# Help me help my daughter. (Calculus Question) **URL:** \ **Category:** Factual Questions\ **Created:** [January 17, 2010, 8:41pm UTC](https://boards.straightdope.com/t/help-me-help-my-daughter-calculus-question/525351 "2010-01-17T20:41:20Z")\ **Posts on this page:** 13\ **Page:** 1
**Author:** ![Pduol](https://avatars.discourse-cdn.com/v4/letter/p/6bbea6/32.png) [@Pduol](https://boards.straightdope.com/u/Pduol)\ **Post date:** [January 17, 2010, 8:41pm UTC](https://boards.straightdope.com/t/help-me-help-my-daughter-calculus-question/525351/1 "2010-01-17T20:41:20Z")
The question is: For the ellipse x^2 - 2xy + 3y^4 = 16, find any points with a horizontal tangent line and any points with a vertical tangent line. How would we do this? Any help is certainly appreciated Thanks in advance. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [January 17, 2010, 8:58pm UTC](https://boards.straightdope.com/t/help-me-help-my-daughter-calculus-question/525351/2 "2010-01-17T20:58:12Z")
Here’s a hint: what would it mean algebraically for the tangent line to be horizontal or vertical? ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [January 17, 2010, 9:03pm UTC](https://boards.straightdope.com/t/help-me-help-my-daughter-calculus-question/525351/3 "2010-01-17T21:03:20Z")
One caveat: the equation you’ve given does not define an ellipse. I assume you meant to write “y^2” rather than “y^4”. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [January 17, 2010, 9:08pm UTC](https://boards.straightdope.com/t/help-me-help-my-daughter-calculus-question/525351/4 "2010-01-17T21:08:33Z")
Given that assumption, I don’t know if the following is the method you’re expected to use (I suppose it depends on what you’ve been taught), but here’s one way: Given a curve defined by f(x, y) = some constant c, the horizontal tangents occur at the points on the curve where df/dx = 0 (since this means you can move an infinitesimal amount horizontally without beginning to depart from the curve) and the vertical tangents analogously occur where df/dy = 0. In this case, c = 16 and f(x, y) = x^2 - 2xy + 3y^2. With this starting prod, your daughter should hopefully be able to do the rest on her own. However, in case you need a little more help, read as much of the following as necessary: Therefore, by basic differentiation rules, df/dx = 2x - 2y and df/dy = -2x + 6y. So, the horizontal tangents are the solutions to the curve equation where furthermore 2x - 2y = 0, which is to say, where x = y. And the vertical tangents are the solutions to the curve equation where furthermore -2x + 6y = 0, which is to say, x = 3y. So, for the horizontal tangents: when x = y, our curve equation becomes y^2 - 2y^2 + 3y^2 = 16; i.e., y^2 = 8. Thus, the two points with horizontal tangents (keeping in mind that x = y) are \ and \<-sqrt(8), -sqrt(8)\>. As for the vertical tangents: when x = 3y, our curve equation becomes 9y^2 - 6y^2 + 3y^2 = 16; i.e. y^2 = 8/3. Thus, the two points with horizontal tangents (keeping in mind that x = 3y) are \<3 \* sqrt(8/3), sqrt(8/3)\> and \<3 \* -sqrt(8/3), -sqrt(8/3)\>. (I hope I haven’t made some small mistake somewhere above, but I may well have. Also, there are, of course, other ways to do it, as well as other ways to express the end result.) ---
**Author:** ![Pduol](https://avatars.discourse-cdn.com/v4/letter/p/6bbea6/32.png) [@Pduol](https://boards.straightdope.com/u/Pduol)\ **Post date:** [January 17, 2010, 9:08pm UTC](https://boards.straightdope.com/t/help-me-help-my-daughter-calculus-question/525351/5 "2010-01-17T21:08:46Z")
> [@ultrafilter](#): > > Here’s a hint: what would it mean algebraically for the tangent line to be horizontal or vertical? Welll I know that if the tangent line is horizontal the slope is 0 and if the tangent line is vertical the slope is undefined. She has had a few questions that involved finding horizontal tangent lines with something like f(x) = x^3 - 2x^2 +3x -2. (that’s not the exact problem, but you get the idea.) In that case, we found the derivative and set it equal to zero and solved for x. We can’t figure out how to make it work for this question, though. ---
**Author:** ![John\_Mace](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/john_mace/32/185_2.png) [@John\_Mace](https://boards.straightdope.com/u/John_Mace)\ **Post date:** [January 17, 2010, 9:10pm UTC](https://boards.straightdope.com/t/help-me-help-my-daughter-calculus-question/525351/6 "2010-01-17T21:10:21Z")
Start by taking the derivative. Then think about what \*\*ultrafilter \*\*said. ---
**Author:** ![Pduol](https://avatars.discourse-cdn.com/v4/letter/p/6bbea6/32.png) [@Pduol](https://boards.straightdope.com/u/Pduol)\ **Post date:** [January 17, 2010, 9:10pm UTC](https://boards.straightdope.com/t/help-me-help-my-daughter-calculus-question/525351/7 "2010-01-17T21:10:45Z")
> [@Indistinguishable](#): > > One caveat: the equation you’ve given does not define an ellipse. I assume you meant to write “y^2” rather than “y^4”. No, the question says 3y^4, I just double checked. ---
**Author:** ![John\_Mace](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/john_mace/32/185_2.png) [@John\_Mace](https://boards.straightdope.com/u/John_Mace)\ **Post date:** [January 17, 2010, 9:15pm UTC](https://boards.straightdope.com/t/help-me-help-my-daughter-calculus-question/525351/8 "2010-01-17T21:15:07Z")
**Pdual** : Just FYI, since you’re new. We don’t generally like to do people’s homework for them here. Don’t expect a step-by-step solution. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [January 17, 2010, 9:18pm UTC](https://boards.straightdope.com/t/help-me-help-my-daughter-calculus-question/525351/9 "2010-01-17T21:18:24Z")
> [@Pduol](#): > > No, the question says 3y^4, I just double checked. Well, if they really intended an ellipse, then that must be a typo. But, either way, the ideas I outlined still apply; the final algebra just may end up a little trickier. ---
**Author:** ![BigT](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/bigt/32/12044_2.png) [@BigT](https://boards.straightdope.com/u/BigT)\ **Post date:** [January 18, 2010, 11:33am UTC](https://boards.straightdope.com/t/help-me-help-my-daughter-calculus-question/525351/10 "2010-01-18T11:33:17Z")
> [@Indistinguishable](#): > > (I hope I haven’t made some small mistake somewhere above, but I may well have. Also, there are, of course, other ways to do it, as well as other ways to express the end result.) I checked your math by using an [online solver](http://www.quickmath.com/webMathematica3/quickmath/page.jsp?s1=equations&s2=solve&s3=basic) and [online grapher](http://www.coolmath.com/graphit/). The only “mistake” you made was forgetting to simplify: 3√(8/3) = 2√6. But, like you said, that’s only “[an]other way to express the end result.” ---
**Author:** ![Thudlow\_Boink](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/thudlow_boink/32/320_2.png) [@Thudlow\_Boink](https://boards.straightdope.com/u/Thudlow_Boink)\ **Post date:** [January 18, 2010, 9:40pm UTC](https://boards.straightdope.com/t/help-me-help-my-daughter-calculus-question/525351/11 "2010-01-18T21:40:35Z")
**Indistinguishable** ’s approach uses partial derivatives, which the OP’s daughter may not have learned about yet. A question like the OP’s might have come up in the context of implicit differentiation. One way to solve it is to use “implicit differentiation” to find dy/dx (= the slope of the tangent line to the graph of the equation). Then horizontal tangents occur when dy/dx = 0, and vertical tangents occur when the reciprocal of dy/dx is 0. Although, in my experience, once students do learn about partial derivatives, they like that approach better than this one. ---
**Author:** ![CookingWithGas](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/cookingwithgas/32/485_2.png) [@CookingWithGas](https://boards.straightdope.com/u/CookingWithGas)\ **Post date:** [January 18, 2010, 9:49pm UTC](https://boards.straightdope.com/t/help-me-help-my-daughter-calculus-question/525351/12 "2010-01-18T21:49:24Z")
> [@John\_Mace](#): > > **Pdual** : Just FYI, since you’re new. We don’t generally like to do people’s homework for them here. Don’t expect a step-by-step solution. True but this is kind of a special case since it’s a father asking for support in helping his daughter; he’s not asking us to do her homework for her. He is using good judgment in teaching her the material without flat-out doing it for her. I have seen lots of other threads that get shut down when some high-schooler basically asks us to write a paper or solve a math problem for him. ---
**Author:** ![Pedro](https://avatars.discourse-cdn.com/v4/letter/p/919ad9/32.png) [@Pedro](https://boards.straightdope.com/u/Pedro)\ **Post date:** [January 18, 2010, 10:12pm UTC](https://boards.straightdope.com/t/help-me-help-my-daughter-calculus-question/525351/13 "2010-01-18T22:12:38Z")
> [@Pduol](#): > > Welll I know that if the tangent line is horizontal the slope is 0 and if the tangent line is vertical the slope is undefined. > > She has had a few questions that involved finding horizontal tangent lines with something like f(x) = x^3 - 2x^2 +3x -2. (that’s not the exact problem, but you get the idea.) In that case, we found the derivative and set it equal to zero and solved for x. > > We can’t figure out how to make it work for this question, though. You need the concept of [implicit functions](http://en.wikipedia.org/wiki/Implicit_function).