What is the slope of #f(t) = (t^2-tsqrtt,t/2)# at #t =3#?

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Answer 1 · 2017-06-15T19:37:39

#m=-8.0386#

This parametric equation says that the #x# value of the function #f# is

#x(t)=t^2-tsqrt(t)#

When #t=3#

#x(3)=3^2-3sqrt(3)=9-3sqrt(3)~~3.8038#

The derivative is

#dx/dt=2t-(t(1/2t^(-1/2))+sqrt(t))#

When #t=3#, the derivative of #x# becomes

#dx/dt=2(3)-(3(1/2(3)^(-1/2)+sqrt(3))~~-0.0622#

It also says the #y# value of the function is

#y(t)=t/2#

When #t=2#, #y(2)=1#

The derivative of #y# is

#dy/dt=1/2~~0.5#

The slope of the function is

#m=(dy)/dx=(dy/dt)/(dx/dt)# where #dx/dt != 0#

Plugging in our derivatives from above at #t=3#, we get

#m=(dy)/dx= 0.5/-0.0622~~-8.0386#

We now have #x_1=3.8038#, #y_1=1#, and #m=-8.0386#