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EXAMPLE 1 (a) If x2 + y? = 169, find dy, (b) Find an equation of the tangent to the circle x2 + y? = 169 at the point (5, 12). SOLUTION 1 (a) Differentiating both sides of the equation x2 + y? = 169: )= (169 ) ) + (x 2) = 0. Remembering that y is a function of x and using the Chain Rule, we have jux = 2v Thus 2x + 2y EX = 0. Now we solve this equation for dy/dx: dy (b) At the point (5, 12) we have x = 5 and y = 12, so dy = -5/12 An equation of the tangent to the circle at (5, 12) is therefore y - 12 = -5/12 ( x - 5 ) or SOLUTION 2 (b) Solving the equation x2 + y? = 169, we get y = ty 169 - x2. The point (5, 12) lies on the upper semicircle y = v 169 - x2 and so we consider the function f(x) = v 169 - x2. Differentiating Fusing the Chain Rule, we have F'(x) = 1(169 - x2)-1/2-9 169 -x2 = 1(169 - x2)-1/2( -2x -x( 169 + x2) ( ?) S '(5) = 12 and, as in Solution 1, an equation of the tangent isEXAMPLE 2 (a) Find y' if x3 + y3 = 6xy. (b) Find the tangent to the folium of Descartes x + y = 6xy at the point (3, 3). (3, 3) (c) At what point in the first quadrant is the tangent line horizontal? SOLUTION (a) Differentiating both sides of x3 + y = 6xy with respect to x, regarding y as a function of x, and using the Chain Rule on the y term and the Product Rule on the 6xy term, we get 3 x 2 + 3( 1 2 or x 2 + 32 y' = 2( x Jy' + 2 3 We now solve for y': y ? y' - 2 xy' = zy - x2 V2 - 2x Jy' = 2y - x2 2y - x2 y' = v2 - 2x (b) When x = y = 3, y' = T 3 -1 and a glance at the top figure to the left confirms that this is a reasonable value for the slope at (3, 3). So an equation of the tangent to the folium at (3, 3) is y - 3 v = -1 ( x - 3 0 ) y= -x+6 Video Example () (c) The tangent line is horizontal if y' = 0 . Using the expression for y' from part (a), we see that y' = 0 when 2y -x- = 0 (provided that y? - 2x = 0). Substituting y = =x2 in the equation of the curve, we get x3 . which simplifies to x6 1673 Since x # 0 in the first quadrant, we have x3 = 16 v . If x = 16 ) -2 2 . then y = Thus the tangent is horizontal at (0, 0) and at (x, y) = ( ). Looking at the bottom figure to the left, we see that our answer is reasonable. X XWe now have -[x5 + )4] = 5x4 + 4yBy': [5] = 0. dx dx Rearranging this, we get 4 Solving for y, we conclude that the derivative is X