Derivatives, Differentiation

Question

Find the equation of the line tangent to y = sqrt of (10 - x^2 ) and perpendicular to the line y = 3x -10

Answer 1

2 tangent lines

y = -x/3 +10/3

and

y = -x/3 -10/3

or

x+3y = 10 and x+3y =-10

one in quadrant I, and one in quadrant III

tangent line to circle y = sqr(10-x^2)

and perpendicular to y=3x+10

has slope = negative inverse = -1/3 = y'= 1/sqr(10-x^2)

sqr(10-x^2) = -3

square both sides

10-x^2 = 9

x^2 = 1

x =+/-1

y = +/-3

slope = -1/3

in point slope form

tangent line is

y+3 = (-1/3)(x+1) and 2nd Tangent line

y -3 = (-1/3)(x-1)

it helps to draw a sketch of the graph

Answer 2

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