Implicit differentiation - not defined?

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June 4th 2012, 12:28 PM
infernalmich

Implicit differentiation - not defined?

I have the problem:

Find the linear approximation about (-1,1) to the implicit function given by the relation: 2xy-x⁸+y²=-2

My problem is that trying implicit differentiation, I always get a slope which is not defined, because the denumerator is always (x+y) and therefore 0.

Did I do an error doing my calculation? If my calculation was right, can I answer that the tangent line is not defined at (-1,1)?
June 4th 2012, 01:42 PM
HallsofIvy

Re: Implicit differentiation - not defined?

No, there is no error. You can, however, say that the "linear approximation" is the vertical line x= -1.

Another way of looking at it is to treat this as an implicitly defining x as a function of y. Then, differentiating both sides with respect to y you would get x'= 0. That gives x= 0(y-1)- 1 or, again, x= -1.
June 5th 2012, 01:26 PM
infernalmich

Re: Implicit differentiation - not defined?

Hm, I always used the formula:

x-x₀= x'/y' (x-x₀)

In that case for this example I get:
y-1=0*(x+1)
so
y=1

what was my mistake? Did I use the wrong formula in order to get the tangent line?
June 5th 2012, 02:04 PM
skeeter

1 Attachment(s)

Re: Implicit differentiation - not defined?

the tangent line is x = -1
June 5th 2012, 05:10 PM
HallsofIvy

Re: Implicit differentiation - not defined?

Quote:

Originally Posted by infernalmich

Hm, I always used the formula:

x-x₀= x'/y' (x-x₀)

What do x' and y' mean here? (And surely, you don't mean $x- x_0$ on both sides?)

Quote:

In that case for this example I get:
y-1=0*(x+1)
so
y=1

what was my mistake? Did I use the wrong formula in order to get the tangent line?

I can't answer that until I know what you mean by x' and y'. Normally, I would interpret them as derivatives, but with respect to what variables? If x and y were given as functions of some other variable, and x' and y' were the derivatives with respect to that third variable, the tangent line would be given by $y- y_0= (y'/x')(x- x_0)$ where the slope is the reciprocal of what you give.
June 6th 2012, 11:23 AM
infernalmich

Re: Implicit differentiation - not defined?

The function in my example is given implicitly:
2xy-x⁸+y²=-2
find the tangent line at the point: (-1,1) - I refer to this points as x₀ and y₀

hence y'_(x)=- (af/ax) / (af/ay)
so y=y₀-y'_(x) (x-x₀)

In this case y=1-0
so y=1

____

Trying to do the same with:
x=x₀-x'_(y)(y-y₀)
x=-1- 0*(y-1)

now my stupid question:
Why is x=-1 the tangent line and not y=1?
June 6th 2012, 12:11 PM
skeeter

Re: Implicit differentiation - not defined?

Quote:

Originally Posted by infernalmich

The function in my example is given implicitly:
2xy-x⁸+y²=-2
find the tangent line at the point: (-1,1) - I refer to this points as x₀ and y₀

now my stupid question:
Why is x=-1 the tangent line and not y=1?

first, $2xy-x^8+y^2=-2$ is not a function ... why?

second, all points on a vertical line have the same x-value, hence, the equation of the vertical line passing thru the point (-1,1) is x = -1

fyi, all points on a horizontal line have the same y-value , hence the equation of a horizontal line would be y = a constant.
June 6th 2012, 12:52 PM
infernalmich

Re: Implicit differentiation - not defined?

Now I try with a second one:

Find the linear approximation at (1,-1) for the implicit function y(x) given by the following relation:
2x³y³-3x²y³=5

af/ax=-5
af/ay=10

than I can calculate:
y=y⁰-(-2)(x-x⁰)
y=-1+2*(x-1)
y=2x+2
0,5y=x

and for x:
x=1-(-0,5)(y--1))
x=0,5y

am I right?
June 6th 2012, 01:34 PM
skeeter

Re: Implicit differentiation - not defined?

Quote:

Originally Posted by infernalmich

Find the linear approximation at (1,-1) for the implicit function y(x) given by the following relation:
2x³y³-3x²y³=5

the point $(1,-1)$ is not on the curve $2x^3y^3 - 3x^2y^3 = 5$

did you copy the problem correctly?
June 6th 2012, 01:45 PM
infernalmich

Re: Implicit differentiation - not defined?

no I didn't, sorry.

The correct problem is:
2x³y²-3x²y³=5
June 6th 2012, 02:04 PM
skeeter

1 Attachment(s)

Re: Implicit differentiation - not defined?

Quote:

Originally Posted by infernalmich

no I didn't, sorry.

The correct problem is:
2x³y²-3x²y³=5

I get ...

$\color{red}{y + 1 = \frac{12}{13}(x - 1)}$
June 7th 2012, 01:03 PM
infernalmich

Re: Implicit differentiation - not defined?

Oh I understood my error. I messed up one of the - signs :-)

Thank you!
June 13th 2012, 09:30 AM
infernalmich

Re: Implicit differentiation - not defined?

I have another question. Could someone check this for me please:

find the tangent line to the equation:

y²+e^2x+xln(y)=4 through (0,2)

First I take the derivatives:

f_y_(x,y)=2e²+y²+ln(y)
f_x[SUB](x,y)[/SUB=2y*e^2x+x/y

through the point (0,2) I have:
f_y=ln(2)+2
f_x=4

so my tangent line is:
y-2=-(ln(2)+2)/4*x
y=x/2-ln(2)+2

Is this true? I really hope so, because I am trying this examples since 2 hours :-)
June 13th 2012, 10:05 AM
HallsofIvy

Re: Implicit differentiation - not defined?

Quote:

Originally Posted by infernalmich

I have another question. Could someone check this for me please:

find the tangent line to the equation:

y²+e^2x+xln(y)=4 through (0,2)

First I take the derivatives:

f_y_(x,y)=2e²+y²+ln(y)

This is wrong to start with. f_y= 2y+ x/y.

Quote:

f_x(x,y)=2y*e^2x+x/y

No, f_x= 2e^2x+ ln(y).

Quote:

through the point (0,2) I have:
f_y=ln(2)+2
f_x=4

No f_y(0, 2)= 4 and f_x(0, 2)= 2+ ln(2). You have "x" and "y" reversed.

Quote:

so my tangent line is:
y-2=-(ln(2)+2)/4*x

Oddly enough, you have again mixed up "x" and "y" so that this is correct!

Quote:

y=x/2-ln(2)+2

This, however, is not correct. You should have -((ln(2)+ 2)/4)x+ 2. The x is not just multiplied by the "2/4".

Quote:

Is this true? I really hope so, because I am trying this examples since 2 hours :-)

You really need to review basic differentiation rules and algebra.
June 16th 2012, 11:39 AM
infernalmich

Re: Implicit differentiation - not defined?

I have another question:

Find all values for which the tangent line of the following equation is horizontal:
$2x^2+xy+y^2-4=-2$

first I get the derivatives with respect to x and y:
af/ax=4x+y
af/ay=x+2y

Than I use implicit differentiation: - (4x+y)/(x+2y)

As I remember, if the numerator is 0 the tangent line is horizontal, so the tangent line is horiziontal for y=4x or x=y/4 - if I would be asked for the vertical tangent line than the denumerator would have to be 0.

Is this true?