CALCULUS I ASSIGNMENT #3 SOLUTIONS

A/ Find y'. Do not simplify beyond basic algebra.

1) With implicit differentiation and product rule where needed, we get

2xy 3 + 3x 2 y 2 y' - 5x 4 sin 2y - (2x 5 cos 2y)y' + y 2 + 2xyy' = 0

2) Let's first put the "y" terms together:

y + 4 cos 3y = 2 tan x
y'(1 - 12 sin 3y) = 2 sec 2 x

3) Again, we'll put all the terms together on one side:

x 2 sin y - y 3 cos 2 x = 0
2x sin y + x 2 y' cos y - 3y 2 y' cos 2 x + 2y 3 cos x sin x = 0

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B/ 1) With implicit differentiation and product rule where needed, we get

3x 2 y + x 3 y' + 28y 3 y' + 1 = 0

The tangent equation then is

2) f (-1) = -6, and f (2) = 6, also, f '(c) = 3c 2 + 1.
So, 3c 2 + 1 = 12/3 = 4, so c 2 = 1 and therefore c = 1.

3) Set r = 2 inches, and dr = 0.23 inches.
Since V = 4/3
o r 3, then dV = 4o r 2 dr u dV = 4o (4 in2)(0.23 in) = 3.68 o in3.

4) dx/dt = 4 cm/sec, dy/dt = -2 cm/sec. We want dz/dt at x =5 cm. and y = 3 cm.
Since it's a right triangle, we know x² + y² = z².
Differentiating with respect to time we get 2x dx/dt + 2y dy/dt = 2z dz/dt

We divide through by 2, then by z to get

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C/

1) Curve Sketching (M2I2ACIDS explanation)

This function has a hole in the graph at x = - 3 since both numerator and denominator = 0.
We do the analysis on , then find the limit as x
t - 3 for the original function.

Step 1: intercepts & asymptotes
When x = 0, g(0) = -1/3 -- so (0, -1/3) is y-intercept, there is no x-intercept or zero.
VA: x = 3, HA: y = 0

Step 2: first derivative

g '(x) < 0 ¼ x since the numerator is a negative constant, denominator is a perfect square.
g(x) is decreasing
¼ x, since g '(x) ! 0, no max, no min.

Step 3: second derivative

g "(x) ! 0, no points of inflection.
g "(x) < 0 when x < 3, concave down.
g "(x) > 0 when x > 3, concave up.

Hole in the graph at x = -3

, hole at (-3, 1/6) -- only 1 hole -- so no symmetry.

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2) Find the vertical, horizontal and oblique asymptotes (if any) for:
 a)Vertical: x = 7/3 Horizontal: y = 0 b) Vertical: x = -8/3 Horizontal: y = 2/3 c) Vertical: Horizontal: y = 0 d) Vertical: none Horizontal: y = 3/5 e) Vertical: x = -2 Oblique: y = x - 4 f) Vertical: x = 0 Oblique: y = x + 3

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M2 I2 ACIDS stands for:

Maximum, Minimum, Intercepts, Inflection points, Asymptotes,
Concavity, Increasing intervals, Decreasing intervals and Symmetry.
For notes on this topic see lesson on curve sketching.