Using Differentials Problem: Can't Separate x and y

Question

I've been asked to estimate a y coordinate by using differentials. This normally isn't overly difficult, however, I'm not sure what to do in a case like this when y cannot be separated and used as a function of x. Can anyone point me in the right direction? I suspect I'll have to use implicit differentiation but I can't quite formulate how I'd approach it.

Solve for the y coordinate of point P near (1,2) on the curve $2x^3 + 2y^3 = 9xy$, given that the x-coordinate of P is 1.1

Answer 1

Use the Leibniz formula, namely $d(fg)=gdf+fdg$. We have $$d(2x^3+2y^3-9xy)=6x^2dx+6y^2dy-d(9xy)=6x^2dx+6y^2dy-9ydx-9xdy=(6x^2-9y)dx+(6y^2-9x)dy$$ so since this is equal to $0$ we have $$(6x^2-9y)dx=(9x-6y^2)dy$$ hence $$dy=\frac{6x^2-9y}{9x-6y^2}dx$$

Using the standard routine in assuming $dy\approx \Delta y$, we have $$x=1.1,y\approx 2+0.1\frac{6(1)-9(2)}{9(1)-6(4)}=2.08$$ Maple says that the real answer is $\approx 2.076$.

Answer 2

Make the assumption that $x=x(t)$ and $y=y(t)$. We don't care what those functions are, it just allows us to differentiate with respect to $t$. We do this because of this definition: $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$ So your equation looks like this: $$2x(t)^3+2y(t)^3=9x(t)\cdot y(t)$$ Now differentiate both sides with respect to $t$. $$\frac{d}{dt} \left(2x(t)^3+2y(t)^3 \right)=\frac{d}{dt} \left(9x(t)\cdot y(t) \right)$$ Apply power and product rules. I am dropping the $(t)$ for clarity: $$6x^2 \cdot \frac{dx}{dt}+6y^2 \cdot \frac{dy}{dt}=9 \left(\frac{dx}{dt}\cdot y + x \cdot \frac{dy}{dt} \right)$$ Notice how we have a $\frac{dx}{dt}$ or $\frac{dy}{dt}$ on every term. We now can separate $\frac{dx}{dt}$ to one side, and $\frac{dy}{dt}$ to the other: $$6x^2 \cdot \frac{dx}{dt}-9\frac{dx}{dt}\cdot y= 9x \cdot \frac{dy}{dt}-6y^2 \cdot \frac{dy}{dt}$$ Now take the ratio to obtain the slope. $$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{9x-6y^2}{6x^2-9y}=\frac{dy}{dx}$$ Now plug into the slope formula to obtain the slope. $$\frac{dy}{dx}=\frac{9*1-6*2^2}{6*1^2-9*2}=\frac{-15}{-3}=5$$ As you know, the point-slope formula of the line is: $$y=mx+b$$ In this case, $m=\frac{dy}{dx}=5$. Take the given point to obtain the constant $b$. $$2=5*1+b \rightarrow b=-3$$ Use the final equation to estimate the value: $$y=5*x-3$$ With $x=1.1$, we have $$y=2.5$$ If I use a numerical solver routine, the actual value is: $y \approx -2.35$. So the relative error is about 6%. That's really quite good! :)