Find the equation of a tangent line to the curve y=e^(x^2) that also passes through the point (1,0).
Thanks in advance !
wow this was all so useless...
y=e^(x^2)
dy/dx = 2x(e^(x^2))
when x = 1, dy/dx = 2(e^1) = 2e
so you now have the slope, m = 2e and a point (1,1)
use your usual way to find the equation of the straight line.
Thanks !
y = m x + b
when y = 0, x=1
so b = -m
that m is the slope where they hit
slope of curve = dy/dx = 2x e^(x^2) = m = -b at crash
at that point
y = m x - m
and y = e^(x^2) = m x - m
and
m = 2 x e^(x^2)
so
e^(x^2) = 2 x^2 e^(x^2) - 2 x e^(x^2)
1 = 2 x^2 - 2 x
2 x^2 - 2 x - 1 =0
x = 1.366 for + root
m = 2 x e^(x^2) = 2.732 * 6.46 = 17.65
b = -17.65
They do not hit at (1,0)
or at (1,1)
as above, we want a line through (h,k) on the curve, and
2h^2-2h-1=0
h = (1+√3)/2
So now we have a slope m = e^h^2/(h-1) (x-1)
That gives us the line
y = e^((1+√3)/2)^2/( (1+√3)/2-1) (x-1)
See the graphs at
https://www.wolframalpha.com/input/?i=plot+y%3De%5E%28x%5E2%29%2C+y%3De%5E%28%281%2B%E2%88%9A3%29%2F2%29%5E2%2F%28+%281%2B%E2%88%9A3%29%2F2-1%29%28x-1%29
argggh
For these type of questions, I always told my students that the first thing they should do is test if the given point actually lies on the function.
I did not do that, and the point (1,0) was not on y = e^(x^2).
Looks like oobleck, our dependable tutor saw that.
My bad!
I'm a bit stuck on the same problem. I'm not sure how to solve it bc (1,0) isn't on the graph of y.
If the point (1,0) is not on the graph of y=e^(x^2), then there does not exist a tangent line passing through that point.
However, we can still find the closest point on the graph to the point (1,0), and then find the tangent line at that point.
To do this, we need to minimize the distance between the point (1,0) and a point on the curve y=e^(x^2).
The distance formula between two points (x1,y1) and (x2,y2) is given by:
d = sqrt((x2-x1)^2 + (y2-y1)^2)
So, we want to minimize:
d = sqrt((x-e)^2 + (e^(x^2))^2)
where (e,1) is a point on the tangent line to the graph y=e^(x^2) at the point (x,e^(x^2)).
To do this, we take the derivative of d with respect to x and set it equal to 0:
d' = (x-e) / sqrt((x-e)^2 + (e^(x^2))^2) + 2x e^(x^2) = 0
Solving for x will give us the x-coordinate of the closest point on the graph to (1,0). We can then find the y-coordinate and the slope of the tangent line at that point.