y
1
x
=
=
=
=
y
=
y
=
𝑑𝑦
𝑑𝑡
x2
𝑑𝑥
2x
𝑑𝑡
2x
If
𝑑𝑦
𝑑𝑡
=
𝑑𝑥
𝑑𝑡
1
2
1
( )2
2
1
4
x
200
h
ℎ
tan 200
=
h
=
x tan(200)
=
tan(200)
=
tan(200) (10)
=
3.6 km/hr
𝑑ℎ
𝑑𝑡
𝑑ℎ
𝑑𝑡
𝑑ℎ
𝑑𝑡
𝑥
𝑑𝑥
𝑑𝑡
19.(a)
Since when t = 250 minutes we only need concern ourselves with the triangle shaped part of
the pool
50 m
2m
b
Using similar triangles
50
2
𝑏
ℎ
=
b =
So
Volume of tank at height h will be
After 250 min V = 250 m3
V
=
V
V
=
=
19.(b)
(𝑏𝑎𝑠𝑒)(ℎ𝑒𝑖𝑔ℎ𝑡)(20)
(25ℎ)(ℎ)(20)
250ℎ2
2
250ℎ2
250ℎ2
h
V
250ℎ2
500ℎ𝑑ℎ
𝑑𝑡
500(1)𝑑ℎ
𝑑𝑡
𝑑𝑡
500
2
1
25h
V =
250 =
1 =
𝑑𝑉
1
1
=
=
1
=
=
𝑑ℎ
𝑑𝑡
h
Volume of pool = 2000 m3 so it will take 2000 min to fill the pool at the rate of 1 m3 per min
13
y
x
x2 + y2
Formula connecting x and y is
Differentiating with respect to t
When x = 5 , y = 12 and
𝑑𝑥
𝑑𝑡
= 0.5
2𝑥
𝑑𝑥
𝑑𝑡
+ 2𝑦
𝑑𝑡
𝑑𝑦
2(5)(0.5) + 2(12)
5 + 24
𝑑𝑦
𝑑𝑡
𝑑𝑦
24
𝑑𝑡
𝑑𝑦
𝑑𝑡
𝑑𝑦
𝑑𝑡
=
132
=
0
=
0
=
0
=
–5
=
−
5
24
ft/sec
We are told
𝑑𝑥
𝑑𝑡
= 8 ft/sec and that x = 15 ft
We are being asked to find
𝑑𝑦
𝑑𝑡
at this point.
Using similar triangles we find a formula using x,y
20
20 ft
5
5 ft
y
x
=
𝑥+𝑦
𝑦
20y =
5x + 5y
15y =
5x
y
𝑑𝑦
𝑑𝑡
𝑑𝑦
𝑑𝑡
𝑑𝑦
𝑑𝑡
=
=
=
=
1
3
𝑥
1 𝑑𝑥
3 𝑑𝑡
1
3
8
3
Differentiating with respect to t
(8)
𝑓𝑡/𝑠
z
y
𝑑𝑥
𝑑𝑦
𝑑𝑧
x
We know = 58.76 , = 10 and we are being asked to find
when t = 10 sec.
𝑑𝑡
𝑑𝑡
𝑑𝑡
2
2
2
Formula connecting x,y and z is
z = x +y
At t = 10 seconds x =10(58.67) = 586.7 ft and y = 150 + 100ft =250 ft
z=√586.72 + 2502 = 637.7
z2
2𝑧
𝑑𝑧
𝑑𝑡
2(637.7)
1275.4
𝑑𝑧
𝑑𝑡
𝑑𝑦
𝑑𝑡
𝑑𝑦
𝑑𝑡
=
=
x2 + y 2
𝑑𝑥
𝑑𝑦
2𝑥 + 2𝑦
=
2(586.7)(58.67) + 2(250)(10)
=
73,843.4
=
57.9 𝑓𝑡/𝑠𝑒𝑐
𝑑𝑡
𝑑𝑡
z
y
x
We know
𝑑𝑥
𝑑𝑡
= 500 𝑚𝑝ℎ ,
we are being asked to find
𝑑𝑦
𝑑𝑡
𝑑𝑧
𝑑𝑡
= 550 𝑚𝑝ℎ and
when y is travelling for t = 2.5 hours and x for 1.5 hours
Formula connecting x,y and z is
z2
2𝑧
𝑑𝑧
𝑑𝑡
2(1566.2)
2995.6
𝑑𝑧
𝑑𝑡
𝑑𝑧
𝑑𝑡
𝑑𝑧
𝑑𝑡
z2
=
x2 + y 2
x
=
(1.5)(500) = 750 miles
y
=
(2.5)(550) = 1375 miles
z
=
√7502 + 13752 = 1566.2 miles
=
x2 + y 2
=
2𝑥
=
2(750)(500) + 2(1375)(550)
=
2,262,500
=
722 𝑚𝑝ℎ (approx.)
𝑑𝑥
𝑑𝑡
+ 2𝑦
𝑑𝑦
𝑑𝑡