MATH 180 FALL 1995

ASSIGNMENT #27

4.8, #1. Differentiating x^{2} + xy − y^{3} = xy^{2} with respect to x gives

2x + y + x \frac{dy}{dx} − 3y^{2} \frac{dy}{dx} = y^{2} + 2xy \frac{dy}{dx}.

Thus \frac{dy}{dx} = \frac{2x + y − y^{2}}{3y^{2} + 2xy − x}.

4.8, #2. Differentiating x^{2} + y^{2} = 7 with respect to x gives 2x + 2y \frac{dy}{dx} = 0 and \frac{dy}{dx} = −\frac{x}{y}.

4.8, #3. Differentiating x^{1/2} + y^{1/2} = 25 with respect to x gives

\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0

or \frac{dy}{dx} = −\begin{matrix}√ & \overline{y} \\ √ & \overline{x} \end{matrix}.

4.8, #9. Differentiating xy^{2} = 1 with respect to x gives y^{2} + 2xy \frac{dy}{dx} = 0. At the point

P = (1. − 1) on the graph of xy^{2} = 1, we have 1 − 2 \frac{dy}{dx} = 0 or dy/dx = 1/2. Thus the

tangent line is y = −1 + \frac{1}{2}(x − 1) = \frac{1}{2}x − \frac{3}{2}.

4.8, #15. Differentiating x^{2}/25 + y^{2}/9 = 1 with respect to x gives dy/dx = −\frac{9x}{25y}. This

is the slope of the tangent line to the ellipse at the point (x, y). The slope is undefined when y = 0, that is, at the points (5, 0) and (−5, 0) of the ellipse where the tangent line is vertical.

4.8, #18. Differentiating xe^{5y} = 3y with respect to x gives e^{5y} + xe^{5y} · 5 \frac{dy}{dx} = 3 · \frac{dy}{dx}. At the point (0, 0) on the graph, this relation implies dy/dx = 1/3. Thus the tangent line at (0, 0) is y = \frac{1}{3}x. In particular, when x = 0.1, y is approximately \frac{1}{3} · (0.1) = 0.03333.