Suppose we have a subatomic particle that follows a path des

Suppose we have a subatomic particle that follows a path described by x = e^t, y = (t - l)^2 for all t Elementof R. Find the equation of the tangent line to the curve when t = 2. When would the tangent line to the curve be horizontal, if ever? When would it be vertical, if ever? When will the path of the particle be concave up?

Solution

t=ln(x)

y=(ln(x)-1)^2

a)

dy/dx=2(ln(x)-1)/x

At, t=2, x=e^2

dy/dx=2(ln(e^2)-1)/e^2=2/e^2

dy/dx=2/e^2

x=e^2,y=1

y-1=2(x-e^2)/e^2

e^2(y-1)=2(x-e^2)

b)

dy/dx=2(ln(x)-1)/x

So tangent is horizontal at

ln(x)=1 ie x=e

x=e ie t=1, for horizontal tangent

For tangent to be vertical we need dy/dx=infinity which happens at

x=0 which never happens because e^t never takes the value 0

c)

y\'=2(ln(x)-1)/x

y\'\'=2/x^2-2(ln(x)-1)/x^2

For concave up we need : y\'\'>0

2-2(ln(x)-1)>0

1-ln(x)+1>0

ln(x)<2

x<e^2

ie t<2

So the path of the particle is concave up for t<2