AboutAlon Mandes Expertise Kind of questions I can answer : Limits, Derivatives, Integration, Implicit functions, continuousity, differentiation ,Extremum problems, Lagrange multipliers, Gradients, Surface integrals, Multi variables functions ,Multi variables Integrals,Complex variables ,Complex functions, Curves, Trajectory integrals & Vector analyse,Divergence,Rotor & word problems.
Kind of question I can't answer : Economics,Combinatorics,infinite series & convergence ,Statistics & Probabilities .
Experience 1. I'm a team member of mathnerds (math site for answering questions)
2. I'm a team member in the Student's Union of the Technion, helping
students who have problems in mathematics.
3. 2 years of experience as a math teacher in college.
4. I give free homework help for high school students in
Mathematics & Physics.
5. I teach part time in collage the subjects : "Digital Signal Processing" , "Random Signals & Noise" ,
"Complex Functions".
Question QUESTION: Question given:
The curve y=lnx^3 has one tangent line that passes through the point (0,1). Find the equation of this tangent line.
My first step was to take the derivative of the curve's function, in which I got y'=3/x. From there, to get the slope, substituting in the x value of 0 gives me an undefined slope.
I'm not sure how to solve this question if I'm given an undefined slope. (Note: This question is supposed to be worth 9 marks on an assignment)
Any help would be greatly appreciated. :)
Thank you in advance
ANSWER: Hi Julia ,
The 1st step you did was right, the 2nd not .
The point (0,1) is not a point on the curve !!! its an external
point that the tangent line passes through.
So, The facts that we know :
1. The tangent line has slope of xo/3
2. The tangent line passes through the point (xo,3Lnxo)
3. The tangent line passes through the point (0,1)
Suppose that the equation of the tangent line is y=mx+n , then
m=3/xo .
According to (3) : n=1
According to (2) : 3lnxo=3xo/xo + 1 -> lnxo=4/3 -> xo=e^(4/3) .
Therefore the equation of the Tangent Line is :
y=[3/e^(4/3)]x+1
Alon.
---------- FOLLOW-UP ----------
QUESTION: I understand the path that you took to solve the question
now, but I have a few questions regarding some of the
values.
When the derivative is 3/x, how did you receive xo/3 for the
slope of the tangent line? Where did the o come from? (You
note that the slope of the tangent line is xo/3 in point 1,
but below point 3, note that m=3/xo). Please clarify.
Last question would be how did you find that the tangent
line passes through the point (xo,3Lnxo).
Thanks again for your help on this question!
Answer Ok Julia, first I made assumption that:
The Tangent Line touch the curve at point A which has an x
coordinates of x=xo.
Thus :
1. this point has y coordinate of yo, that can be written as
function of xo : yo=3Lnxo. Because this point exists on the
curve.
2. The point of touching is common point for both the curve
& the line.
3. The slope of the line is the derivative of the curve at the
touching point (xo,3Lnxo). That means that m=3/xo .
Now,
Our line passes through 2 points : (0,1) & (xo,3Lnxo) .
All that left is simple algabric equation to know that value of xo.
