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

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

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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,

3) Set *r = 2 inches*, and *dr =* *0.23 inches*.

Since V = 4/3o r^{ 3}, then *dV = 4**o** r ^{ 2} dr *

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 = 7/3Horizontal: y = 0 |
b) Vertical: x = -8/3Horizontal: y = 2/3 |
c) Vertical: Horizontal: y = 0 |

d) Vertical: noneHorizontal: y = 3/5 |
e) Vertical: x = -2Oblique: y = x - 4 |
f) Vertical: x = 0Oblique: y = x + 3 |

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Back to Calculus I MathRoom Index

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**M _{2 }I_{2 }ACIDS** stands for:

**M**aximum, **M**inimum, **I**ntercepts, **I**nflection points, **A**symptotes,

**C**oncavity, **I**ncreasing intervals, **D**ecreasing intervals and **S**ymmetry.

For notes on this topic see lesson on curve sketching.