# Math 226 Quiz #5 Sample Problems

```Math 226
Quiz #5 Sample Problems
1. A stone dropped into a still pond sends out a circular ripple whose radius increases at a constant
rate of 3 ft/s. At time t = 0, the radius is 0 feet. How rapidly is the area enclosed by the ripple
increasing at time t = 10 seconds?
dr
, since it’s the rate at which the radius is increasing. We want to find the rate at
dt
dA
, so we want a relationship between A and r. We can use the fact
which the area is increasing,
dt
2
that A = πr , since the ripple is circular. Thus, we know
3 ft/s is our
A =
dA
dt
πr2
=
2πr
dr
dt
So now the last piece of the puzzle will be finding r. Now since we are concerned with what
happens at the end of 10 seconds, we know that the radius will be 3 ft/s ·10 s = 30 ft at t = 10 s.
So, we have
dA
dt
= 2π(30)(3)
dA
dt
= 180π ft2 /s ≈ 565.5 ft2 /s
2. A spherical balloon is to be deflated so that its radius decreases at a constant rate of 15 cm/min.
At what rate is air being removed when the radius is 9 cm?
dr
dV
= −15 cm/min, and we want to find
when r = 9 cm. The relationship we
dt
dt
have between r and V is:
Here, we have
=
4
dV
dt
=
4
dV
dt
=
4πr2
=
4π (9) (−15)
=
−4860π cm3 /min
V
3
3
πr3
π 3r
2 dr
!
dt
dr
dt
2
3. A 17 ft ladder is leaning against a wall. If the bottom of the ladder is pulled along the ground
away from the wall at a constant rate of 5 ft/s, how fast is the top of the ladder moving down the
wall when it is 8 ft above the ground?
dy
, when y = 8 ft. Thus we need a relationship
dt
between x and y. Since there is a right triangle, we can use the pythagorean theorem, and then
differentiate with respect to t. So,
Using the picture above, we are trying to find
x2 + y2
2x
dy
dx
+ 2y
dt
dt
dy
dt
= 172
= 0
=
−x dx
y dt
Now, since we have that y = 8 ft, we can solve for x. Using the pythagorean theorem, we have
√
x = 172 − 82 = 15. So,
dy −15
75
=
·5=−
ft/s
dt
8
8
2
4. Water is being poured into a conical water tank so that the height of the water is increasing at
a constant rate of 3 ft/min. If the top of the conical tank has a radius of 12 ft, and the height of
the tank is 28 ft, find the rate at which water is being poured into the tank, in terms of the height
h.
dV
1
, and the relationship we have between V, r, h, is that V = πr2 h. In
dt
3
dh
this instance, we want only r or h. Since we are given that
= 3 ft/min, we would rather use h.
dt
Thus, using the similar triangle argument, for the ratio of the radius of the height of the water at
any time is
r
12
3
3
=
=
⇒ r= h
h 28 7
7
We are trying to find
Now we can substitute this value for r and differentiate with respect to t. Sew,
V
=
=
dV
dt
=
=
dV
dt
=
1
3
π
3π
49
9π
49
3
7
2
h h
h3
h2
9π 49
27π
49
dh
dt
h2 (3)
h2 ft3 /min
(Notice that our answer is in terms of h, as requested in the question)
3
5. Find the local linear approximation of f at x0 = 1.
a) f (x) =
1
2+x
Noting that f 0 (x) =
−1
, we have
(2 + x)2
f (x) ≈
=
=
=
f (x0 ) + f 0 (x0 ) (x − x0 )
1
2 + x0
1
2+1
1
−
3
1
9
−
+
1
(2 + x 0 )2
(x − x0 )
−1
(x − 1)
(2 + 1)2
(x − 1)
b) f (x) = (x + 4)3
Noting that f 0 (x) = 3(x + 4)2 , we have
f (x) ≈
(x0 + 4)3 + 3 (x0 + 4)2 (x − x0 )
=
(1 + 4)3 + 3(1 + 4)2 (x − 1)
=
125 + 75(x − 1)
c) f (x) = tan−1 (x)
Here, f 0 (x) =
1
1 + x2
. Hence,
f (x) ≈
=
=
tan−1 (x0 ) +
tan−1 (1) +
π
4
+
1
2
1
1 + x20
1
1 + 12
(x − 1)
4
(x − x0 )
(x − 1)
6. Use an appropriate local linear approximation to estimate the value of the given quantity.
√
a) 65
√
Here, we should use the function f (x) = x, and we should use x0 = 64, since we’ll get a nice
number there. So, note that f 0 (x) =
1
2
√ . Using the formula for approximating f (x) near x0 = 64,
x
we have
√
f (x) ≈
64 +
= 8+
1
16
!
1
√
2
64
(x − 64)
(x − 64)
Now, since 65 is close to 64, we can conclude that
f (65) ≈ 8 +
1
16
(65 − 64) =
129
16
b) 1.983
This time, we are going to use the function f (x) = x3 , where x0 = 2. Note that f 0 (x) = 3x2 , so
we have
Now, we can conclude that
f (x) ≈
(2)3 + 3(2)2 (x − 2)
f (x) ≈
8 + 12(x − 2)
f (1.98) ≈ 8 + 12(1.98 − 2) = 7.76
c) ln(1.01)
This time, our function will be f (x) = ln x, and we will use x0 = 1. f 0 (x) =
f (x) ≈
ln(1) +
f (x) ≈
x−1
1
1
(x − 1)
Therefore f (1.01) ≈ 1.01 − 1 = .01.
7. Find
dy
dx
a) y = ex (x3 + 5)
dy
= ex x3 + 5 + 3x2 ex
dx
5
1
x
. Thus,
√
b) y = sin−1 ( x)
dy
= q
dx
1
!
1
1
√ 2 2 √x = p
2 x(1 − x)
1−
x
c) y = ln (cos (ex ))
dy
1
(− sin (ex )) (ex ) = −ex tan (ex )
=
dx cos (ex )
d) y =
p
cos−1 (x2 )



dy 1 −1 2 −1/2 
x
1
 (2x) = − p

=
cos x
− p

2
−
1
dx 2
cos (x2 ) (1 − x4 )
1 − (x2 )
e) y = etan(x) (1 + x2 )
dy
= etan(x) sec2 (x) · 1 + x2 + 2x etan(x) = etan(x) 2x + sec2 (x) + x2 sec2 (x)
dx
f) y =
1
π
tan−1 (2x)
!
dy
1
1
2
(2) =
=
dx π 1 + (2x)2
π (1 + 4x2 )
8. Find a point on the graph of y = e3x at which the tangent line passes through the origin.
Here, we have that
dy
= 3e3x . So we want the (x, y) where y − 0 = 3e3x (x − 0), or y = 3xe3x .
dx
Now, we have y = e3x and y = 3xe3x , telling us that
3xe3x
=
e3x
3x
=
0
e (3x − 1) =
0
=
1
3xe
3x
−e
3x
x
3
Notice that e3x is never equal to zero, so the point is when x = 1/3, which means y = e3(1/3) = e.
So,
1
(x, y) = , e
3
6
```