Consider the relation $4x^2 - y^2 = -2$

(a) Use implicit differentiation to calculate $dy/dx$ and find all critical points of the curve.

(b) Calculate the second derivative and determine the function's concavity at each critical point.

(c) Graph and clearly label the relation using (a) and (b). What type of curve is described by the relation?

So this seems like an easy question but there are somethings that confuse me:

I did part (a) getting the derivative as $\large\frac{4x}{y}$ and to find the critical points I guess its just at $x = 0$.

For part (b): the second derivative I calculated to be $(4y^2 - 16x^2)/y^3$ but I don't know how to determine the functions concavity at the critical point cause when I plug in $x = 0$ I still have another variable... so help me with that please...

After doing that I wouldnt mind some help on how to draw the graph :p


For part (a) , a critical is any point where the derivative is 0 or undefined. Therefore, y = 0 is also a critical point. However, it doesn't exist because there is no value of x that satisfies that value of y.

$$-4x^2 =2 $$ $$x^2 = \frac{-1}{2}$$

Which cannot exist.

When x = 0

$$-y^2 = -2$$ $$y = \sqrt{2}, -\sqrt{2}$$

For part (b) then, $$\frac{4y^2 - 16x^2}{y^3}$$ When $x = 2, y = \sqrt{2}$ $$\frac{4*2 - 0}{2\sqrt{2}}$$ $$\frac{4}{\sqrt{2}}$$ $$2\sqrt{2}$$

When $x = 2, y = -\sqrt{2}$ $$\frac{4*2 - 0}{-2\sqrt{2}}$$ $$\frac{4}{-\sqrt{2}}$$ $$-2\sqrt{2}$$

When drawing the graph, use values of x and y that yield integer values. From (a) and (b), you can guess the concavity at the critical points of the functions, or the point where the slope is 0. However, please note that if the derivative is undefined it yields a vertical tangent line, if it is 0 then it yields a horizontal tangent line or a slope of 0.

If you need more help, please ask as a comment.

  • $\begingroup$ where are you getting -4x^2 = -2 from?? $\endgroup$ – Panthy Jul 16 '14 at 2:23
  • $\begingroup$ tHat's when you substitute y = 0 into the original equation. It's $-4x^2 = 2$ by the way $\endgroup$ – Varun Iyer Jul 16 '14 at 2:28
  • $\begingroup$ i am so confused... can we take this to a chat? $\endgroup$ – Panthy Jul 16 '14 at 2:31

Write $y = y(x)$ to keep track of what is a variable and what is a function. Then use the chain rule. $$\begin{align} 4x^2 - (y(x))^2 &= -2 \\ 8x - 2y(x)y'(x) &=0 \\ y'(x) &= \frac{4x}{y(x)} \end{align}$$ Notice that in the points such that $y(x) = 0$, the implicit function is not well-defined. Indeed, $y'(x) = 0 \iff x = 0 $. For the second derivative, we can take off from where we left, and get $$y''(x) = \frac{4y(x) - 4xy'(x)}{(y(x))^2}$$ by using the quotient rule. The matter here, is that we have two possible points for $x = 0$, $(0, \sqrt{2})$ and $(0,-\sqrt{2})$.

  • $y = \sqrt{2}$: we get $y''(0) = 2\sqrt{2} > 0$.

  • $y = -\sqrt{2}$: we have $y''(0) = -2\sqrt{2} < 0$.

The calculation was straightforward because $x$ being zero kills half the fraction.

Hint: look at the hyperbola $x^2 - y^2 = 1$. What is different from what you have now? The graphic is rotated? It has a different scale? It is a bit deformed?


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