Thread: Find Points if horizontal or vertical tanget line

1. Find Points if horizontal or vertical tanget line

I was given an implicit function to solve for $\frac{dy}{dx}$

Which I have done here $f(x)=x^2+y^2-6x+10y=20$

$2x+2y\frac{dy}{dx}-6+10\frac{dy}{dx}=20$

Which I have solved $\frac{dy}{dx}$ for

$\frac{dy}{dx}=\frac{6-2x}{2y+10}$

I know the vertical tangent line point can be found when $\frac{dy}{dx}$ numerator is equal to zero which in this case is 3 and for the horizontal tangent line the denom set to 0 is -5 in this case

However both the numerator and denom. have to satisfy the originial equation?

for the num I have just the x variable and so all I can plug is the x value so what I do is solve for y? and see if it satifies or what

and vice versa for the denom

2. Originally Posted by sk8erboyla2004
I was given an implicit function to solve for $\frac{dy}{dx}$

Which I have done here $f(x)=x^2+y^2-6x+10y=20$

$2x+2y\frac{dy}{dx}-6+10\frac{dy}{dx}=20$

Which I have solved $\frac{dy}{dx}$ for

$\frac{dy}{dx}=\frac{6-2x}{2y+10}$

I know the vertical tangent line point can be found when $\frac{dy}{dx}$ numerator is equal to zero which in this case is 3 and for the horizontal tangent line the denom set to 0 is -5 in this case

However both the numerator and denom. have to satisfy the originial equation?

for the num I have just the x variable and so all I can plug is the x value so what I do is solve for y? and see if it satifies or what

and vice versa for the denom
Note that $x^2+y^2-6x+10y=20 \implies x^2-6x+9-9+y^2+10y+25-25=20$ $\implies (x-3)^2+(y+5)^2=54$

So the function you are dealing with is a circle of radius $3\sqrt{6}$ centered at $(3,-5)$. This may be a helpful picture to have in your mind while doing this.

Given that $\frac{dy}{dx}=\frac{6-2x}{2y+10}$, the horizontal tangent occurs when the numerator is zero (because then the derivative is 0), which is at $x=3$. Use the function to find that $y=-5\pm 3\sqrt{6}$.

Similarly, the vertical tangent occurs when the denominator is zero (because then the derivative is "infinity"), which is at $y=-5$. Use the function to find that $x=3\pm 3\sqrt{6}$.

So the horizontal tangent lines occur at $(3,-5+3\sqrt{6})$ and $(3,-5-3\sqrt{6})$ and the vertical tangent lines occur at $(3+3\sqrt{6},-5)$ and $(3-3\sqrt{6},-5)$.

If you think about the picture of the circle, this makes a lot of sense.

3. Originally Posted by sk8erboyla2004
I was given an implicit function to solve for $\frac{dy}{dx}$

Which I have done here $f(x)=x^2+y^2-6x+10y=20$

$2x+2y\frac{dy}{dx}-6+10\frac{dy}{dx}=20$

Which I have solved $\frac{dy}{dx}$ for

$\frac{dy}{dx}=\frac{6-2x}{2y+10}$

I know the vertical tangent line point can be found when $\frac{dy}{dx}$ numerator is equal to zero which in this case is 3 and for the horizontal tangent line the denom set to 0 is -5 in this case

However both the numerator and denom. have to satisfy the originial equation?

for the num I have just the x variable and so all I can plug is the x value so what I do is solve for y? and see if it satifies or what

and vice versa for the denom
You didnt differeniate Differentiate the right hand side
d/dx(20) = 0

4. At first I thought so too, but it seems like even though he didn't write it, he still got the correct derivative. $2x+2y\frac{dy}{dx}-6+10\frac{dy}{dx}=0 \implies \frac{dy}{dx}(2y+10)=6-2x \implies \frac{dy}{dx}=\frac{6-2x}{2y+10}$

5. Originally Posted by redsoxfan325
Note that $x^2+y^2-6x+10y=20 \implies x^2-6x+9-9+y^2+10y+25-25=20$ $\implies (x-3)^2+(y+5)^2=54$

So the function you are dealing with is a circle of radius $3\sqrt{6}$ centered at $(3,-5)$. This may be a helpful picture to have in your mind while doing this.

Given that $\frac{dy}{dx}=\frac{6-2x}{2y+10}$, the horizontal tangent occurs when the numerator is zero (because then the derivative is 0), which is at $x=3$. Use the function to find that $y=-5\pm 3\sqrt{6}$.

Similarly, the vertical tangent occurs when the denominator is zero (because then the derivative is "infinity"), which is at $y=-5$. Use the function to find that $x=3\pm 3\sqrt{6}$.

So the horizontal tangent lines occur at $(3,-5+3\sqrt{6})$ and $(3,-5-3\sqrt{6})$ and the vertical tangent lines occur at $(3+3\sqrt{6},-5)$ and $(3-3\sqrt{6},-5)$.

If you think about the picture of the circle, this makes a lot of sense.

What if i plug x and y in the original equation unimplifed like you did my answer will not be the same such as when i plug in y i get

$\frac{-10+\sqrt{128}}{2}$

6. What are you using for x and what are you using for y when you plug them into the original equation?

7. Well for the denom I solved y for -5 and for the num. x for 3 so i am trying to plug one in and then solve for the other unknown variable

8. When I plug $x=3$ into the original equation, I get $y=-5\pm3\sqrt{6}$, and when I plug $y=-5$ into the original equation, I get $x=3\pm3\sqrt{6}$. Can you show your steps to how you ended up with $\frac{-10+\sqrt{128}}{2}$?

9. Yeah sure thing I am sure it is something stupid I am doing

Ok so here is the original function

$x^2+y^2-6x+10y=20$

Ok the num when equal to zero is x=3

$3^2+y^2-6(3)+10y=20$

So I get

$y^2+10y-9=20$

and then somewhere I think is where I make my mistake and plugged in incorrect numbers

10. Originally Posted by sk8erboyla2004
Yeah sure thing I am sure it is something stupid I am doing

Ok so here is the original function

$x^2+y^2-6x+10y=20$

Ok the num when equal to zero is x=3

$3^2+y^2-6(3)+10y=20$

So I get

$y^2+10y-9=20$

and then somewhere I think is where I make my mistake and plugged in incorrect numbers
$y^2+10y-29=0 \implies x=\frac{-10\pm\sqrt{100-(4\cdot-29)}}{2} =\frac{-10\pm\sqrt{216}}{2}$ $= \frac{-10\pm6\sqrt{6}}{2} = -5\pm3\sqrt{6}$

11. Ah thanks Ill re-do my work now

I forgot how to rewrite or do the sqrt of 216 the way u did :x mind showing that too