PRACTICE PROBLEMS: SET 3
MATH 100: PROF. DRAGOS GHIOCA
1. Problems
( 11π ))
Problem 1. Find arcsin sin 3 .
(
Problem 2. If
y 3 − x2 + 2xy 2 = 1
find y ′′ at x = 0.
Problem 3. A roast turkey is taken from an oven when its temperature has
reached 75 degrees Celsius and is placed on a table in a room where the temperature
is 25 degrees Celsius. If the turkey is cooling at the rate of 1 degree per minute
at the moment when it reached 50 degrees Celsius, determine when will the turkey
reach the temperature of 40 degrees.
2
Problem 4. Find the inverse function for f (x) = 3x where f : [0, +∞) −→
[1, +∞).
(
( 11π ))
2. Solutions.
is an angle θ with the properties:
Problem 1. arcsin sin 3
• − π2 ≤ θ ≤ π2(; and)
• sin(θ) = sin 11π
3 .
We know that sin(x) is a function periodic of period 2π; so,
(
)
(
)
(
)
(
)
11π
11π
11π
−π
sin
= sin
− 2π = sin
− 4π = sin
.
3
3
3
3
11π
So, θ = − π3 . Note that[11π
3 or ] 3 − 2π =
π π
are not in the interval − 2 , 2 .
5π
3
are not the correct answer since they
Problem 2. First we compute the value of y when x = 0. So, substituting x = 0
in the equation of the curve yields
y 3 = 1 and so, y = 1.
Then we differentiate both sides of the equation of the curve with respect to x –
this is the method of implicit differentiation for computing y ′ . So,
3y 2 y ′ − 2x + 2y 2 + 2x · 2yy ′ = 0.
Thus
3y 2 y ′ + 4xyy ′ = 2x − 2y 2 ,
1
2
MATH 100: PROF. DRAGOS GHIOCA
and therefore (after factoring out y ′ on the left hand side) we get:
2x − 2y 2
.
3y 2 + 4xy
So, substituting x = 0 and y = 1 yields
−2
2
y ′ (0) =
=− .
3
3
In order to compute y ′′ (0) we first find y ′′ using again implicit differentiation for
the equation:
3y 2 y ′ − 2x + 2y 2 + 4xyy ′ = 0.
We obtain
y′ =
6y · y ′ · y ′ + 3y 2 y ′′ − 2 + 4yy ′ + 4yy ′ + 4x(y ′ · y ′ + yy ′′ ) = 0.
Next we substitute x = 0, y = 1 and y ′ = − 23 . We get
(
)
(
)
4
2
2
′′
6 · + 3y − 2 + 4 · −
+4· −
+ 0 = 0.
9
3
3
In conclusion,
8
16
+ 3y ′′ (0) − 2 −
= 0,
3
3
and thus
y ′′ (0) =
14
.
9
Problem 3. We apply the Newton’s Law of Cooling for the function
f (t) = T (t) − 25,
where T (t) is the temperature of the turkey measured after t minutes. Since
dT
= k(T − 25),
dt
and f ′ (t) =
dT
dt
, we get
f ′ (t) = kf (t).
Hence
f (t) = f (0)ekt .
However, f (0) = T (0) − 25 = 75 − 25 = 50. So,
f (t) = 50ekt .
We find the constant k by using the information that the turkey is cooling at the
rate of 1 degree per minute at the time when its temperature is 50 degrees. So, at
the time t when
f (t) = 50 − 25 = 25,
we also know that
f ′ (t) = −1.
Note that f ′ is negative since the turkey is cooling. So, using that
f ′ (t) = kf (t),
we get
−1 = k · 25,
PRACTICE PROBLEMS: SET 3
and thus
k=−
3
1
.
25
Finally, we are asked to find t such that
f (t) = 40 − 25 = 15.
So, we need to solve the equation:
15 = 50e− 25 ·t ,
1
or
t
15
= e− 25 .
50
We take the natural logarithm of both sides and compute:
( )
50
t
− ln
=− ,
15
25
and so,
( )
10
t = 25 ln
minutes.
3
Problem 4. We have that
2
y = 3x
and now we solve for x in terms of y in order to find the inverse function. So, we
have (after taking logarithms of both sides)
log3 (y) = x2 .
Then we take square-roots and obtain
x=
√
log3 (y).
So, the inverse function for f (x) is the function g : [1, +∞) −→ [0, +∞) given by
the formula
√
f (x) = log3 (x).