What is the slope of #f(t) = (t-2t+4,t^2)# at #t =0#?

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Answer 1 · 2016-11-22T00:28:51

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Define #x(t), y(t)# as follows

#{ (x(t)=t-2t+4=4-t), (y(t)=t^2) :} => f(t) = (x(t), y(t)) #

Differentiating #x(t)# wrt #t# we get:

# dot x(t) = dx/dt = -1 #

Differentiating #y(t)# wrt #t# we get:

# dot y(t) = dy/dt=2t #

Then the applying the chain rule;

# f'(t) = dy/dx = dy/dt*dt/dx = (dy/dt)/(dx/dt) #
# :. f'(t) = ( 2t ) / ( -1 ) #
# :. f'(t) = -2t #

So when #t=0# we get:

# f'(0) = 0 #

Hence, The slope of #f(t)# is #0# at #t=0#