Download Page 1 MCV4U0 PROBLEM SET V: Derivatives II In order to receive

MCV4U0
PROBLEM SET V: Derivatives II
In order to receive FULL MARKS for this problem set (as well as future problem sets), you must show ALL your work.
1) The figure to the right shows the graphs of f ( x ) , f ′( x ) and f ′′( x ) . Match the three functions with the letters P, Q
and R. Briefly JUSTIFY your selection.
SOLUTION:
f ( x ) is P
f ′( x ) is Q
f ′′( x ) is R
If you compare the slopes from graph P (i.e. relative extrema and regions where the function is increasing and
decreasing), they coincide with the relative values of graph Q. Now if you compare the slopes from graph Q (i.e.
relative extrema and regions where the function is increasing and decreasing), they coincide with the relative values of
graph R.
2) Let f ( x ) = 10 x + ( arctan x ) + 2 . If g( x ) = f
2
−1
( x) , i.e. if g is the inverse function of f, then find the value of g ′( 2) .
SOLUTION:
Let y = 10 x + ( arctan x ) + 2 . Then the inverse of this function, by definition is x = 10 y + ( arctan y ) + 2 . By implicit
2
2
(
)
( )


differentiation, 1 = 10 y ′ + 2 arctan y  1+1y 2  y ′ . We need to find the value of y when x = 2 on the inverse function
in order to evaluate g ′( 2) {i.e. the point of tangency on the inverse is ( 2, y ) } . Observe that y = 2 must be on the
original function {i.e. the point ( y,2) must lie on the original function} . Hence, 2 = 10 x + ( arctan x ) + 2 which upon
2
trial and error yields x = 0 . Thus, the required point of tangency on the inverse function is ( 2,0) which upon
( 1+10 )( y ′ ) or
substitution yields 1 = 10 y ′ + 2(arctan 0)
y ′ y=0 =
1
10
= g ′( 2 ) .
3) Let f ( x ) = x 3 − 8 x 2 + 8 x + 17 , x ∈[ − 1,5] . Use calculus to answer the following questions.
a)
Find and classfiy all relative extrema. JUSTIFY your answers.
SOLUTION:
Observe that f ( x ) is a polynomial which is continuous over the closed interval and differentiable over the open
interval. Hence, an absolute maximum and absolute minimum are guaranteed to exist by the extreme value
theorem. Any relative extrema will occur at critical points of f ( x ) . Hence, f ′( x ) = 3x 2 − 16 x + 8 and the critical
points of f ( x ) occur at x =
8± 2 10
3
. A first or second derivative test is necessary to classify the critical points.
For sake of brevity, a second derivative test will be used. Hence, f ′′( x ) = 6 x − 16 and f ′′
f ′′
(
x=
8− 2 10
3
8+ 2 10
3
) < 0 . Therefore, there is a relative maximum at x =
8− 2 10
3
(
8+ 2 10
3
) > 0 while
and there is a relative minimum at
. The approximate coordinates of the relative maximum and relative minimum are ( 0.5585, 19.1468)
and ( 4.7749, − 18.332) respectively.
b)
Find and classify all absolute extrema. JUSTIFY your answers.
SOLUTION:
Observe that f ( − 1) = 0 , f
(
8− 2 10
3
) ≈ 19.1468 , f (
8+ 2 10
3
) ≈ −18.332 and f (5) = −18 .
the absolute maximum of f ( x ) is approximately 19.1468 which occurs when x =
of f ( x ) is approximately −18.332 which occurs when x =
8+ 2 10
3
.
8− 2 10
3
As mentioned in part a),
and the absolute minimum
c)
Find all values of c that satisfy the conclusion of the Mean Value Theorem of Derivatives.
SOLUTION:
If the MVT for derivatives exists for this function, then f ′ (c) =
f (5) − f ( − 1)
5− ( − 1)
Simplification and use of the quadratic formula reveal that c =
8± 31
3
. Therefore, 3c 2 − 16c + 8 =
− 18− 0
6
= −3 .
, both of which are in the stated interval
x ∈[ − 1,5] .
d)
Sketch the graph of f ( x ) .
SOLUTION:
4) In general, the strength S of an animal is proportional to the cross–sectional area of its muscles and this area is
proportional to the square of its height H, so the strength S = aH 2 (a is a positive proportionality constant). Similarly,
the weight W of the animal is proportional to the cube of its height, so W = bH 3 (b is a positive proportionality
constant). Finally, the relative strength R of an animal is the ratio of its strength to its weight. As the animal grows,
show that its strength and weight increase but that the relative strength decreases.
Hint: You need to find
dS
dt
,
dW
dt
and dR
and somehow connect each quantity to
dt
dH
dt
.
SOLUTION:
From the given information, S = aH 2 , W = bH 3 and R =
dW
dt
dS
dt
( dHdt ) and
= (3bH 2 )
> 0 and
dW
dt
dR
dt
= −
> 0 however
( ab ) H −1 . As time progresses,
( ab )( H −2 )( dHdt ) . Since, a, b, H and
dR
dt
dH
dt
dS
dt
= ( 2aH )
( dHdt ) ,
are positive (the animal is getting larger), then
< 0 . Hence, the strength and weight increase however the relative strength
decreases.
5) You are walking along a sidewalk toward a 15 m long wall with graffiti which is adjacent to the sidewalk and
perpendicular to it (see diagram to the right).
a)
If your viewing angle at a particular instant is θ =
π
6
rads. , then approximately how far are you from the nearest
corner of the wall?
SOLUTION:
Let x represent the distance from the nearest corner of the wall. Basic trigonometry should reveal that at the
particular instant that θ =
π
6
rads. , x =
15
tan π6
( )
≈ 25.981 m.
b)
π
6
If your viewing angle at a particular instant is θ =
rads. and you are walking at 8
m
min
, then approximately how
fast is your viewing angle changing?
SOLUTION:
Let x represent the horizontal distance traveled by the person on the sidewalk. Thus, tan θ =
dθ
dt
15
x
. We need to find
. Observe that the wall is the reference point for the person’s motion. Thus, their displacements are
θ = π6
getting smaller as they approach the wall which in turn means that their velocities should be negative.
Method 1: We could simply differentiate the above equation implicitly with respect to time. Hence,
(sec2 θ )( ddtθ ) = − ( 15x )( dxdt ) .
2
dθ
dt
=
2
15
rads.
min
dθ
dt
Using the given information,
= −
(
15
25.9812
)(cos( )) (− 8)
π
6
2
rads.
min
which reduces to
.
Method 2: Using the inverse trigonometric function, arctangent, we can isolate for the angle θ . Hence,
( 15x ) .
θ = arctan
dθ
dt


=
 1+
(
1
which implies that
)


15 2 
25.981 
(−
15
25.9812
)(− 8)
dθ
dt
rad.
min

1

=
15
 1+ x
( )
2


 −

( )( ) . Using the given information,
dx
dt
15
x2
which reduces to
dθ
dt
=
2
15
rads.
min
.
Using either method should reveal that the viewing angle is increasing at a rate of
c)
If your viewing angle at a particular instant is θ =
π
6
2
15
rads.
min
rads. and is increasing at a rate of
π
90
.
rads.
min
, then
approximately how fast are you walking at this particular instant?
SOLUTION:
The basic formula tan θ =
15
x
does not change although we are now looking for
established in part b) should reveal that when θ =
π
6
rads. and
person is walking towards the wall at approximately 2.1
m
min.
dθ
dt
=
π
90
rads.
min
dx
dt
that
. Using either method
dx
dt
≈ − 2.094
m
min.
. Hence, the
at this particular instant.
6) One corner of a rectangle is at ( 4,0) and the opposite corner is on the graph of y = x 2 (see diagram to the right). What
are the dimensions of the rectangle that maximizes the area of the rectangle and what is the maximum area? JUSTIFY
your answer.
Hint: The base of the rectangle is NOT x.
SOLUTION:
Let the base of the rectangle be u and let the height of the rectangle be y. Clearly from the diagram, the base of the
rectangle can be written in terms of x as well as the height y can be written in x. From the diagram provided,
x + u = 4 which implies that u = 4 − x . Furthermore, y lies on the curve y = x 2 . Let A( x ) represent the area
( )
function. Thus, A( x ) = ( 4 − x ) x 2 = 4 x 2 − x 3 , x ∈[ 0,4] .
For critical points, we need A ′( x ) = 0 . Hence, A ′( x ) = 8 x − 3x 2 which equals 0 when x = 0 and x = 83 . Since the
function is continuous and differentiable over the given interval, then any absolute maximum or minimum must come
from the endpoints and/or critical points as guaranteed by the extreme value theorem.
. Hence, the maximum area of the inscribed rectangle is
( 83 ) = 256
27
Checking, A( 0) = A( 4) = 0 and A
which occurs when the rectangle has dimensions
4
3
units by
64
9
units.
256
27
square units