What is the minimum distance from the origin to the curve #y=e^x#?
1 Answer
Minimum distance
Explanation:
Let
By Pythagoras;
# OP^2 = x^2 + y^2 #
# " "= x^2 + (e^x)^2 #
# " "= x^2 + e^(2x) #
Out aim is to minimise
# l=x^2 + e^(2x) #
Our aim is to minimize by
# (dl)/dx = 2x + 2e^(2x) #
At a critical point we require the derivative to be zero:
# (dl)/dx = 0 => 2x + 2e^(2x) = 0#
# :. 2(x + e^(2x)) = 0#
# :. x + e^(2x) = 0#
If we look at the graph of this function
graph{x+e^(2x) [-5, 5, -10, 10]}
we can see there is one solution. Using Newton-Rhapson we can numerically calculate the solution, which gives
We should check if this value of
Differentiating again wrt
# (d^2l)/dx^2 = 2 + 4e^(2x) #
When
So the minimum distance occurs with
# OP^2 = x^2 + e^(2x) #
# " " = 0.60803679 #
# :. OP = 0.779767 # (6dp)