What is the minimum distance from the origin to the curve #y=e^x#?

1 Answer
Apr 12, 2017

Minimum distance # = 0.779767 # (6dp)

Explanation:

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Let #P(x,y)# be a generic point on the given curve

By Pythagoras;

# OP^2 = x^2 + y^2 #
# " "= x^2 + (e^x)^2 #
# " "= x^2 + e^(2x) #

Out aim is to minimise #OP# which is the same as minimizing #OP^2#. Let use denote #OP^2# by #l#, that is:

# l=x^2 + e^(2x) #

Our aim is to minimize by #l# , ie find a critical point. Differentiating wrt #x# we get:

# (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 #x=-0.42630275#

We should check if this value of #x# is associated with a minimum or maximum, by performing the second derivative test:

Differentiating again wrt #x# we get:

# (d^2l)/dx^2 = 2 + 4e^(2x) #

When #x=-0.42630275 => (d^2l)/dx^2 >0 # confirming a minimum

So the minimum distance occurs with #x=-0.42630275# which gives us:

# OP^2 = x^2 + e^(2x) #
# " " = 0.60803679 #

# :. OP = 0.779767 # (6dp)