Find 'a' so tangent to y=x^2 e^x at x = a passes thru origin

[jwpaine]

Find the value of a =/ 0 such that the tangent line to the graph of \(\displaystyle \L y = x^{2}e^{x}\) at x=a passes through the origin.

This is my thought process so far:

For m = y' = (a^2 + 2a)e^a, Find a such that y=mx+b contains the point (0,0)

I just don't know how to do it.... but the above quote I came up with and I believe it's the path to the answer.

John

 

[pka]

\(\displaystyle \L y = x^2 e^x \quad \Rightarrow \quad y' = 2xe^x + x^2 e^x\)

 

[galactus]

Hey there JW.

The thing is about this problem is that we don't know where it's tangent.

You can use the old line formula: \(\displaystyle \L\\y-y_{1}=m(x-x_{1})\)

\(\displaystyle \L\\y=x^{2}e^{x}, \;\ y_{1}=0, \;\ x_{1}=0, \;\ m=(x^{2}+2x)e^{x}\)

Then we have:

\(\displaystyle \L\\a^{2}e^{a}-0=(a^{2}+2a)e^{a}(a-0)\)

Now, solve for a.

 

[jwpaine]

EDIT: OK - thanks Galactus, I knew I would be needing point-slope... I just didn't know how to apply it!

\(\displaystyle \L y = x^2 e^x \quad \Rightarrow \quad y' = 2xe^x + x^2 e^x\)

Yes, I know that the derivative of y' = \(\displaystyle \frac{d}{dx}x^{2}e^{x}\,+\, x^{2}\frac{d}{dx}e^{x}\,\,\,=\,\,\, 2xe^{x} + x^{2}e^{x}\)

I already found the derivative when I said for x = a, y' = \(\displaystyle (a^2\,+\,2a)e^{a}\)

I just don't know how to find the value for x = a, such that a line that passes through (0,0) has y'(a) as it's slope.

John.

 

[pka]

I just don't know how to find the value for x = a, such that a line that passes through (0,0) has y'(a) as it's slope.

My goodness, a=0. Or x=0 at (0,0).
What is the problem?

 

[skeeter]

you have two points ...

\(\displaystyle \L (a, a^2 e^a)\) and \(\displaystyle \L (0,0)\)

since \(\displaystyle \L a \neq 0\), slope between these two points is \(\displaystyle \L \frac{a^2 e^a}{a} = a e^a\)

slope at any point on the curve is \(\displaystyle y' = e^x(2x+x^2)\).

at \(\displaystyle \L x=a\), \(\displaystyle \L y'(a) = e^a(2a + a^2)\)

set the slopes equal ...

\(\displaystyle \L a e^a = e^a(2a + a^2)\)

since \(\displaystyle \L e^a \neq 0\), \(\displaystyle \L a = 2a + a^2\)

\(\displaystyle \L 0 = a + a^2\)

\(\displaystyle \L 0 = a(1 + a)\)

since \(\displaystyle \L a \neq 0\) ... \(\displaystyle \L a = -1\).

 

[galactus]

Here's your graph JW.

 

[jwpaine]

Thanks guys! I've only been in my University's calculus 1 for a few weeks.....and I know this is real basic stuff....but it's still so much fun to learn.

 

[galactus]

Yes, it is fun. When you get to related rates and the max/min problems, it'll get 'funner'. In Calc II, integration is a hoot.