What is the slope of #f(t) = (t^2+2t,2t-3)# at #t =-1#?

Question

Ezekiel Alexander · Accepted Answer

The slope of #f(t)# is infinite when #t=-1# (ie the tangent is vertical).

We have # f(t)=(x(t), y(t)) # Where #x(t)=t^2+2t#, #y(t)=2t-3#

The by the chain rule, # dy/dx = dy/dt * dt/dx = (dy/dt) / (dx/dt) #

# x(t)=t^2+2t => dx/dt=2t+2 #
# y(t)=2t-3 => dy/dt = 2 #

So,

# dy/dx = 2 / (2t+2) #
# :. dy/dx = 1 / (t+1) #

When # t=-1 => dy/dx = 1/0 = oo #, so the slope of #f(t)# is infinite when #t=-1# (ie the tangent is vertical).

This can be visualised by looking at the graph of #f(t)#

Adam Adkins · Answer

undefined To find the slope of the function $ f(t) = (t^2 + 2t, 2t - 3) $ at $ t = -1 $, we need to calculate the derivative of each component function with respect to $ t $, and then evaluate it at $ t = -1 $.

The derivative of $ t^2 + 2t $ with respect to $ t $ is $ 2t + 2 $, and the derivative of $ 2t - 3 $ with respect to $ t $ is $ 2 $.

Evaluating these derivatives at $ t = -1 $, we get $ 2(-1) + 2 = 0 $ for the first component and $ 2 $ for the second component.

Therefore, the slope of the function $ f(t) = (t^2 + 2t, 2t - 3) $ at $ t = -1 $ is $ \frac{\text{change in } y}{\text{change in } x} = \frac{2}{0} $, which is undefined.