Can you help me with these basic derivative Calculus Problems?

Question

1. The equation of motion of a particle, where s is in meters and t is in seconds, is given below.
s = 7 t ^3 + 3 t ^2 + 8 t + 3
(a) Find the velocity and acceleration as functions of t.
b) Find the acceleration after 3 s.

2. Find the points on the curve given below, where the tangent is horizontal.
y = 7 x ^3 + 2 x ^2 - 4 x + 7
P1 = ( , ) (smaller x-value)
P2 = ( , ) (larger x-value)

3. Find an equation of the line that is both normal to the given parabola and also parallel to the given line.
yp = 6 x ^2 - 4 x + 5
8 x - 5 y = 2

Really, thank you to eternity and back!

By the way, please include an actual answer, so that I can check my own frazzled mess of work!

Answer 1

a) The velocity is the first derivative.
The acceleration is the second derivative.

v = s' = 21t^2 + 6t + 8
a = s'' = 42t + 6

b) Find the acceleration after 3 s means you need to replace t with 3 in the acceleration function.

2. Take the derivative
y' = 21x^2 + 4x - 4
Set equal to zero and solve for x.

3. Parallel to 8x - 5y = 2 means it would have the same slope.
The slope of 8x - 5y = 2 is m = 8/5
If it is normal to the curve, then the derivative will be the negative reciprocal of the slope.
Find y' and set if equal to -5/8

Answer 2

1) f(x) = motion
f'(x) = instantaneous velocity
f''(x) = acceleration

2) The tangent line is horizontal when f'(x) = 0.

3) normal = perpendicular
Find the slope of the line, then get the slope of the perpendicular line, then set f'(x) = perpendicular slope and solve for x to find out where this happens.