What is the slope of #f(t) = (t+1,t)# at #t =0#?

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Answer 1 · 2017-12-10T23:03:56

The slope of the gradient of #f(t)# is #1# when #t=0#

We have:

# f(t) = { (x(t)=t+1), (y(t)=t) :} #

Differentiating wrt #t# we have:

# dx/dt = 1 # and # dy/dt=1 #

Then by the chain rule we have:

# dy/dx =(dy/dt) // (dy/dt)#
# \ \ \ \ \ =1#

So the slope of the gradient of #f(t)# is #1# when #t=0#

We can also see this is the case if we consider isolate the parameter #t# giving

# x = y+1 => y = x-1 #

Which represents a straight line with gradient #1#.