How do you find the equation of the tangent to the curve #x=t^4+1#, #y=t^3+t# at the point where #t=-1# ?

Question

How do you find the equation of the tangent to the curve #x=t^4+1#,  #y=t^3+t# at the point where #t=-1# ?

Christopher Agresta · Accepted Answer

To find the equation of the tangent line, we need the slope 
#m = dy/dx# and the point of tangency #(x_o,y_o)#.  Then the equation is the usual # y-y_o = m(x - x_o).# We have the parametric curve #x = t^4+1, y = t^3+t#, so we compute #dx/dt = 4t^3 and dy/dt = 3t^2+1.# The chain rule #dy/dt = dy/dx*dx/dt# says that #dy/dx = (dy/dt)/(dx/dt).#  So we use the derivatives of the parametric equations: #dy/dx = (3t^2+1)/(4t^3).# Now put in #t = -1# and find: #m = dy/dx = (3*(-1)^2+1)/(4*(-1)^3) = (3+1)/(-4) = -1.# Also at #t=-1# the original equations give 
#(x_o,y_o) = ((-1)^4+1,(-1)^3+(-1))
=(1+1,(-1)+(-1))=(2,-2)# Now we put in the info for the tangent line:
#y-y_o=m(x-x_o)#
#y-(-2)=(-1)(x-2)#  or
#y+2=-x+2#  or just plain old #y=-x#. \ Another great answer from the modest dansmath! /

Emma Aldridge · Answer

undefined To find the equation of the tangent to the curve $ x = t^4 + 1 $ and $ y = t^3 + t $ at the point where $ t = -1 $, you first find the coordinates of the point of tangency by substituting $ t = -1 $ into the equations for $ x $ and $ y $. This gives $ x = (-1)^4 + 1 = 2 $ and $ y = (-1)^3 + (-1) = -2 $.

Next, you find the derivative of $ y $ with respect to $ x $, $ \frac{dy}{dx} $, by differentiating $ y $ with respect to $ t $ and dividing by the derivative of $ x $ with respect to $ t $, $ \frac{dy}{dt} \div \frac{dx}{dt} $.

$$ \frac{dy}{dt} = 3t^2 + 1 $$
$$ \frac{dx}{dt} = 4t^3 $$

Now, substitute $ t = -1 $ into these derivatives:
$$ \frac{dy}{dt} = 3(-1)^2 + 1 = 4 $$
$$ \frac{dx}{dt} = 4(-1)^3 = -4 $$

Then, find the slope of the tangent line by evaluating $ \frac{dy}{dx} $ at $ t = -1 $:
$$ \frac{dy}{dx} = \frac{4}{-4} = -1 $$

Finally, use the point-slope form of the equation of a line $ y - y_1 = m(x - x_1) $, where $ (x_1, y_1) $ is the point of tangency and $ m $ is the slope:
$$ y - (-2) = -1(x - 2) $$

Simplify to get the equation of the tangent line:
$$ y + 2 = -x + 2 $$
$$ y = -x $$

Therefore, the equation of the tangent to the curve at the point where $ t = -1 $ is $ y = -x $.