Yr 12 Maths 1 Help - Get Assistance for Horizontal Tangents & Normals

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Homework Help Overview

The discussion revolves around finding horizontal tangents and normals for various mathematical curves, specifically focusing on calculus concepts related to derivatives and tangent lines. The original poster expresses difficulty with these types of problems and seeks assistance in understanding how to approach them.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to find points where the derivative equals zero for horizontal tangents and explore how to derive equations for tangents that are parallel to given points. There are attempts to clarify the relationship between the derivative and the slope of the tangent line.

Discussion Status

Some participants have offered insights into the process of finding derivatives and the implications for tangent lines, while others express confusion about specific steps and concepts, indicating that the discussion is ongoing and that clarification is still needed.

Contextual Notes

The original poster mentions a time constraint due to an upcoming test and expresses a sense of urgency in needing help, which may influence the depth of exploration in the discussion.

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Im havin some real problems with my maths :( I am behidn coz I went on a holiday for easter. I have a test tomorrow and I really need help :frown:

If anyone could teach me how to do these sort of questions it would be much appreciated.

1. Find all points of contact of horizontal tangents to the curve
y=2√x+1/√x

2. Find equation of tangent to
y=1-3x+12x²-8x³
which is parallel to the tangent at (1,2)

3. The normal to the curve
y=a√x+b/√x
where a and b are constants, has equation 4x+y=22 at the point where x=4. Find the values of a and b.

I haven't had much trouble with other work in the chapter, but I have continously got questions similar to this wrong. Some help would be much appreciated.
 
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1. [tex]y'=0[/tex]

The solve for x.

2. [tex]y'(1) = y' (x)[/tex]

Find x.

3. 4x+y=22, from which k=-4, from which [itex]k_t=1/4[/itex] From which [itex]y' (4) = 0.25.[/itex]

Edit:

In 2. After you find the point of interest. Find the value of the derivative there, this will equal k (y=kx+l) of the tangent. It is simple then to find l.
 
Last edited:
I don't quite understand 2 :(

and where do u get k from?!
 
I don't quite understand 2 :(

Edit:oops soz, ment to edit and get rid of the part where I said "where do u get k from"

Edit2: aw crap, I still have no idea for those questions
 
Last edited:
The k (y=kx+l) of a tangent of a curve at a certain point is equal to the value of the derivative of the curve at that point. After solving the equation, you will get an x or two that satisfy it. Find the value of the derivative at those points. Find the l's by satisfying that the tangents intersect with the curve at that point.
 

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