Message Boards Message Boards

8 Replies
0 Total Likes
View groups...
Share this post:

Homework Help

Posted 10 years ago

Ok, so this is my homework and I'm stuck on Exercise 2 part d). I've spent the past 2 hours trying an array of different ways to solve this but I can't figure it out and my professor isn't answering her email. Please help! It's due at 11:59pm.

8 Replies

Well, I had to go to bed....

But.... a couple of confusions in your notebook. First, note that when your professor wrote

enter image description here

the quantities in parentheses after the f and the f' (i.e., the xo) is not multiplication. This is the traditional notation for a function argument. Thus you should not have the 7 in the following

enter image description here

Also note that the function name is L[x], not l[x] in your plotting function (I am not sure how you got a plot but perhaps you had defined l[x] somewhere else)--Mathematica parameter names are case-sensitive.

enter image description here

So if you make these changes you get the following plot:

enter image description here

And, my advice rom earlier still stands! Start your homework well before the deadline--perhaps the day it is assigned so that you can bring some of your (valid and smart) questions to your professor, TAs, or other classmates to give your learning process a bit more slack. (And, yes, I know you are busy with lots of classes and homework--it;s not easy!)

POSTED BY: David Reiss

Why does the tangent line cross at 2 different places?


Use the equation that your professor supplied. Then, as you did, compute the derivative of f and substitute the value of x0 in that expression. Also put that same value into the expression for f itself and then use these in the linearization equation for L[x]. You do not need to use Solve. Think of the reason for (i.e., the idea behind) the expression for L[x]. And (fatherly advice) try to get your homework started well before the midnight deadline ;-) !

POSTED BY: David Reiss

I can't find L(x)! I only have 30 min to figure this out. I don't know what's going on. Everything just comes out as giant complicated functions. Please help!


In your latest case you will notice that the point at the top does correspond to a horizontal tangent ((just taking a look at it you see that the slope of the curve there is zero), but your other points are not vertical tangents since the slopes of the curve there is not vertical, it has a slope in both cases. The problem is that you substituted into the x values and solved for y rather than substituted for the y values and solved for x in each case,

Here are those two cases, corrected:

NSolve[(84 x + 6 x^2 + 28 x^3 + 3 x^4 - 288 y - 12 y^2 + 4 y^3 ==  5) /. y -> -4, x]


NSolve[(84 x + 6 x^2 + 28 x^3 + 3 x^4 - 288 y - 12 y^2 + 4 y^3 ==  5) /. y -> 6, x]

These give the following points (confirm for yourself):

{{-9.14477, -4}, {-2.97489, -4}}


{-9.86928, 6}

so your ListPlot is

ListPlot[{{-9.144774626644757`, -4}, {-2.974889393649998`, -4}, \
{-9.869278763054162`, 6}}, PlotStyle -> {Black, PointSize[Large]}]
POSTED BY: David Reiss

Wait! I think I did that wrong. Because when I plotted them the points weren't on the original contour plot. I think I had the x and y values switched. This is right now, yes?


Ok, this is what I got. Would you say that is right or do I need to switch the x and y values in the coordinates?

Thank you so much for getting back to me! It was extremely helpful!!!


The implicit differentiation gave you the function

(-7 - x - 7 x^2 - x^3)/(-24 - 2 y + y^2)

as the result for dy/dx. Notice that the numerator only depends on x and the denominator only depends on y. The horizontal tangents are where the expression is zero and this in turn is where the numerator is zero. So solve for the values of x where this is true:

Solve[-7 - x - 7 x^2 - x^3 == 0, x]

Once you have the x values from this (there is only one that is real in this case) you can find the corresponding y value by going back to the original implicit equation

84 x + 6 x^2 + 28 x^3 + 3 x^4 - 288 y - 12 y^2 + 4 y^3 == 5

substitute each x value and then solve for the corresponding y value for that x value. So now you have the {x,y} points where there are horizontal tangents.

Similarly the vertical tangents are where the expression goes to + or - infinity and this is where the denominator becomes zero (so long as it also is not one of the points where the numerator is also zero). You can determine this from

Solve[-24 - 2 y + y^2 == 0, y]

And like in the horizontal case, use the implicit equation to now solve for the x values corresponding to each of these y-values.

When you are done you can plot all of these values using ListPlot and then combine them with your original contour plot using Show.

You can make the points in your ListPlot show nicely by using something like (I am using made up points here):

ListPlot[{{1, 2}, {3, 5}}, PlotStyle -> Directive[Red, AbsolutePointSize[12]]]

If this ListPlot is called plot4 then you would combine this with your contour plot (which was called plot3) using


P.S. Give Professor Read our best regards ;-)

POSTED BY: David Reiss
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Reply Preview
or Discard

Group Abstract Group Abstract