2c. Let P = (x, y) be a point on the graph of y =x - the origin as a function of the point's x- coordinate. - 4. Express the distance, d, from P to function of the point's x-coordinate. S. Since P is on the graph y=x, the y-coordinate of P of can be replaced by x. Thus, P= (x, y) = (x, x³) Recall that the origin is the point with coordinates (0, 0). We use the distance formula to find the distance between (x, x') and (0, 0). d = J(x - 0)² +(x³ - 0)² = /(x)² +(x³)² = x² +x° %3D Now write as a function. .2 d(x) = Vx +x° or d(x)= \x° +x²

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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Section 1.10
2c. Let P= (x, y) be a point on the graph of
y =x - 4. Express the distance, d, from P to
the origin as a function of the point's x-
coordinate.
function of the point's x-coordinate.
Since P is on the graph of y =x, the y-coordinate of P
can be replaced by x’. Thus,
P = (x, y) = (x, x³)
Recall that the origin is the point with coordinates (0, 0).
We use the distance formula to find the distance between
(x, x') and (0, 0).
d = (x - 0)² +(x³ - o)² = /cx)² +(x³y² = ft o
%3D
Now write as a function.
d(x) = Vx² +x° or d(x)= Vx° +x?
%3D
n narentheses):
Transcribed Image Text:Section 1.10 2c. Let P= (x, y) be a point on the graph of y =x - 4. Express the distance, d, from P to the origin as a function of the point's x- coordinate. function of the point's x-coordinate. Since P is on the graph of y =x, the y-coordinate of P can be replaced by x’. Thus, P = (x, y) = (x, x³) Recall that the origin is the point with coordinates (0, 0). We use the distance formula to find the distance between (x, x') and (0, 0). d = (x - 0)² +(x³ - o)² = /cx)² +(x³y² = ft o %3D Now write as a function. d(x) = Vx² +x° or d(x)= Vx° +x? %3D n narentheses):
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