Limit problem with work so far

[sarahjohnson]

Let P be a point on the graph of

y=e^−x2/3


with NONZERO x-coordinate a. The normal line to the graph through P will have y-intercept b. (Note: The normal line at P is perpendicular to the tangent line at P.)

Diagram: (a) Calculate

lim_{a→ 0} b=

So far I have found the derivative of the equation: (-e^(-x^2/3)((-2/3)(x)) and take the negative reciprocal to find the slope of the normal line. Then I found the equation of the normal to be y=-1/(-e^(-a^2/3)((-2/3)(a))(X)+b
so if taking the limit wouldn't the slope of 1/0 which DNE?

Thanks for any help

 

[DrPhil]

Let P be a point on the graph of

y=e^−x2/3


with NONZERO x-coordinate a. The normal line to the graph through P will have y-intercept b. (Note: The normal line at P is perpendicular to the tangent line at P.)

Diagram: View attachment 3119(a) Calculate

lim_{a→ 0} b=

So far I have found the derivative of the equation: (-e^(-x^2/3)((-2/3)(x)) and take the negative reciprocal to find the slope of the normal line. Then I found the equation of the normal to be y=-1/(-e^(-a^2/3)((-2/3)(a))(X)+b
so if taking the limit wouldn't the slope of 1/0 which DNE?

Thanks for any help

Yes. Just looking at the figure should confirm your conclusion: as P approaches the y-axis, the slope of the the normal becomes vertical and b is completely undetermined.

There is ambiguity in where the "3" goes in the exponent. You need more parentheses or curly brackets:
e^{-x^2/3} or e^{-x^{2/3}}
when expressed as LaTeX code, becomes
\(\displaystyle \displaystyle e^{-x^2/3} \text{ or } e^{-x^{2/3}} \)

 

[sarahjohnson]

Yes. Just looking at the figure should confirm your conclusion: as P approaches the y-axis, the slope of the the normal becomes vertical and b is completely undetermined.

There is ambiguity in where the "3" goes in the exponent. You need more parentheses or curly brackets:
e^{-x^2/3} or e^{{-x^2}/3}
when expressed as LaTeX code, becomes
\(\displaystyle \displaystyle e^{-x^2/3} \text{ or } e^{-x^{2/3}} \)

Yes sorry It is e^(-x^2)/3 with the x squared. I tried DNE but it says it is the wrong answer...

 

[pka]

I tried DNE but it says it is the wrong answer...

You may try \(\displaystyle -\infty\). I don't like that as an answer. But it may work.

 

[sarahjohnson]

You may try \(\displaystyle -\infty\). I don't like that as an answer. But it may work.

So so far I have tried DNE, infinity, 0 and -infinity but all those are wrong. I don't understand what it can be...I thought it would be zero at first because X=0 when Y=b and that would make the equation y=0+b which is 0?

 

[DrPhil]

Let P be a point on the graph of

y=e^−(x2)/3


with NONZERO x-coordinate a. The normal line to the graph through P will have y-intercept b. (Note: The normal line at P is perpendicular to the tangent line at P.)

Diagram: View attachment 3119
(a) Calculate lim_{a→ 0} b=

So far I have found the derivative of the equation:
(-e^(-x^2/3)((-2/3)(x))
and take the negative reciprocal to find the slope of the normal line. Then I found the equation of the normal to be y=-1/(-e^(-a^2/3)((-2/3)(a))(X)+b
so if taking the limit wouldn't the slope of 1/0 which DNE?

Thanks for any help

\(\displaystyle \displaystyle y = e^{-x^2/3} \)

\(\displaystyle \displaystyle y\prime = e^{-x^2/3} \times (-2/3)\ x \) [I think you have an extra minus, The slope of the function is - if x is +.]

At point \(\displaystyle \displaystyle P = (a, e^{-a^2/3})\), the slope of the normal \(\displaystyle m\) is

\(\displaystyle \displaystyle m = -\dfrac{1}{y\prime} = \dfrac{3\ e^{a^2/3}}{2\ a } \)

Have you calculated the intercept, \(\displaystyle b\)? That is what the question asked for. Running the line backwards from point \(\displaystyle P\), that is, from \(\displaystyle x=a\), to \(\displaystyle x=0\),

\(\displaystyle \displaystyle b = e^{-a^2/3} - m\ a\)

That looks promising! perhaps, even though \(\displaystyle m\) is not defined, \(\displaystyle b\) IS defined?

 

[fcabanski]

Dr. Phil has the answer below.

 

[DrPhil]

So so far I have tried DNE, infinity, 0 and -infinity but all those are wrong. I don't understand what it can be...I thought it would be zero at first because X=0 when Y=b and that would make the equation y=0+b which is 0?

I would like to continue from my previous post, with some editing.

\(\displaystyle \displaystyle y(x) = e^{-x^2/3}\)

\(\displaystyle \displaystyle y\prime(x) = e^{-x^2/3} \times \dfrac{-2x}{3}\)

At point \(\displaystyle \displaystyle P = (a,\ y(a)) = (a,\ e^{-a^2/3})\), the slope of the normal \(\displaystyle \ m\) is

\(\displaystyle \displaystyle m(a) = - \dfrac{1}{y\prime (a)} = + e^{+a^2/3} \times \dfrac{3}{2a}\)

Now it is clear that \(\displaystyle m(a)\) becomes infinite (with the same sing as \(\displaystyle a\))as \(\displaystyle a \to 0\), but that should not stop us from trying to find \(\displaystyle b\). Running the normal line backwards from point \(\displaystyle P\) to the y-axis, that is from \(\displaystyle x=a\) to \(\displaystyle x=0\), we have

\(\displaystyle \displaystyle m(a) = \dfrac{\Delta y}{\Delta x} \implies \Delta y = (-a)\ m(a) = -e^{a^2/3} \times \dfrac{3}{2}\)

Thus \(\displaystyle \displaystyle b(a) = e^{-a^2/3} - e^{a^2/3} \times \dfrac{3}{2}\)

\(\displaystyle \displaystyle \lim_{a\to 0} b(a) = 1 - \dfrac{3}{2} = -\dfrac{1}{2} = b(0) \)

Note that limit is the same as \(\displaystyle a\) approaches \(\displaystyle 0\) from either side.