Question

You are given the following equation. 16x2 + 25y2 = 400 (a) Find dy / dx by implicit differentiation. dy / dx = Correct: Your answer is correct. (b) Solve the equation explicitly for y and differentiate to get dy / dx in terms of x. (Consider only the first and second quadrants for this part.) dy / dx = Incorrect: Your answer is incorrect. (c) Check that your solutions to parts (a) and (b) are consistent by substituting the expression for y into your solution for part (a). (Do this on paper. Your teacher may ask you to turn in this work.)

Answer 1

a) $\frac{dy}{dx}=-\frac{16x}{25y}$

b) $\frac{dy}{dx}=-\frac{4x}{25\sqrt{1-\frac{x^2}{25}}}$

c) The two expressions match

Answer:

a)

The equation in this problem is

$16x^2+25y^2=400$

Here, we want to find $\frac{dy}{dx}$ by implicit differentiation.

To do that, we apply the operator $\frac{d}{dx}$ on each term of the equation. We have:

$\frac{d}{dx}(16 x^2)=32x$

$\frac{d}{dx}(25y^2)=50y \frac{dy}{dx}$ (by applying composite function rule)

$\frac{d}{dx}(400)=0$

Therefore, the equation becomes:

$32x+50y\frac{dy}{dx}=0$

And re-arranging for dy/dx, we get:

$50\frac{dy}{dx}=-32x\\\frac{dy}{dx}=-\frac{32x}{50y}=-\frac{16x}{25y}$

b)

Now we want to solve the equation explicitly for y and then differentiate to find dy/dx. The equation is:

$16x^2+25y^2=400$

First, we isolate y, and we find:

$25y^2=400-16x^2\\y^2=16-\frac{16}{25}x^2$

And taking the square root,

$y=\pm \sqrt{16-\frac{16}{25}x^2}=\pm 4\sqrt{1-\frac{x^2}{25}}$

Here we are told to consider only the first and second quadrants, so those where y > 0; so we only take the positive root:

$y=4\sqrt{1-\frac{x^2}{25}}$

Now we differentiate this function to find dy/dx; using the chain rule, we get:

$\frac{dy}{dx}=4[\frac{1}{2}(1-\frac{x^2}{25})^{-\frac{1}{2}}\cdot(-\frac{2x}{25})]=-\frac{4x}{25\sqrt{1-\frac{x^2}{25}}}$ (2)

c)

Now we want to check if the two solutions are consistent.

To do that, we substitute the expression that we found for y in part b:

$y=4\sqrt{1-\frac{x^2}{25}}$

Into the solution found in part a:

$\frac{dy}{dx}=-\frac{16x}{25y}$

Doing so, we find:

$\frac{dy}{dx}=-\frac{16x}{25(4\sqrt{1-\frac{x^2}{25}})}=-\frac{4x}{25\sqrt{1-\frac{x^2}{25}}}$ (1)

We observe that expression (1) matches with expression (2) found in part b: therefore, we can conclude that the two solutions are coeherent with each other.