1. The combined electrical resistance of two resistances, R
1 is, in ohms,
1
R
=
1
R
1
+
1
R
2
.
and R
2
, connected in parallel
(1)
If R
1 and R
2 are increasing at rates 0 at which R is changing when R
1
.
5 and 0
= 100 and R
2
.
4 ohms/sec, respectively, find the rate
= 200.
solution: Take derivatives, using the chain rule:
1
−
R 2 dR dt
1
= −
R 2
1 dR dt
1
1
+ −
R 2
2 dR
2
.
dt
(2)
We are given that dR
1
/dt = 0 .
5 and dR
2
/dt = 0 .
4 and asked to compute dR/dt when
R
1
= 100 and R
2
= 200. For these values of R
1 and R
2
, equation (1) yields
R =
1
100
1
+
1
200
=
200
,
3 from which equation (2) gives
1
−
(200 / 3) 2 dR dt
1
= −
(100) 2
1
(0 .
5) + −
(200) 2
(0 .
4) .
Solving for dR/dt , we get dR
= dt
4
15
.
2. A light house is located on a small island 3 km away from the nearest point P on a straight shoreline. The light makes 4 revolutions/minute. How fast is the beam moving along the shore when the point at which the beam meets the shore is 1 km away from P ?
solution:
Here is a picture:
3
P
θ x
1

We see that tan θ = x
,
3
We are given that dθ/dt = 4(2 π ) = 8 π rad / min, and are asked to compute dx/dt when x = 1. Differentiating the above equation, we find
(sec
2
θ ) dθ dt
=
1
3 dx
.
dt
When x
√
= 1, the hypotenuse of the triangle in the picture is sec θ = 10 / 3. Plugging into the above equation, we find
√
3 2 + 1 2 =
√
10, so
√
10
!
2
(8 π ) =
3
1
3 dx
.
dt
Thus, when x = 1, we have dx/dt = 80 π/ 3 ≈ 83 .
8 km / min, moving towards P , at least in our interpretation. Another possibility is that the light is rotating away from point P , in which case the point on shore is moving with the same speed but away from P .
3. If the rate at which a snowball melts is proportional to its surface area, what can you say about the rate of change of the snowball’s radius?
solution: The volume of a sphere is V = (4 / 3) πr
3
, where r is the radius. Therefore, dV dt
= 4 πr
2 dr dt
.
We are given that dV /dt is proportional to the surface area. So there is a constant C such that dV
= C (4 πr
2
) .
dt
Combining the two displayed equations and solving for dr/dt gives dr
= C.
dt
So the radius decreases at a constant rate.
4. There are 50 apple trees in an orchard. Each tree produces 800 apples. For each additional tree planted in the orchard, the output per tree drops by 10 apples. How many trees should be added to the existing orchard in order to maximize the total output of trees?
solution: The formula for the total output of apples given t additional trees is a ( t ) = (50 + t )(800 − 10 t ) .
2

We will worry about the interval for t in a moment. To find the local extrema, set the derivative equal to 0: a
0
( t ) = (800 − 10 t ) − 10(50 + t ) = 300 − 20 t = 0 ⇒ t = 15 .
We may assume that − 50 ≤ t ≤ 80 since a ( t ) is clearly negative otherwise. On this closed interval, the extreme value theorem guarantees a minimum and a maximum.
The extrema occur either at the endpoints of the interval or at a point strictly inside the interval where the derivative is 0. Since the value of a at the endpoints is 0, the maximum is a (15) = 42250.
5. Find the point ( x, y ) on the graph of y =
√ x nearest the point (4 , 0).
solution: The square distance between (4 , 0) and an arbitrary point ( x, graph is s ( x ) = ( x − 4)
2
+ (
√ x )
2
= x
2 − 7 x + 16 .
√ x ) on the
Minimizing the square of the distance will minimize the distance. There is one critical point: s
0
( x ) = 2 x − 7 = 0 ⇒ x =
7
2
.
To assure this gives an absolute minimum, we apply the extreme value theorem. Choose a huge number L and consider the closed interval I = [ − L, L ]. By the extreme value theorem, we know that if we restrict s to the interval I it has an absolute minimum. We know s ( ± L ) is very large, so the minimum must occur at the critical point, x = 7 / 2.
Outside of I , the function s is also very large. This x = 7 / 2 gives the global minimum on the whole real number line.
3