

3. Suppose that lim x→c f (x) ~= 0 and lim x→c g(x) = 0. Show that lim x→c f (x) g(x)

does not exist.

Solution: Assume that lim x→c f (x) g(x)

exists, and let L = lim x→c f (x) g(x)

. Then by the product rule for limits we obtain

lim x→c f (x) = lim x→c Œ f (x) g(x) ⋅ g(x)‘ = lim x→c f (x) g(x) ⋅ lim x→c g(x) = L ⋅ 0 = 0.

This contradicts the fact that lim x→c f (x) ~= 0. Therefore our assumption cannot be

true: lim x→c f (x) g(x) does not exist.

4. Suppose that lim x→c f (x) = 0 and there exists a constant K such that Sg(x)S ≤ K for all x ~= c in some open interval containing c. Show that lim x→c (f (x)g(x)) = 0.

Solution: We have Sf (x)g(x)S = Sf (x)S ⋅ Sg(x)S ≤ Sf (x)S ⋅ K in some open interval around c. Therefore −KSf (x)S ≤ f (x)g(x) ≤ KSf (x)S. Now applying the Sandwich Theorem and using the fact that lim x→c Sf (x)S = 0 we obtain the result.

Solution: Subtracting x^2 + y^2 = r^2 from (x − 1 )^2 + y^2 = 1 we obtain x = r^2 ~ 2 , and substituting this back in x^2 + y^2 = r^2 gives us Q(r^2 ~ 2 ,

r^2 − r^4 ~ 4 ).

Let R(a, 0 ) be the coordinates of R and let S be the foot of the perpendicular from Q to the x-axis. Since the triangles RSQ and ROP are similar we have

»^ a^ −^ r^2 ~^2 r^2 − r^4 ~ 4

= a r

and hence a = r^3 ~ 2 r −

r^2 − r^4 ~ 4

Then

r^ lim→ 0 +^ a^ =^ rlim→ 0 +

r^3 ~ 2 r −

r^2 − r^4 ~ 4

= lim r→ 0 +^ Œ r

r^2 − (r^2 − r^4 ~ 4 )

⋅ (r +

r^2 − r^4 ~ 4 )‘

= 2 lim r→ 0 +( 1 +

1 − r^2 ~ 4 ) = 2 ⋅ ( 1 +

Therefore R approaches the point ( 4 , 0 ) as r → 0 +.

7. Use the formal definition of the limit to show that lim x→ 1 ~ 2

x

Solution: Given ε > 0 we want to find δ > 0 such that

0 < Vx −

V < δ Ô⇒ V

x − 2 V < ε. (⋆)

We will do this in two different ways.

“Solve the Inequality” Method : First we solve V

x − 2 V < ε for x.

V

x − 2 V < ε ⇐⇒ 2 − ε <

x < 2 + ε

The next step depends on whether 2 − ε is positive, zero or negative.

If 2 − ε > 0 , that is if ε < 2 , then

2 − ε <

x < 2 + ε ⇐⇒

2 − ε

x >

2 + ε

If 2 − ε = 0 , that is if ε = 2 , then

2 − ε <

x < 2 + ε ⇐⇒ x >

2 + ε

If 2 − ε < 0 , that is if ε > 2 , then

2 − ε <

x < 2 + ε ⇐⇒ x >

2 + ε or −

ε − 2

x.

SxS < 4. Therefore, using the Triangle Inequality, we obtain Sx^3 − 3 x^2 + 9 x − 20 S ≤ SxS^3 + 3 SxS^2 + 9 SxS + 20 < 43 + 3 ⋅ 42 + 9 ⋅ 4 + 20 = 168. Now if we choose δ to satisfy 0 < δ ≤ min{ε~ 168 , 1 }, then we have

Sx^4 + 7 x − 17 − 43 S = Sx^4 + 7 x − 60 S = Sx + 3 S ⋅ Sx^3 − 3 x^2 + 9 x − 20 S < δ ⋅ 168 ≤ ε 168 ⋅ 168 = ε

whenever 0 < Sx − (− 3 )S < δ. We are done.

9. Suppose that for all 0 < ε < 1 ,

Sx − 1 S < ε^2 4 Ô⇒ Sf (x) − 3 S < ε (⋆)

and Sx − 1 S < ε 35 Ô⇒ Sg(x) − 4 S < ε. (⋆⋆)

Find a real number δ > 0 such that

Sx − 1 S < δ Ô⇒ Sf (x) + g(x) − 7 S <

Solution: If we take ε =

in (⋆) we get

Sx − 1 S <

Ô⇒ Sf (x) − 3 S <

If we take ε =

in (⋆⋆) we get

Sx − 1 S <

Ô⇒ Sg(x) − 4 S <

Therefore if Sx − 1 S <

, then we have

Sf (x) + g(x) − 7 S = Sf (x) − 3 + g(x) − 4 S ≤ Sf (x) − 3 S + Sg(x) − 4 S <

by the Triangle Inequality. Hence we can take δ = 1 400

Remark: In fact, any δ ≤

works.

10. Let f (x) =

1 if x =

n where n is a positive integer,

0 otherwise.

a. Show that if c =~ 0 then lim x→c f (x) = 0.

b. Show that lim x→ 0 f (x) does not exist.

Solution: a. Assume c > 0. Then there is a positive integer m such that 1 ~m is the closest to c among all real numbers different from c and of the form 1 ~n where n is a positive integer. (Why?) Let δ = Vc −

m

V > 0. Then for any ε > 0 , we have

0 < Sx − cS < δ Ô⇒ x =~

n

for any positive integer n Ô⇒ f (x) = 0 Ô⇒ Sf (x) − 0 S = S 0 − 0 S = 0 < ε.

Therefore lim x→c f (x) = 0.

Assume c < 0. Take δ = ScS. Then for any ε > 0 , we have

0 < Sx − cS < δ Ô⇒ x < 0 Ô⇒ x =~

n for any positive integer n Ô⇒ f (x) = 0 Ô⇒ Sf (x) − 0 S = S 0 − 0 S = 0 < ε.

Therefore lim x→c f (x) = 0.

b. Let L be a real number and assume that lim x→ 0 f (x) = L. Then for every ε > 0 , there exists a δ > 0 such that for all x,

0 < Sx − 0 S < δ Ô⇒ Sf (x) − LS < ε.

If L is not 0, let ε = SLS~ 2 > 0. Then there is a δ > 0 such that

0 < SxS < δ Ô⇒ Sf (x) − LS < SLS~ 2.

Take x = −δ~ 2. Then 0 < SxS < δ is true, but SLS = S 0 − LS = Sf (x) − LS < SLS~ 2 is not true. We have a contradiction.

On the other hand, if L = 0 , let ε = 1 ~ 2. Then there is a δ > 0 such that

0 < SxS < δ Ô⇒ Sf (x)S < 1 ~ 2.

If n is a positive integer satisfying n > 1 ~δ, take x = 1 ~n. Then 0 < SxS < δ is true, but 1 = S 1 S = Sf (x)S < 1 ~ 2 is not true. Again we have a contradiction.

Hence lim x→ 0 f (x) cannot exist.

11. Show that the equation x^2 − 10 = x sin x has a real solution.

Solution: Consider the function f (x) = x^2 − 10 − x sin x. Then f ( 0 ) = − 10 < 0 and f ( 10 ) = 102 − 10 − 10 sin( 10 ) = 90 − 10 sin( 10 ) ≥ 90 − 10 = 80 > 0. Note that f is continuous on [ 0 , 10 ]. Therefore we can apply the Intermediate Value Theorem to the function f on the interval [ 0 , 10 ] for the value 0 to conclude that there is a point c in ( 0 , 10 ) such that f (c) = 0. This c is also a solution of the given equation.

14. Find all tangent lines to the graph of y = x^3 that pass through the point ( 2 , 4 ).

Solution: As dy~dx = d(x^3 )~dx = 3 x^2 , the equation of the tangent line through a point (x 0 , x^30 ) on the graph is y − x^30 = 3 x^20 (x − x 0 ). This line passes through ( 2 , 4 ) exactly when 4 − x^30 = 3 x^20 ( 2 − x 0 ), or in other words, x^30 − 3 x^20 + 2 = 0. We observe that x 0 = 1 is a root of this polynomial. Therefore we have the factorization x^30 − 3 x^20 + 2 = (x 0 − 1 )(x^20 − 2 x 0 − 2 ). The roots of the quadratic factor are x 0 = 1 ±

Therefore the tangent lines to y = x^3 at the points ( 1 , 1 ), ( 1 +

3 ), and ( 1 −

3 ) pass through ( 2 , 4 ). The equations of these lines are y = 3 x − 2 , y = ( 12 + 6

3 )x − ( 20 + 12

3 ), and y = ( 12 − 6

3 )x − ( 20 − 12

3 ), respectively.

15. Evaluate the limit lim x→ 0

1 + sin^2 x^2 − cos^3 x^2 x^3 tan x

Solution:

lim x→ 0

1 + sin^2 x^2 − cos^3 x^2 x^3 tan x = lim x→ 0

1 + sin^2 x^2 − 1 x^4

1 − cos^3 x^2 x^4

x tan x

= lim x→ 0 ŒŒ‹ sin x^2 x^2

2 ⋅

1 + sin^2 x^2 + 1

1 − cos^3 x^2 x^4

x tan x

Now we observe that: lim x→ 0 sin x^2 x^2

lim x→ 0

1 + sin^2 x^2 + 1

1 + sin^2 02 + 1

lim x→ 0 x tan x

and

lim x→ 0 1 − cos^3 x^2 x^4 = lim x→ 0 ‹ 1 − cos x^2 x^4 ⋅ ( 1 + cos x^2 + cos^2 x^2 )

= lim x→ 0 Œ 2 sin^2 (x^2 ~ 2 ) x^4 ⋅ ( 1 + cos x^2 + cos^2 x^2 )‘

⋅ Œlim x→ 0 sin(x^2 ~ 2 ) x^2 ~ 2

2 ⋅ lim x→ 0 ( 1 + cos x^2 + cos^2 x^2 )

=

Therefore: lim x→ 0

1 + sin^2 x^2 − cos^3 x^2 x^3 tan x

16. Find the equation of the tangent line to the graph of y = sin^2 (πx^3 ~ 6 ) at the point with x = 1.

Solution: y = sin^2 (πx^3 ~ 6 ) ⇒ dy dx

= 2 sin(πx^3 ~ 6 ) ⋅ cos(πx^3 ~ 6 ) ⋅ 3 πx^2 ~ 6. Therefore, dy dx

V

x= 1

= 2 sin(π~ 6 ) ⋅ cos(π~ 6 ) ⋅ π~ 2 =

3 π 4

Since ySx= 1 = 1 ~ 4 , using the point-slope formula we find the equation of the tangent line as y −

3 π 4 (x − 1 ) or, after some reorganization,

y =

3 π 4 x +

3 π 4

17. Let

f (x) =

2 x + x^2 sinŒ

x ‘ if x =~ 0 ,

0 if x = 0. a. Find f ′(x) for all x. b. Show that f ′^ is not continuous at 0.

18. Find d^2 y dx^2

V

(x,y)=( 2 , 1 )

if y is a differentiable function of x satisfying the equation x^3 + 2 y^3 = 5 xy.

Solution:

x^3 + 2 y^3 = 5 xy Ô⇒ d~dx

3 x^2 + 6 y^2 dy dx = 5 y + 5 x dy dx

Ô⇒

x = 2 , y = 1

12 + 6 dy dx

dy Ô⇒ dx

dy dx

Ô⇒

dy dx

at (x, y) = ( 2 , 1 )

Now we differentiate the equation marked (⋆) with respect to x to find the second

derivative.

3 x^2 + 6 y^2 dy dx = 5 y + 5 x dy Ô⇒ dx d~dx

6 x + 12 y Πdy dx

2

• 6 y^2 d

(^2) y dx^2

= 5 dy dx

• 5 dy dx

• 5 x d

(^2) y dx^2 Ô⇒ x = 2 , y = 1 , dydx = (^74)

2

• 6 d^2 y dx^2

d^2 y Ô⇒ dx^2

d^2 y dx^2

at (x, y) = ( 2 , 1 )

Remark: An alternative approach is to solve y′^ from (⋆), viz. y′^ = 5 y − 3 x^2 6 y^2 − 5 x , and then

differentiate this with respect to x to find y′′.

19. A piston is connected by a rod of length 14 cm to a crankshaft at a point 5 cm away from the axis of rotation of the crankshaft. Determine how fast the crankshaft is rotating when the piston is 11 cm away from the axis of rotation and is moving toward it at a speed of 1200 cm/sec.

Solution: Let P (x, y) and Q(a, 0 ) be the ends of the connecting rod as shown in the picture. The axis of rotation of the crankshaft passes through the origin of the xy-plane and is perpendicular to it. The point P where the rod is connected to crankshaft moves on a circle with radius 5 cm and center at the origin. The point Q where the rod is connected to the piston moves along the positive x-axis. θ is the angle between the ray OP and the positive x-axis.

P (x, y)

θ Q(a, 0 ) x

y

x^2 + y^2 = 25

(^14) cm

− 5 5

The question is:

a = 11 cm and da dt = − 1200 cm/sec Ô⇒ dθ dt

Remark:

radian/sec is

2 π

rpm or approximately 3105 rpm.

Remark: This problem has a shorter solution if we use the law of cosines. Start with x^2 + 52 − 2 ⋅ 5 x cos θ = 142 and differentiate with respect to t to obtain

dθ dt

5 cos θ − 11 5 sin θ v.

Put x = 11 cm in the first equation to find cos θ = − 5 ~ 11 and then sin θ = 4

6 ~ 11. Now substituting these in the second equation gives the answer.

20. Determine how fast the length of an edge of a cube is changing at the moment when the length of the edge is 5 cm and the volume of the cube is decreasing at a rate of 100 cm^3 /sec.

Solution: Let a denote the length of an edge of the cube, and V denote the volume of the cube. Then we have V = a^3. Differentiating with respect to time t gives dV dt = 3 a^2 da dt

. Substituting dV dt = − 100 cm^3 /sec and a = 5 cm for the moment in

question, we obtain da dt

cm/sec. Therefore the length of the edge is decreasing at a rate of

cm/sec at that moment.

21. We measure the radius and the height of a cone with 1% and 2% errors, respectively. We use these data to compute the volume of the cone. Estimate the percentage error in volume.

Solution: Let r, h, and V be the radius, the height and the volume of the cone, respectively.

V = π 3

r^2 h Ô⇒ dV = 2 π 3

rhdr + π 3

r^2 dh Ô⇒ dV V

dr r

dh h

Since the error in r is 1% we have V dr r V ≤ 1%. Similarly V dh h V ≤ 2%. Now using the triangle inequality we obtain

V dV V

V = V 2

dr r

dh h

V ≤ 2 V

dr r

V + V

dh h

V ≤ 2 ⋅ 1% + 2% = 4%.

The error in volume is 4%.

22. A cone of radius 2 cm and height 5 cm is lowered point first into a tall cylinder of radius 7 cm that is partially filled with water. Determine how fast the depth of the water is changing at the moment when the cone is completely submerged if the cone is moving with a speed of 3 cm/s at that moment.

Solution: Let r and h be the radius and the height of the part of the cone that is under the water level, respectively. Let L be the depth of the water in the cylinder and let y be the vertical distance from the tip of the cone to the bottom of the cylinder. Let V 0 be the volume of the water.

Then V 0 = π ⋅ 72 ⋅ L − π 3 r^2 h = π ⋅ 72 ⋅ L − π 3

h‘

2 h = 49 πL − 4 π 75 h^3

where we used the fact that r~h = 2 ~ 5.

Now differentiating this with respect to time t gives

0 = d dt V 0 = 49 π dL dt

4 π 25 h^2 dh dt

In particular at the moment when the cone is completely submerged we have h = 5 cm and 49 dL dt

dh dt

On the other hand, at the same moment

h = L − y Ô⇒ dh dt

dL dt

dy dt

Ô⇒

dh dt

dL dt

because dy~dt = − 3 cm/s as the cone is being lowered at a speed of 3 cm/s.

From these two equations we obtain dL~dt = 4 ~ 15 cm/s. In other words, the depth of the water is increasing at a rate of 4/15 cm/s at that moment.