Find dy/dx and d2y/dx2,
and find the slope and concavity (if possible) at the given value of the parameter. (If an answer does not exist, enter DNE.)
Parametric Equations Point
x = 5t, y = 6t − 1
t = 2

Step-by-step explanation:

Given the parametric equation points x = 5t, y = 6t − 1  when t = 2

From x = 5t, t = x/5. Substituting t = x/5 into the second equation y = 6t − 1 we will have;

y = 6(x/5) - 1

y = 6/5 x - 1

The derivative of y with respect to x i.e dy/dx = 6/5 - 0. (Note that differential of any constant is zero).

dy/dx = 6/5

d²y/dx² = d/dx(dy/dx)

d²y/dx² = d/dx(6/5)

Since 6/5 is a constant, the derivative of 6/5 with respect to x will be zero.

d²y/dx² = 0.

Since the first derivative and the second derivative are both constant then, the slope m at the given parameter will be 6/5.

m = dy/dx = 6/5

The concavity is the value of the second derivative at the given value of the parameter.

The concavity d²y/dx² = 0.

the correct answer is 828.07