Find parametric equations for the tangent line to the curve with parametric equations at the point with

Question

Find parametric equations for the tangent line to the curve with parametric equations    at the point with

Quizplus · Accepted Answer

The derivative of the parametric equations x(t) and y(t) can be found by differentiating each equation with respect to t and finding the ratio dy/dx:

x(t) = 11e^(t), y(t) = t^2 - 1
dx/dt = 11e^(t), dy/dt = 2t
dy/dx = (dy/dt)/(dx/dt) = (2t)/(11e^(t))

At the point (1,0), t=0, so dy/dx = 0/11 = 0.
The tangent line at this point will be horizontal, so we can set dy/dt = 0 and solve for t:

dy/dt = 2t = 0  ==>  t=0

Substituting t=0 into the x and y equations gives the point (11,-1).

To find the parametric equations for the tangent line, we can use the point-slope form:

y - y0 = m(x - x0), where (x0, y0) is the point on the line and m is the slope.

Since the slope at the point (1,0) is 0, the equation becomes:

y + 1 = 0(x - 11)  ==>  y = -1

This gives us the equation for the horizontal line passing through (11,-1). To write this in parametric form, we can use x(t) = 11e^(t) as the parameter:

x(t) = 11e^(t), y(t) = -1

Therefore, the answer is C.

Find Parametric Equations for the Tangent Line to the Curve