5.6 Implicit Applications: Using Implicit Derivatives

© 2002 by Karl Hahn
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Finding the Slope at a given Point

There is a long tradition in calculus courses that they give the students various equations in x and y, and then ask that the student use implicit differentiation to answer some questions about the curve formed by those equations. Let's jump right in and do an example.
   x2y + y3 + x  =  11                                            eq. 5.6-1
One of the things they might ask you, for example, is what is the slope of this curve at the point,  (1,2)?  Recall that implicit differentiation does not give you the derivative of something outright (that is, as a function of x), but it does tell you something about the derivative of a curve like this. So lets take the implicit derivative of this equation.

The first term is  x2y.  Clearly you have to use the product rule to take its derivative. The derivative of x2 is 2x. The derivative of y is y'. So applying the product rule to the first term only, you get  2xy + x2y'

The second term is y3. Using what you learned about implicit differentiation, you can see that the derivative of this term is  3y2y'

The third term is x, and you already know that the derivative of x is 1. And the last term (the one to the right of the equal) is a constant, so its derivative is zero. Putting it all together, when you take the derivative of equation 5.6-1, you get

   2xy + x2y' + 3y2y' + 1  =  0                                   eq. 5.6-2a
So how do you use this to answer the question of what the slope of the curve is at  (1,2)?  First, solve for y' in terms of x and y. Do this using a little algebra. Move all the terms of the derivative equation (eq. 5.6-2a) that have a y' factor over to one side of the equation.
   x2y' + 3y2y'  =  -2xy - 1                                      eq. 5.6-2b
Now factor the y' out of that side of the equation:
   y'(x2 + 3y2)  =  -2xy - 1                                      eq. 5.6-2c
and then divide both sides by the other factor:
          -2xy - 1
   y'  =                                                          eq. 5.6-2d
          x2 + 3y2
The point in question was  (1,2).  So in equation 5.6-2d, substitute 1 for x and 2 for y:
                     -4 - 1        5
   at (1,2):  y'  =          =  -                                 eq. 5.6-3
                     1 + 12       13
Once you have the implicit derivative and you have isolated y', this kind of problem is pretty easy. It's just plugging in the numbers.

In the figure, you can see the curve along with the tangent line at the point, (1,2). See if you can figure out how I determined the equation of that line.

Finding Horizontal and Vertical Tangents

The other kind of problem they like to give you is to ask you where the horizontal and vertical tangents are to the curve. We will do this with

   x2 + 5xy + 20y2  =  100                                        eq. 5.6-4
Of course, we will need to do the implicit differentiation:
   2x + 5y + 5xy' + 40yy'  =  0                                   eq. 5.6-5
First we find all the horizontal tangents. How do we do that? When the tangent line is horizontal, its slope is zero. That means that y' at the point of tangency must also be zero. So we set  y' = 0,  and see what happens to our derivative equation. Notice that all the terms that have a y' as a factor drop out and you are left with:
   2x + 5y  =  0                                                  eq. 5.6-6a
or equivalently
   y  =  -   x                                                    eq. 5.6-6b
Now substitute this expression for y back into the original equation (eq 5.6-5) and solve for x.
   x2  -  2x2  +     x2  =  100                                   eq. 5.6-7a
Simplifying you get
   11                            500
      x2  =  100    or    x2  =                                    eq. 5.6-7b
    5                             11
Take the square root to find x.
   x  =  ±                                                        eq. 5.6-7c
That gives you the x coordinates for two points of tangency. Put those x's back into equation 5.6-6b to find the corresponding y's. That's it

So now, how do you find the vertical tangents? The slope of a vertical line doesn't exist because when you try to find it, you get a zero in the denominator. As you draw lines that are closer and closer to the vertical (with positive slope), the slope goes to infinity. This offers a method for finding vertical points of tangency. We take the implicit derivative equation, and take the limit as y' goes to infinity.

      lim  2x + 5y + 5xy' + 40yy'  =  lim 0                       eq. 5.6-8a
    y' → ∞
What happens here is that as y' gets huge, all the terms that have no y' factor in them become insignificant, including those to the right of the equal. Another way of looking at it is if you divide the whole thing through by y':

   y' → ∞
  2x     5y
      +      +  5x  +  40y  =  lim 0                     eq 5.6-8b
  y'     y'
Can you see how, as y' gets huge, the only terms that are left being significant are the ones without the y' in their denominators? The effect of all this is that when you go to find vertical points of tangency, look at the implicit derivative (equation
5.6-5 in this case), and eliminate all the terms that don't have a factor of y'

   5xy' + 40yy'  =  0                                             eq. 5.6-9a
And now divide out the y', and you're left with
   5x + 40y  =  0                                                 eq. 5.6-9b

   y  =  -    x                                                   eq. 5.6-9c
Now simply substitute this expression for y back into equation 5.6-4, and solve for x.
          25         500
   x2  -     x2  +       x2  =  100                               eq. 5.6-10a
          40        1600

   11                                    1600
      x2  =  100        or        x2  =                           eq. 5.6-10b
   16                                     11

   x  =  ±  __                                                    eq. 5.6-10c
That gives you the x coordinates of the two vertical points of tangency. To get the corresponding y values, just take these x's, and plug them back into equation 5.6-9c.

Now suppose they ask you to find the tangent lines whose slopes are 1 (or any other value for that matter). Simply go back to equation 5.6-5, and plug in 1 (or whatever other slope value they are asking for) for y'.

   2x + 5y + 5x + 40y  =  7x + 45y  =  0                          eq. 5.6-11a

   y  =  -    x                                                   eq. 5.6-11b
I'll leave it to you to finish this part of the problem in the same way as I did the other two parts -- that is, substitute this expression for y back into equation 5.6-4, and solve for x.

You can see the graph of all this to the right. The blue labels a vertical tangency point. The brown labels a horizontal tangency point. You get to fill in a black tangency point (one with slope of 1) yourself.

Exercises for this section are still under construction

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