(a) If x2 + y2 = 400, find dy/dx. (b) Find an equation of the tangent to the cir

Question

(a) If x2 + y2 = 400, find dy/dx. (b) Find an equation of the tangent to the circle x^2 + y^2 = 400 at the point (16, 12). (a) Differentiating both sides of the equation x^2 + y^2 = 400: d/dx(x^2 = y^2) = d/dx (400) d/dx(x^2) + d/dx(y^2) = 0. Remembering that y is a function of x and using the Chain Rule, we have d/dx(y^2) = d/dy(___) dy/dx = _____ dy/dx. Thus 2x + _____ dy/dx = 0. Now we solve this equation for dy/dx: dy/dx = -x/y. (b) At the point (16, 12) we have x = 16 and y = 12, so dy/dx = An equation of the tangent to the circle at (16, 12) is therefore y - = (x -) or ______. (b) Solving the equation x^2 + y^2 = 400, we get y = plusminus Squareroot 400 - x^2. The Point (16, 12) lies on the upper semicircle y = Squareroot 400 - x^2 and so we consider the function f(x) = Squareroot 400 - x^2. Differentiating f using the Chain Rule, we have f'(x) = 1/2 (400 - x^2)^-1/2 d/dx(____) = 1/2 (400 - x^2)^-1/2 (____) = ____. so f'(16) = ____ and, as in Solution 1, an equation of the tangent is ____.

Explanation / Answer

( 1)

( a )

given x^2 +y^2 = 400

apply differentiation w.r to ' x'

d/dx( x^2 + y^2 ) = d/dx(400)

2x + 2y dy/dx = 0

2ydy/dx = - 2x

ydy/dx = -x

dy/dx = - x/y

( b) given point ( 16, 12) , x= 16 ,y = 12

slope m = dy/dx at point ( 16 ,12)

dy/dx = - x/y ==> -16/12 ==> - 4/3

slope-point formula

y - y1 = m ( x - x1 )

y - 12 = ( -4/3) ( x - 16 )

3y - 36 = - 4x + 64

3y = - 4x + 100

y = ( -4/3)x + 33.3333

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