a) [tex]\frac{dy}{dx}=-\frac{16x}{25y}[/tex]
b) [tex]\frac{dy}{dx}=-\frac{4x}{25\sqrt{1-\frac{x^2}{25}}}[/tex]
c) The two expressions match
Answer:
a)
The equation in this problem is
[tex]16x^2+25y^2=400[/tex]
Here, we want to find [tex]\frac{dy}{dx}[/tex] by implicit differentiation.
To do that, we apply the operator [tex]\frac{d}{dx}[/tex] on each term of the equation. We have:
[tex]\frac{d}{dx}(16 x^2)=32x[/tex]
[tex]\frac{d}{dx}(25y^2)=50y \frac{dy}{dx}[/tex] (by applying composite function rule)
[tex]\frac{d}{dx}(400)=0[/tex]
Therefore, the equation becomes:
[tex]32x+50y\frac{dy}{dx}=0[/tex]
And re-arranging for dy/dx, we get:
[tex]50\frac{dy}{dx}=-32x\\\frac{dy}{dx}=-\frac{32x}{50y}=-\frac{16x}{25y}[/tex]
b)
Now we want to solve the equation explicitly for y and then differentiate to find dy/dx. The equation is:
[tex]16x^2+25y^2=400[/tex]
First, we isolate y, and we find:
[tex]25y^2=400-16x^2\\y^2=16-\frac{16}{25}x^2[/tex]
And taking the square root,
[tex]y=\pm \sqrt{16-\frac{16}{25}x^2}=\pm 4\sqrt{1-\frac{x^2}{25}}[/tex]
Here we are told to consider only the first and second quadrants, so those where y > 0; so we only take the positive root:
[tex]y=4\sqrt{1-\frac{x^2}{25}}[/tex]
Now we differentiate this function to find dy/dx; using the chain rule, we get:
[tex]\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}}}[/tex] (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:
[tex]y=4\sqrt{1-\frac{x^2}{25}}[/tex]
Into the solution found in part a:
[tex]\frac{dy}{dx}=-\frac{16x}{25y}[/tex]
Doing so, we find:
[tex]\frac{dy}{dx}=-\frac{16x}{25(4\sqrt{1-\frac{x^2}{25}})}=-\frac{4x}{25\sqrt{1-\frac{x^2}{25}}}[/tex] (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.
