Derivatives, Equation of the Line

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

y = -x/3 + 10/3

or 3y = -x +10

or

x+3y =10 is the line tangent to y =sqr(10-x^2) at the point (1,3), perpendicular to the line y =3x-10

y=sqr(10-x^2) is the top half of a circle with equation x^2+y^2 = 10

the slope of a line tangent to the semi-cirle = -x/y

at the point (1, 3) the tangent line has slope =-x/y = -(1)/3 = -1/3

y =(-1/3)x + 10/3 has slope = -1/3 which is the negative inverse of the slope of y=3x-10 which has slope = the coefficient of the x term = 3. 3 and -1/3 are negative inverses

to be perpendicular, a line must have a negative inverse slope

it helps to graph the semi-circle. It's the top half of a circle with center at the origin and radius =sqr10. The point of tangency is a little to the right side of the top half of the circle