Answer to Question #166974 in Calculus for Hiba

Question

i) Find all points on the curve x(x + y2) = y where the tangent line is parallel to the x-axis.

ii) Find all points on the curve x(x + y2) = y where the tangent line is parallel to the y-axis

Answer 1

i) Use implicit differentiation, then set dy dx = 0.

$\dfrac{d}{dx}(x(x+y^2))=\dfrac{dy}{dx}$

$2x+y^2+2xy\dfrac{dy}{dx}=\dfrac{dy}{dx}$

$\dfrac{dy}{dx}=\dfrac{2x+y^2}{1-2xy}$

$\dfrac{dy}{dx}=0=>\dfrac{2x+y^2}{1-2xy}=0$

$2x+y^2=0, 1-2xy\not=0$

$x=-\dfrac{y^2}{2}$

Substitute

$-\dfrac{y^2}{2}(-\dfrac{y^2}{2}+y^2)=y$

$-\dfrac{y^4}{4}=y$

$y_1=0, x_1=0, Point(0, 0)$

$y^3=-4$

$y_2=-\sqrt[3]{4}, x_2=-\sqrt[3]{2}, Point (-\sqrt[3]{2}, -\sqrt[3]{4})$

$Point(0, 0), Point (-\sqrt[3]{2}, -\sqrt[3]{4})$

ii) ) The points with vertical tangents are those where the denominator of $\dfrac{dy}{dx}$ is zero (making the slope undefined). From part (i), we have

$\dfrac{dy}{dx}=\dfrac{2x+y^2}{1-2xy}=>1-2xy=0$

$x=\dfrac{1}{2y}, y\not=0$

Substitute

$\dfrac{1}{2y}(\dfrac{1}{2y}+y^2)=y$

$\dfrac{1}{4y^2}=\dfrac{y}{2}$

$y=\dfrac{\sqrt[3]{4}}{2}$

$x=\dfrac{\sqrt[3]{2}}{2}$

$Point(\dfrac{\sqrt[3]{2}}{2}, \dfrac{\sqrt[3]{4}}{2})$