Answer
$ a_N = \frac{ 4e^x}{\left[1 + e^{2x}\right]^{3/2}}$
Work Step by Step
Step 1 Given: \[ x = \ln y, \quad \text{with fixed speed 2} \] Step 2 Since: \[ y = e^x \] \[ \kappa =\frac{ \left| \frac{d^2y}{dx^2} \right|}{ \left[1 + \left(\frac{dy}{dx}\right)^2\right]^{3/2}} = \frac{\left| e^x \right|}{ \left[1 + e^{2x}\right]^{3/2}} \] Then: \[ a_N = \kappa\left(\frac{ds}{dt}\right)^2 = \frac{ 4e^x}{\left[1 + e^{2x}\right]^{3/2}} \] Result \[ a_N = \frac{ 4e^x}{\left[1 + e^{2x}\right]^{3/2}} \]
