MATH 136
Minimizing the Distance
from a Point to Function
Exercises
1. Let f (x) = ln x and P = (2, 3).
(i) Show a graph of f (x) and point P .
(ii) Give the distance function d(x) from the graph of f to point P .
€ distance function and determine the minimum distance from the graph
€ (iii) Graph the
of f to point P .
(iv) Specify€the point(s) on €
the graph of f at which the minimum distance occurs.
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2. Let f (x) = x 2 and P = (0, 5).
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(i) Show a graph of f (x) and point P .
(ii) Give the distance function d(x) from the graph of f to point P .
€ (iii) Evaluate€ d ′(x) , determine the critical points, and determine which of these critical
points yields the minimum of d(x) . Give this minimum value.
(iv) Specify€the point(s) on€the graph of f at which the minimum distance occurs.
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€ and P = (–3, –1/2).
3. Let f (x) = sin(π x /2)
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(i) Show a graph of f (x) and point P .
(ii) Give the distance function d(x) from the graph of f to point P .
€
€ (iii) Graph the distance
function and determine the minimum distance from the graph
P
of f to point .
(iv) Specify€the point(s) on €
the graph of f at which the minimum distance occurs.
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Solutions
1. (ii) For y = ln x and point (2, 3), the distance function is
d (x ) = (x − 2)2 + (ln x − 3)2 , defined for x > 0,
which is graphed in Y2 after y = ln x is graphed in Y1.
€
(i) y = ln x and the
point (2, 3)
(iii) Minimizing
the dist. function
Evaluate Y1 to find
point on graph
where min. dist.
occurs
The minimum of the distance function occurs when x ≈ 2.7307312 and the minimum
distance to the point (2, 3) is approximately 2.125. (iv) The point on the graph of
y = ln x where the min. distance occurs is (2.7307312, 1.0045694).
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2. For f (x) = x 2 and P = (0, 5), the distance function is
d(x) = (x − 0) 2 + (x 2 − 5) 2 = x 2 + (x 2 − 5) 2 for all x .
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Then d ′(x) =
2x + 2(x 2 − 5)(2x)
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2 x 2 + (x 2 − 5) 2
=
x + (x 2 − 5)(2x)
2x 3 − 9x
=
.
x 2 + (x 2 − 5) 2 € x 2 + (x 2 − 5) 2
Then d ′(x) = 0 when x (2x 2 − 9) = 0 , which is at x = 0 and x = ±
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€
€
€
f (x)
€
€
3
2
€
d(x)
The minimum distance occurs at x = ± 3/ 2 ≈ ± 2.12132 . and the minimum distance is
2.17945. The points on the
d(3/ 2) = 4.75 ≈ €
€ graph where the min. dist. occurs are
( ± 3/ 2 , 4.5).
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3. (ii) For y = sin(π x / 2) and point (–3, –1/2), the distance function is
d (x ) = (x + 3)2 + (sin(π x / 2) + 0.5)2 , for all x ,
which is graphed in Y2 after y = sin(π x / 2) is graphed in Y1.
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y = sin(π x / 2) and
point (–3, –1/2)
(iii) Finding first
occurrence of the
dist. function
minimum
Evaluate Y1 to find
points on graph
where min. dist.
occurs
The minimum of the distance function occurs at two places: when x ≈ –3.9367 and
when x ≈ –2.0633. The minimum distance to the point (–3, –1/2) is approximately
1.112.
(iv) The two points on the graph of y = sin(π x / 2) where the€min. distance occurs are
€ then at (–3.9367, 0.099267) and (–2.0633, 0.099267).