a) Compute the derivative of g(t) = 11t3 + 9t2 + 13 as a function of t. b) What is the slope of the tangent line to this parabola at t = 2? c) Write the equation of the tangent line to g(t) at t = 2.

a) To compute the derivative of the function g(t) = 11t^3 + 9t^2 + 13, we can use the power rule of differentiation. According to the power rule, if we have a term of the form f(t) = ct^n, the derivative is given by f'(t) = nct^(n-1). Let's differentiate each term of g(t) separately: g'(t) = d/dt (11t^3) + d/dt (9t^2) + d/dt (13) g'(t) = 33t^2 + 18t + 0 Simplifying, we get: g'(t) = 33t^2 + 18t So, the derivative of g(t) with respect to t is g'(t) = 33t^2 + 18t. b) To find the slope of the tangent line to the parabola at t = 2, we substitute t = 2 into the derivative we found in part a: g'(2) = 33(2)^2 + 18(2) g'(2) = 33(4) + 18(2) g'(2) = 132 + 36 g'(2) = 168 Therefore, the slope of the tangent line to the parabola at t = 2 is 168. c) To write the equation of the tangent line to g(t) at t = 2, we use the point-slope form of a linear equation: y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope of the line. Given that t = 2 and g(t) = 11t^3 + 9t^2 + 13, we have: x1 = 2 y1 = g(2) = 11(2)^3 + 9(2)^2 + 13 = 88 + 36 + 13 = 137 Using the point-slope form with the slope we found in part b: y - 137 = 168(x - 2) Expanding and simplifying: y - 137 = 168x - 336 y = 168x - 199 Therefore, the equation of the tangent line to g(t) at t = 2 is y = 168x - 199.