I'm currently a student in an online AP BC calc class, and I'm having a little trouble with a question about derivatives and parametric equations.
Anyways, the problem is written as follows:
"The position of an object is described by the parametric equations $x=\ln t$ and $y=5t^2$. What is the acceleration of the object in $m/sec^2$ when $t=2$?"
I've been wondering the whole time what I should be calculating – my sister says I should be calculating $\frac{d^2y}{dx^2}$, but personally I think I should be calculating $\sqrt{\frac{d^2y}{dt^2}+\frac{d^2x}{dt^2}}$ for t=2 – I'm a little confused, because I thought acceleration had direction as well, meaning that a single scalar would not accurately describe the acceleration of a particle with motion described in x and y axis?
Thanks.
Correction:$\sqrt{\frac{d^2y}{dt^2}+\frac{d^2x}{dt^2}}$ should be $\sqrt{(\frac{d^2y}{dt^2})^2+(\frac{d^2x}{dt^2})^2}$
EDIT: After submission and feedback it appears that I have gotten the problem wrong. The answer and explanation given are as follows:
"We know that dy/dx=(dy/dt)/(dx/dt). x=lnt, therefore dx/dt=1/t*y=5t^2, therefore dy/dt=10t.
So, dy/dx=(dy/dt)/(dx/dt)=(10t)/(1/t)=10t^2
and
a=d^2y/dx^2=(d/dx(dy/dx))/(dx/dt)=(20/t)/1/t)=20t^2.
So, when t=2, a=20(2)^2=80m/sec^2."
I do disagree with this answer, which assumes that a=d^2y/dx^2. At the time we were learning about parametric differentiation – I guess I should have known better, but something bothered me about the intuition of what d^2y/dx^2 represents. I have talked a bit with my teacher about the problem, but I'm guessing that I might want to call her to understand what she's trying to say better.
In short, does a=d^2y/dx^2, or does a=sqrt((d^2y/dt^2)^2+(d^2x/dt^2))? And if a is not equal to d^2y/dx^2, how should I explain that to my teacher?
Thanks.
Best Answer
You are right on all counts!
Acceleration is indeed a vector: $\vec a = \left(\frac{d^2x}{dt^2},\frac{d^2y}{dt^2}\right)$. But if we only care about the magnitude of the acceleration, then we take the magnitude of the vector, which gives your expression with the square root.
Edit: As Matt pointed out below, your expression for $||\vec a||$ as it stands is actually not quite right, but I assume this is a typo!