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Find the point on the curve y = √ x where the tangent line is parallel to the line y = x 8 .

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If tangent to the curve y = √x is parallel to the line y = 8x, then this implies that the tangent to y = √x has the same slope as the line y = 8x. In other words, the derivative (slope) function of y = √x is equal to the slope of the line y = 8x, which is m = 8. Hence y' = 8 once we find y'

y = √x = x^(1/2)

Applying the power rule and simplifying, we find that the derivative is

y' = 1/(2
√x)

Now remember that y' must equal 8

1/(2√x) = 8

Multiplying both sides by 2
√x, we obtain

1 = 16
√x

Dividing both sides by 16, yields

√x = 1/16

But wait a minute, √x = y. Thus 1/16 must be the y-coordinate of the point at which the tangent to y = √x is drawn.

Squaring both sides, yields

x = 1/256

This is the x-coordinate of the point on the curve where the tangent is drawn.

∴ the required point must be (1/256, 1/16)

GOOD LUCK!!!

User Shehan Ekanayake
by
4.4k points
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