(3) Differentiating both sides of
[tex]2x^{3/2} + y^{3/2} = 29[/tex]
with respect to x gives
[tex]3x^{1/2} + \dfrac32 y^{1/2} \dfrac{\mathrm dy}{\mathrm dx} = 0[/tex]
Solve for dy/dx :
[tex]\dfrac32 y^{1/2} \dfrac{\mathrm dy}{\mathrm dx} = -3x^{1/2} \\\\ \dfrac{\mathrm dy}{\mathrm dx} = \dfrac{-3x^{1/2}}{\frac32y^{1/2}} = \dfrac{-2x^{1/2}}{y^{1/2}} = -2\sqrt{\dfrac xy}[/tex]
Then the slope of the tangent line to the curve at (1, 9) is
[tex]\dfrac{\mathrm dy}{\mathrm dx} = -2\sqrt{\dfrac19} = -\dfrac23[/tex]
The equation of the tangent line would then be
y - 9 = -2/3 (x - 1) ==> y = -2/3 x + 29/3
(4) The slope of the tangent line to
[tex]y=\dfrac{ax+1}{x-2}[/tex]
at a point (x, y) on the curve is
[tex]\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{a(x-2)-(ax+1)}{(x-2)^2} = -\dfrac{2a+1}{(x-2)^2}[/tex]
When x = -1, we have a slope of 2/3, so
-(2a + 1)/(-1 - 2)² = 2/3
Solve for a :
-(2a + 1)/9 = 2/3
2a + 1 = -18/3 = -6
2a = -7
a = -7/2
