Sounds good. Why have you not done it?2. Find the equation of the tangent to the curve y = x3 - 9x2 + 20x - 8 at the point (1, 4). At what points of the curve is the tangent parallel to the line y = -4x + 3?
First, I used the method from the previous question to obtain the equation for the parallel line. For the next part of the question, I equated the gradient function to the gradient of the parallel line and solved for x, substituting the newly found x values into the original curve equation to get my y-coordinates.
No, it's not. Putting h= 3 into h(t) gives h(3)= 82 meters- not "meters per second", because, as you were told, h(t) is the height of the plane at t seconds, NOT the rate at which that height is changing.3. A model plane flying level at 250 m above the ground suddenly dives... blahablahhal... h(t) = 8t2 - 80t + 250. Find the rate at which the plane is losing height at t = 3.
I simply substituted t = 3 into the original equation and solved, giving my answer as 82 ms-1. Is this question as simple as that?
Yes, that is correct. Why didn't you do that for the previous question?4. A car starts from rest and moves a distance s (m) in t (s), where s = (1/6)t3 + (1/4)t2. What is the acceleration when t = 2?
I know that velocity = ds/dt and acceleration = dv/dt, so I found that s' = (1/2)t2 + (1/2)t and v' = t + (1/2). Then, to answer the question, I simply substituted t = 2 into the equation for acceleration and solved, giving my (5/2) ms-2.
Thanks in advance. Your feedback is greatly appreciated.