Implicit Differentiation (2 Viewers)

[nrlwinner]

Hi. I've finally figured out how to graph composite functions and now I'm up to implic differentiation and graphing it. The textbook has a small explanation but it isn't clear enough. Can anyone explain to me. Hopefull Jetblack you can help me out again.

Eg

x^0.5 + y^0.5 = a^0.5

 

[nrlwinner]

Actually, all I need help with is graphing these types of functions. Can someone show me an example and give me some tips.

 

[madsam]

To draw all details of a given equation

First thing first, when x = 0 find the values of y
When y = 0 find the values of x

That gives you all the intercepts

Differentiate, implicit or not
Sometimes you have to force the equation in terms of x

ie from your original equation

y = (a^.5 - x^.5)^2

Find all the x values for dy/dx = 0
That tells you stationary points, which you sub into the original equation to find the stationary points


Now you have where the graph cuts all the axis, and high above and below the x axis it goes (Assuming the graph is a function, and not a relation)

Now you also need to find out any asymptotes, or lines the graph does not touch.
There are multiple ways to find these, essentially finding the x and y values that do not exist (eg for the graph y = 1/x x cannot equal zero so x = 0 is an asymptote and so on and so forth)
Draw these as dotted lines

At the end, you need to figure out where the graph ends, both in the positive and negative sides of the graph.
This is achieved by using limits
y value as x approaches infinite

in the example y = x^2, as x approaches infinite, y too approaches infinite, so y =x becomes the asymptote in the positive and negative axis

Thats pretty much the basics of the graphing, however depending on the detail required, simply using the second derivative instead of finding all stationary points will suffice to give the graphs general shape

 

[vafa]

For some reasons, I can not upload files here. I have uploaded them at upload (tex3141592)

If you need explanations, then please shout.

 

[untouchablecuz]

use this info to sketch the graph

 

[nrlwinner]

Hi. I've got a little query here.

For this absolute value equation



I get.

y = 4 for x > 4
y = 2x-4 for 0 < x < 4
y = -4 for x < 0

How do I know where the greater than or equal sign goes?

 

[Iruka]

In this case, it doesn't matter because the graph is continuous. Just make sure that where ever you put it, the function is well defined (i.e., doesn't have 2 y-values specified for one x-value).

 

[nrlwinner]

I'm doing that question and I was wondering why the solution only has it in the first quadrant.

 

[life92]

Because you cant find the square root of a negative number.
If its in any other quadrant than the 1st quadrant, then either x or y is negative, and you cannot evaluate them in the real number system.

 

[nrlwinner]

x^3 + y^3 = 1

Can someone help me graph that because in my working I get results that will make me get something only in the first quadrant

 

[jet]

x^3 + y^3 = 1

Can someone help me graph that because in my working I get results that will make me get something only in the first quadrant

ok, remember the steps I told you?

1) Intercepts
x = 0, y = 1
y = 0, x = 1

2) Asymptotes
There are none.

3) Turning points/vertical tangents (In essence, calculus)
y' = -x²/y²

When y' = 0, x = 0. Since y' ≤ 0 for all x, this is an HPOI. This can be confirmed with the second derivative.

When y --> 0, y' --> infinity. This corresponds to a vertical tangent at that point.

Thus we have:
Intercepts (0,1), (1,0)
HPOI at (0,1)
Vertical tangent at (1,0)

Plotting these elements and realising that the derivative is always negative should lead you to sketch this:

 

[nrlwinner]

Thanks for the help. However my working goes like this:




x^2=0
At (0,1) there is a horizontal tangent

y^2=0
At (1,0) there is a vertical tangent

Intercepts:
x-intercept: 1
y-intercept: 1


With this information, shouldn't I draw like a quadrant?

 

[jet]

No. The fact that there are intercepts means that the graph MUST exit the first quadrant twice. Also, The horizontal tangent is actually a point of inflexion, which gives you even more of a clue.

You can quite easily see from the derivative equation that it is always ≤ 0. This is a strong indicator that it won't just stay in the first quadrant and that it MUST cross the axes (to prevent it being +ve).

Does that help?