Study Guide PDF

Math Study Guide

Part 1: Differentiation and Implicit Differentiation

1. Differentiate the Following Functions:

Function 1:

• Given: ( y = e^{4x^2 + 3x} )

• Process: Apply the chain rule.

• Result: ( \frac{dy}{dx} = e^{4x^2 + 3x} (8x + 3) )

This function represents exponential growth influenced by a quadratic expression in the exponent.

Function 2:

• Given: ( y = e^{\sin x} )

• Process: Apply the chain rule.

• Result: ( \frac{dy}{dx} = e^{\sin x} \cos x )

The sine function in the exponent causes the result to oscillate, reflecting the periodic nature of sine.

Function 3:

• Given: ( y = e^{x^3 - 5x + 2} )

• Process: Apply the chain rule.

• Result: ( \frac{dy}{dx} = e^{x^3 - 5x + 2} (3x^2 - 5) )

This function includes a cubic polynomial, which greatly affects the growth rate of the exponential function.

Function 4:

• Given: ( y = e^{\tan x} )

• Process: Apply the chain rule.

• Result: ( \frac{dy}{dx} = e^{\tan x} \sec^2 x )

The tangent function yields the rapid growth of the exponential function at its asymptotes.

Function 5:

• Given: ( y = e^{\sqrt{x} + x^2} )

• Process: Apply the chain rule.

• Result: ( \frac{dy}{dx} = e^{\sqrt{x} + x^2} \left( \frac{1}{2\sqrt{x}} + 2x \right) )

Here, the combination of a square root and quadratic function influences the rate of change.

2. Differentiate the Following Inverse Trigonometric Functions:

Function 1:

• Given: ( y = \sin^{-1} (2x^3 - 1) )

• Process: Apply the chain rule.

• Result: ( \frac{dy}{dx} = \frac{2 (3x^2)}{\sqrt{1 - (2x^3 - 1)^2}} )

This function shows how the output of the sine inverse function varies with respect to a cubic input.

Function 2:

• Given: ( y = \cos^{-1} (x^4 + 3x) )

• Process: Apply the chain rule.

• Result: ( \frac{dy}{dx} = -\frac{(4x^3 + 3)}{\sqrt{1 - (x^4 + 3x)^2}} )

The negative sign indicates the relationship between cosine and sine derivatives.

Function 3:

• Given: ( y = \tan^{-1} (5x^2 - 7) )

• Process: Apply the chain rule.

• Result: ( \frac{dy}{dx} = \frac{10x}{1 + (5x^2 - 7)^2} )

The arctangent indicates the proportional relationship and growth of its internal function.

Function 4:

• Given: ( y = \sin^{-1} (\sqrt{x}) )

• Process: Apply the chain rule.

• Result: ( \frac{dy}{dx} = \frac{1}{\sqrt{1 - x}} \cdot \frac{1}{2\sqrt{x}} )

This involves the nested function structure which influences the derivative through both square root and sine.

Function 5:

• Given: ( y = \tan^{-1} (3x + e^x) )

• Process: Apply the chain rule.

• Result: ( \frac{dy}{dx} = \frac{3 + e^x}{1 + (3x + e^x)^2} )

The derivative combines polynomial and exponential growth rates impacting the tangent inverse function.

Part 2: Tangent Line Problem

Find the Equation of the Tangent Line for the Given Inverse Function:

Example 1:

• Given: ( f(2) = 8, f'(2) = 6 )

• Tangent Line: ( y - 8 = 6(x - 2) )

• Simplified: ( y = 6x - 4 )

Example 2:

• Given: ( f(4) = 10, f'(4) = 3 )

• Tangent Line: ( y - 10 = 3(x - 4) )

• Simplified: ( y = 3x - 2 )

Example 3:

• Given: ( f(5) = 20, f'(5) = 2 )

• Tangent Line: ( y - 20 = 2(x - 5) )

• Simplified: ( y = 2x + 10 )

Example 4:

• Given: ( f(6) = 15, f'(6) = 4 )

• Tangent Line: ( y - 15 = 4(x - 6) )

• Simplified: ( y = 4x - 9 )

Example 5:

• Given: ( f(3) = 9, f'(3) = 5 )

• Tangent Line: ( y - 9 = 5(x - 3) )

• Simplified: ( y = 5x - 6 )

Part 3: Related Rates

8. Water Tank Problems

Problem 1:

• Scenario: Conical tank drains at a rate of 4 O³/min.

• Dimensions: Height = 20 O, radius = 5 O.

• Depth of interest: ( h = 10 O )

• Key considerations include the shape of the tank influencing the flow rate and volume.

Problem 2:

• Scenario: Spherical tank filled at a rate of 6 O³/min.

• Dimension: Radius = 10 O.

• Depth of interest: ( h = 5 O )

• Spherical volume formulas play a critical role in determining depth with respect to time.

Problem 3:

• Scenario: Cylinder leaks at a rate of 8 O³/min.

• Dimensions: Height = 30 O, radius = 6 O.

• Leaking fluid dynamics affect pressure and species concentration.

Problem 4:

• Scenario: Hemispherical tank filled at a rate of 9 O³/min.

• Dimensions: Radius = 12 O, depth = 6 O.

• Rapid fill rates require monitoring for overflow and pressure changes.

Problem 5:

• Scenario: Cone drains at a rate of 5 O³/min.

• Dimensions: Height = 15 O, base radius = 4 O, depth = 7 O.

• Conical properties impact drainage efficiency and effective depth over time.

9. Ladder Problems

Problem 1:

• Situation: A 10-O ladder with the bottom sliding away at a rate of 2 O/sec, positioned 6 O from a wall.

• Analyzing the angles and lengths is essential for determining the change in height.

Problem 2:

• Situation: A 20-O ladder where the bottom moves away at 3 O/sec, with the bottom positioned 12 O from the wall.

• Assessment of the trajectory reveals critical insights into stability and height alteration.

Problem 3:

• Situation: A 30-O ladder slides at a rate of 4 O/sec with the bottom 16 O from a tree.

• The impact on height requires calculations on the rope length to prevent falling.

Problem 4:

• Situation: A 25-O ladder having its bottom move away at 1 O/sec, situated 15 O from a wall.

• The forces acting on the ladder need careful evaluation to ensure safety.

Problem 5:

• Situation: A 12-O ladder where the bottom slides away at 1.5 O/sec, with a distance of 9 O from a pole.

• Understanding the loading angles is critical for maintaining equilibrium in this setup.