Math Problem Statement

Solve 21 and 16

Solve 21 and 16 from the provided math problem set.

Solution

Let's solve problems 16 and 21:

Problem 16

Given the function: y=2x+52xy = \frac{2x + 5}{2 - x} We are asked to find the derivative and evaluate the slope of the tangent line at x=2x = -2.

Step 1: Find the derivative dydx\frac{dy}{dx}.

This is a quotient, so we use the quotient rule: ddx(uv)=vdudxudvdxv2\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} For our function:

  • u=2x+5u = 2x + 5 and v=2xv = 2 - x
  • dudx=2\frac{du}{dx} = 2 and dvdx=1\frac{dv}{dx} = -1

Applying the quotient rule: dydx=(2x)(2)(2x+5)(1)(2x)2\frac{dy}{dx} = \frac{(2 - x)(2) - (2x + 5)(-1)}{(2 - x)^2} Simplify: dydx=42x+2x+5(2x)2=9(2x)2\frac{dy}{dx} = \frac{4 - 2x + 2x + 5}{(2 - x)^2} = \frac{9}{(2 - x)^2}

Step 2: Evaluate the derivative at x=2x = -2. dydxx=2=9(2(2))2=942=916\frac{dy}{dx} \bigg|_{x = -2} = \frac{9}{(2 - (-2))^2} = \frac{9}{4^2} = \frac{9}{16}

The slope of the tangent line at x=2x = -2 is 916\frac{9}{16}.

Problem 21

Given the function: r=59θr = \frac{5}{\sqrt{9 - \theta}} We are asked to find drdθ\frac{dr}{d\theta} at θ=0\theta = 0.

Step 1: Find the derivative drdθ\frac{dr}{d\theta}.

Rewrite the function: r=5(9θ)12r = 5 \cdot (9 - \theta)^{-\frac{1}{2}}

Use the power rule and the chain rule: drdθ=5(12)(9θ)32(1)\frac{dr}{d\theta} = 5 \cdot \left( -\frac{1}{2} \right) (9 - \theta)^{-\frac{3}{2}} \cdot (-1) Simplify: drdθ=52(9θ)32\frac{dr}{d\theta} = \frac{5}{2} (9 - \theta)^{-\frac{3}{2}}

Step 2: Evaluate the derivative at θ=0\theta = 0. drdθθ=0=52(90)32=5227=554\frac{dr}{d\theta} \bigg|_{\theta = 0} = \frac{5}{2} (9 - 0)^{-\frac{3}{2}} = \frac{5}{2 \cdot 27} = \frac{5}{54}

The value of drdθ\frac{dr}{d\theta} at θ=0\theta = 0 is 554\frac{5}{54}.

Do you want more details or have any questions?

Here are some related questions to expand your understanding:

  1. How would the slope change if we evaluated problem 16 at a different point?
  2. What are the key steps for applying the quotient rule?
  3. How does the chain rule help in problem 21?
  4. What would happen to the derivative if the functions in problems 16 and 21 were squared?
  5. How do we determine if a tangent line is horizontal?

Tip: Remember that the quotient rule requires careful application of the numerator and denominator derivatives—keeping track of signs is crucial!

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Math Problem Analysis

Mathematical Concepts

Differentiation
Quotient Rule
Power Rule
Chain Rule

Formulas

Quotient rule: (d/dx)(u/v) = (v(du/dx) - u(dv/dx)) / v^2
Power rule: d/dx (x^n) = n*x^(n-1)
Chain rule: d/dx (f(g(x))) = f'(g(x)) * g'(x)

Theorems

Derivative properties
Tangent line slope

Suitable Grade Level

Grades 11-12