Thread: Calculus

Im doing some calculus revision for my up coming end of year exam, and ive gotten myself a bit confused on this question:

If x= f(t), y=g(t) and d(squared)y/dx(squared)= 0, prove that dx/dt times d(squared)y/dx(squared)=dy/dt times d(squared)x/dt(squared)

sorry that looks a little confusing. I dont know how to type squared.

I mostly understand how to go about doing the question, i just wonder what it means that d(squared)y/dx(squared)= 0. I know that if the second derrivative is less than zero, the turning point will be a maxium, and that if the second derrivative is more than zero, the turning point will be a minimum, so what will it be when the second derrivative = 0?

Originally Posted by Surreptitous_Smiles

Im doing some calculus revision for my up coming end of year exam, and ive gotten myself a bit confused on this question:

If x= f(t), y=g(t) and d(squared)y/dx(squared)= 0, prove that dx/dt times d(squared)y/dx(squared)=dy/dt times d(squared)x/dt(squared)

sorry that looks a little confusing. I dont know how to type squared.

I mostly understand how to go about doing the question, i just wonder what it means that d(squared)y/dx(squared)= 0. I know that if the second derrivative is less than zero, the turning point will be a maxium, and that if the second derrivative is more than zero, the turning point will be a minimum, so what will it be when the second derrivative = 0?

If the second derivative is 0 it means that you do not know whether it is a minimum of maximum point and you have to test it by other means .

Oh okay. Thanks so much!