Related Rates
LRT
02/27/2017
#64
x2 + y 2 = 202 = 400.
dy
dx
Problem: Find
when x = 12 and
= 5 ft/sec.
dt
dt
2xx0 + 2yy 0 = 0.
We are looking for y 0 so we are done if we know x, y and x0 .
We are given x = 12 and x0 = 5 so we need y.
(12)2 + y 2 = 400; 144 + y 2 = 400; y 2 = 256; y = ±16 and
y > 0 so y = 16.
Hence 2 · 12 · 5 + 2 · 16 · y 0 = 0; 120 + 32y 0 = 0;
120
y0 = −
= −3.75.
32
And the units are? Feet per second.
#44
dx
dp
Problem: Find −
when
= −0.02 $/carton/week and
dt
dt
x = 25.
1250pp0 − 2xx0 = 0.
We still need p: 625p2 − 252 = 100; 625p2 − 625 = 100;
100
= 0.16; p2 = 1.16;
625(p2 − 1) = 100; p2 − 1 =
625
p = ±1.0770329614. Since p > 0; p = 1.0770329614
1250pp0 − 2xx0 = 0
1250 · (1.0770329614)(−0.02) − 2 · 25x0 = 0
−26.9258240357 − 50x0 = 0; −26.9258240357 = 50x0 ;
x0 = −0.5385164807.
Hence the answer is
0.5385164807
Units?
This problem is a bit of a mess since 625 or 1 (or both)
must have units so that we can do the subtraction. The
easiest solution is the units problem is to remember that
dx
always has units (units for x)/(units for t).
dt
Hence in this case the units are in (thousands of cartons)
per week.
Suppose oil spills from a underwater well blowout. If the
radius of the oil spill increases at a constant rate of 1m/s,
how fast is the area of the spill increasing when the radius
is 10km? If the slick has a constant height of 2cm, how fast
is oil escaping from the rupture?
A = πr2 and V = Ah.
dA
dr
= 2πr . We know r = 10. If we go
dt
dt
dr
with this, our units are kilometers so
= 1 · 10−3 km/sec.
dt
For the first part,
dA
Hence
= 2 · π · 10−3 ≈ 0.0062831853 km2 /sec. It is
dt
easier to keep it scientific notation,
Hence
dA
≈ 6.2831853 · 10−3 .
dt
dV
dA
dh
=
·h+A·
. Since the
dt
dt
dt
dh
slick has constant height,
= 0 so
dt
dV
≈ 6.2831853 · 10−3 · 2 · 10−5 km3 /sec =
dt
12.5663706144·10−5 km3 /sec = 1.25663706144·10−4 km3 /sec
For the second question,
#29
Find
dy
if (x + y)3 + x3 + y 3 = 0.
dx
3(x + y)2 (1 + y 0 ) + 3x2 + 3y 2 y 0 = 0;
3(x + y)2 + 3(x +
y)2 y 0 + 3x2 + 3y 2 y 0 = 0;
3(x + y)2 + 3y 2 y 0 + 3(x + y)2 + 3x2 = 0;
3(x + y)2 + 3x2
(x + y)2 + x2
y0 = −
=
−
.
3(x + y)2 + 3y 2
(x + y)2 + y 2
(x + y)3 + x3 + y 3 = 0
y0 = −
(x + y)2 + x2
(x + y)2 + y 2
(x + y)3 + x3 + y 3 = 0
(x + y)2 + x2
y0 = −
(x + y)2 + y 2
(x + y)2 + x2 0
(x + y)2 + y 2 + (x2 − y 2 )
;
y
=
−
;
(x + y)2 + y 2
(x + y)2 + y 2
x2 − y 2
;
y 0 = −1 −
(x + y)2 + y 2
y0 = −
y 0 = −1 +
y 2 − x2
(x + y)2 + y 2
(x + y)3 + x3 + y 3 = 0; (x + y)3 + (x + y)(x2 − xy + y 2 ) = 0.
If x + y 6= 0, (x + y)2 + (x2 − xy + y 2 ) = 0;
x2 + 2xy + y 2 + x2 − xy + y 2 = 0 2x2 + 2y 2 + xy = 0.
p
y ± y 2 − 16y 2
x=
4
so no real solutions if y 6= 0. Hence the only solutions occur
when y = −x so y 0 = −1.
y 0 = −1 +
The weird term
y 2 − x2
(x + y)2 + y 2
y 2 − x2
really is needed since if you
(x + y)2 + y 2
do the problem
(x + y)3 + x3 + y 3 = 10
you get the same formula for y 0 and this time the graph no
longer has constant slope.
y 0 = −1 +
y 2 − x2
(x + y)2 + y 2
Notice for large values
of x, x + y is close to 0 and the curve is close to the line
x + y = 1. When x = y = 1 the slope is also −1.
1
1
But when x = 0 the slope is y 0 = −1 + = − and when
2
2
−1
3
0
y = 0, y = −1 +
=− .
2
2