What is the instantaneous velocity of an object with position at time t equal to # f(t)= (tsqrt(t+2),t^22t) # at # t=1 #?
2 Answers
We have the parametric curve which define the movement of your material point
We need to derive once to have the parametric curve which define the velocity of your material point
Your material point
If you calculate the function of distance in time
Explanation:

The instantaneous velocity of an object is the derivative of distance as a function of time.

The distance traveled by the object can be taken as its the distance between (0,0) and the current position of the object.

If distance as a function of time is
#y=f_d(t)# , velocity is denoted#v# , the function of position in time is#f(t)# then:
#v=(dt)/(dy)#
#f_d(t)= sqrt((t^22t)^2+(tsqrt(t+2))^2)#
#f_d(t)=sqrt((t^44t^3+4t^2)+(t^3+2t^2))#
#f_d(t)=sqrt(t^43t^3+6t^2)=(t^43t^3+6t^2)^(1/2)# 
#v=(dt)/(dy)=d((t^43t^3+6t^2)^(1/2))/(dy)#
#v=1/2[(t^43t^3+6t^2)^(1/2)][4t^39t^2+12t]#
#v=(4t^39t^2+12t)/(2sqrt(t^43t^3+6t^2))#
#v=(4t^29t+12)/(2sqrt(t^23t+6))# 
When
#t=1#
#v=(49+12)/(2sqrt(13+6))#
#v=7/4#
I think?