|Calculus I Final Exam Solutions|
C/ Curve Sketching
Now that we know
Setting V'(r) = 0 will give us the critical points.
When r = 14/3 cm the cylinder will have a maximum volume.
The height will be 5 cm.
E/ Increments and Differentials
F/ Related Rates
G/ Mean Value Theorem
13) Since P is the vertex of f, at (2, 3) and
since we move 36 units horizontally, Q is at .
T is the mean value point -- the point where the tangent is parallel to the chord.
So, since b - a = 36 , and f(b) - f(a) = 3/4, then 36 f '(c) must = 3/4.
When we solve for c, we get c = 11 and f(11) = 27/8. Point T is at .
|a) F(x) = 3x 3 / 2 + 2x 1 / 2 + C||b) F(x) = 2/3 sin 3x + 3/2 cos 2x + C|
|c) F(x) = 24/5 x 5 / 3 + 15/2 x 2 / 3 + C|
|15) a) f " (x) = 3x² + 2||f '(x) = x³ + 2x - 1||f(x) = 1/4 x 4 + x² - x + 4|
16) v(t) = 2t - 3t² - 5, therefore s(t) = t² - t³ - 5t + 4
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