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Calculus/Tangent Lines with Specified Slopes

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Expert: - 10/14/2009


Question
Hi there Scotto, I am having trouble with the following question:

Find an equation of the straight line having slope 1/4 that is tangent to the curve y = sqrt(x).

So far I got this figured out:

1/4 = m = lim h->0 (sqrt(x+h) - sqrt(x)) / h

Answer
t = √x, so y' = 1/(2√x).

This means that at y' = 1/4, 1/4 = 1/(2√x), so 4 = 2√x, so 2 = √x, so x = 4.

At x=4, y = √4 = 2.

The point-slope form of a line is then y-y0 = m(x-x0).  We know x0, y0, and m, so
y - 2 = 0.25(x-4), which is one form of the line.

A more standard equation can be found by carrying out the multiplication and
adding 2 to both sides.  This gives y = 0.25x + 1.

Calculus

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