Why aren't the derivatives equal?

[Calpalned]

Homework Statement

Homework Equations

see above

The Attempt at a Solution

After getting $v = |\frac{dz}{dt}|=\frac{3}{t^2+1}$ why can't I simply take the derivative of that with respect to $t$ to get the acceleration?
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[Mark44]

Homework Statement

View attachment 90178

Homework Equations

see above

The Attempt at a Solution

After getting $v = |\frac{dz}{dt}|=\frac{3}{t^2+1}$ why can't I simply take the derivative of that with respect to $t$ to get the acceleration?

The notation is a little weird. What they're calling v is actually the magnitude of the velocity, not the velocity itself, and $\frac{dz}{dt}$ is the velocity. To get the acceleration, you need to differentiate the velocity (not its magnitude) with respect to t.

 

[Calpalned]

Please see post #3 here https://www.physicsforums.com/threads/physical-applications-of-complex-numbers.837585/

 

[Calpalned]

Okay, so simply put the derivative of the magnitude of velocity is not the magnitude of acceleration? $\frac{d}{dt}(|\frac{dz}{dt}|)\neq|\frac{d^2 z}{dt^2}|$

 

[Calpalned]

The notation is a little weird. What they're calling v is actually the magnitude of the velocity, not the velocity itself, and $\frac{dz}{dt}$ is the velocity. To get the acceleration, you need to differentiate the velocity (not its magnitude) with respect to t.

But the question asks for the magnitude of acceleration

 

[Mark44]

Please see post #3 here https://www.physicsforums.com/threads/physical-applications-of-complex-numbers.837585/

I posted in that thread.

Okay, so simply put the derivative of the magnitude of velocity is not the magnitude of acceleration? $\frac{d}{dt}(|\frac{dz}{dt}|)\neq|\frac{d^2 z}{dt^2}|$

Correct.

 

[Mark44]

But the question asks for the magnitude of acceleration

Right, but they first have to get the acceleration, and then get its magnitude.