Solution 1.43 dy (a) Let be the gradient at a fixed point (x, y) on the curve, with x ≥ 0 and y ≥ 0. Then the equation of the tangent line at this point is da dy Y-y=(X-x), da where X and Y vary along the line. When X = 0, we have Y=y- dz' and the given condition then becomes dy dx 2 dy da +1=y-x- How Come??

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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Solution 1.43
dy
(a) Let be the gradient at a fixed point (x, y) on the curve, with x 20 and
y > 0. Then the equation of the tangent line at this point is
Y-y=
dy
dx
-(X-x),
where X and Y vary along the line. When X = 0, we have
du
Y=y-
dx
and the given condition then becomes.
dy
dx
²
1=y=x
dy
dx
Come??
How Come
(a) Find the family of curves for which the length of the part of the tangent between
the point of contact (x, y), in the first quadrant, and the y-axis is equal to the
y-coordinate of the intercept of this tangent with the y-axis. (In other words,
find curves where the lengths OA and AP shown in Figure 1.10 are equal.)
Figure 1.10
میرا
(x, y)
x
Transcribed Image Text:Solution 1.43 dy (a) Let be the gradient at a fixed point (x, y) on the curve, with x 20 and y > 0. Then the equation of the tangent line at this point is Y-y= dy dx -(X-x), where X and Y vary along the line. When X = 0, we have du Y=y- dx and the given condition then becomes. dy dx ² 1=y=x dy dx Come?? How Come (a) Find the family of curves for which the length of the part of the tangent between the point of contact (x, y), in the first quadrant, and the y-axis is equal to the y-coordinate of the intercept of this tangent with the y-axis. (In other words, find curves where the lengths OA and AP shown in Figure 1.10 are equal.) Figure 1.10 میرا (x, y) x
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