PreCalculus GT - Rates of Change Practice Problems

Rates of Change Practice Problems

Problem 1: Tangent Line Equation

• Objective: Find the equation of the line tangent to $f(x) = x^3 - 5$ at $x = 2$.
• Step 1: Find the derivative (slope of the tangent line).
  • The derivative, $m$, is found using the limit definition:
    $m = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
  • Applying this to the given function:
    $m = \lim_{h \to 0} \frac{(x+h)^3 - 5 - (x^3 - 5)}{h}$
  • Expanding and simplifying:
    $m = \lim<em>{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 - 5 - x^3 + 5}{h}$$m = \lim</em>{h \to 0} \frac{3x^2h + 3xh^2 + h^3}{h}$
    $m = \lim_{h \to 0} 3x^2 + 3xh + h^2$
  • Taking the limit as $h$ approaches 0:
    $m = 3x^2$
• Step 2: Evaluate the derivative at $x = 2$.
  • $m|_{x=2} = 3(2)^2 = 12$
• Step 3: Find the y-coordinate at $x = 2$.
  • $f(2) = (2)^3 - 5 = 8 - 5 = 3$
• Step 4: Write the equation of the tangent line in point-slope form.
  • The point-slope form is given by:
    $y - y<em>1 = m(x - x</em>1)$
  • Plugging in the values, we get:
    $y - 3 = 12(x - 2)$

Problem 2: Perpendicular Tangent Line

• Objective: Find the equation of the line perpendicular to $y = -4x + 1$ and tangent to $f(x) = \sqrt{x} - 2$.
• Step 1: Determine the slope of the perpendicular line.
  • The given line has a slope of $-4$. The slope of a line perpendicular to it is the negative reciprocal, which is $\frac{1}{4}$.
• Step 2: Find the derivative of $f(x) = \sqrt{x} - 2$.
  • Using the limit definition of the derivative:
    $m = \lim<em>{h \to 0} \frac{\sqrt{x+h} - 2 - (\sqrt{x} - 2)}{h}$$m = \lim</em>{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$
  • Multiply by the conjugate to rationalize the numerator:
    $m = \lim<em>{h \to 0} \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})}$$m = \lim</em>{h \to 0} \frac{x+h - x}{h(\sqrt{x+h} + \sqrt{x})}$
    $m = \lim<em>{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})}$$m = \lim</em>{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}}$
  • Taking the limit as $h$ approaches 0:
    $m = \frac{1}{2\sqrt{x}}$
• Step 3: Set the derivative equal to the perpendicular slope and solve for $x$.
  • $\frac{1}{2\sqrt{x}} = \frac{1}{4}$
  • $2\sqrt{x} = 4$
  • $\sqrt{x} = 2$
  • $x = 4$
• Step 4: Find the y-coordinate at $x = 4$.
  • $f(4) = \sqrt{4} - 2 = 2 - 2 = 0$
• Step 5: Write the equation of the tangent line in point-slope form.
  • $y - 0 = \frac{1}{4}(x - 4)$
  • $y = \frac{1}{4}x - 1$

Problem 3: Tangent Line with Given Slope

• Objective: Find the equation of any line with a slope of 2 that is tangent to $y = \frac{8}{x-1}$.
• Step 1: Find the derivative of $f(x) = \frac{8}{x-1}$.
  • Using the limit definition of the derivative:
    $m = \lim_{h \to 0} \frac{\frac{8}{x+h-1} - \frac{8}{x-1}}{h}$
  • Simplifying the expression:
    $m = \lim<em>{h \to 0} \frac{8(x-1) - 8(x+h-1)}{h(x+h-1)(x-1)}$$m = \lim</em>{h \to 0} \frac{8x - 8 - 8x - 8h + 8}{h(x+h-1)(x-1)}$
    $m = \lim<em>{h \to 0} \frac{-8h}{h(x+h-1)(x-1)}$$m = \lim</em>{h \to 0} \frac{-8}{(x+h-1)(x-1)}$
  • Taking the limit as $h$ approaches 0:
    $m = \frac{-8}{(x-1)^2}$
• Step 2: Set the derivative equal to the given slope and solve for $x$.
  • $\frac{-8}{(x-1)^2} = 2$
  • $-8 = 2(x-1)^2$
  • $-4 = (x-1)^2$
  • Since $(x-1)^2$ cannot be negative for real $x$, there are no real solutions. However, if we proceed ignoring this and assuming there was a mistake in the sign, we will continue as the original transcription showed. But, given NO REAL solutions, then there is no tangent line with slope of 2.
  • Assuming a sign change such that $\frac{-8}{(x-1)^2} = -2$:
  • $\frac{4}{(x-1)^2} = 1$
  • $(x-1)^2 = 4$
  • $x-1 = \pm 2$
  • $x = 3, -1$
• Step 3: Find the y-coordinates for each $x$.
  • For $x = 3$: $f(3) = \frac{8}{3-1} = \frac{8}{2} = 4$
  • For $x = -1$: $f(-1) = \frac{8}{-1-1} = \frac{8}{-2} = -4$
• Step 4: Write the equations of the tangent lines in point-slope form.
  • For $x = 3$: $y - 4 = -2(x - 3)$
  • For $x = -1$: $y - (-4) = -2(x - (-1))$ which simplifies to $y + 4 = -2(x + 1)$

Problem 4: Intermediate Value Theorem (IVT)

• Objective: Determine if $g(x) = 5 - 4x - 3x^2$ must equal 3 sometime between $x = -1$ and $x = 1$. Justify the answer.
• Step 1: Check if $g(x)$ is continuous.
  • Since $g(x)$ is a polynomial, it is continuous everywhere.
• Step 2: Evaluate $g(x)$ at $x = -1$ and $x = 1$.
  • $g(-1) = 5 - 4(-1) - 3(-1)^2 = 5 + 4 - 3 = 6$
  • $g(1) = 5 - 4(1) - 3(1)^2 = 5 - 4 - 3 = -2$
• Step 3: Apply the Intermediate Value Theorem (IVT).
  • Since $g(x)$ is continuous on the interval $[-1, 1]$ and $g(-1) = 6$ and $g(1) = -2$, and 3 is between -2 and 6, by the IVT, there must be some $c$ in the interval $(-1, 1)$ such that $g(c) = 3$.

Problem 5: Average Velocity

• Objective: Find the average velocity of a football kicked from the ground at 96 ft/sec from $t = 1$ to $t = 3$. The position function is $s(t) = -16t^2 + 96t$.
• Step 1: Calculate the position at $t = 1$ and $t = 3$.
  • $s(1) = -16(1)^2 + 96(1) = -16 + 96 = 80$
  • $s(3) = -16(3)^2 + 96(3) = -16(9) + 288 = -144 + 288 = 144$
• Step 2: Calculate the average velocity.
  • The average velocity is given by:
    $v_{avg} = \frac{s(3) - s(1)}{3 - 1} = \frac{144 - 80}{3 - 1} = \frac{64}{2} = 32$
  • The average velocity is 32 ft/sec.

Problem 6: Speed of a Falling Plate

• Objective: Find the speed of a plate that fell off a counter 3.5 feet high when it hit the floor.
• Step 1: Define the position function.
  • $s(t) = -16t^2 + 3.5$
• Step 2: Find when the plate hits the ground.
  • Set $s(t) = 0$ and solve for $t$:
    $0 = -16t^2 + 3.5$
    $16t^2 = 3.5$
    $t^2 = \frac{3.5}{16} = 0.21875$
    $t = \sqrt{0.21875} \approx 0.4677$
• Step 3: Find the velocity function.
  • $v(t) = s'(t) = -32t$
• Step 4: Calculate the velocity at $t = 0.4677$.
  • $v(0.4677) = -32(0.4677) \approx -14.97$
• Step 5: State the speed.
  • The speed is the absolute value of the velocity, so the speed is approximately 14.97 ft/sec.

Problem 7: Computer Thrown from Window

• Objective: Analyze the motion of a computer thrown from a 4th-story window 48 feet high with an initial velocity of 32 ft/sec. The position function is given by $s(t) = -16t^2 + 32t + 48$.

  • a. When will the computer hit the ground?

    • Set $s(t) = 0$ and solve for $t$:
      $0 = -16t^2 + 32t + 48$
      Divide by -16:
      $0 = t^2 - 2t - 3$
      Factor:
      $0 = (t - 3)(t + 1)$
      The solutions are $t = 3$ and $t = -1$. Since time cannot be negative, the computer hits the ground at $t = 3$ seconds.

  • b. What is the velocity of the computer after 2.5 seconds?

    • Find the velocity function: $v(t) = -32t + 32$
    • Evaluate $v(2.5)$:
      $v(2.5) = -32(2.5) + 32 = -80 + 32 = -48$
    • The velocity of the computer after 2.5 seconds is -48 ft/sec (downward).

  • c. What is the average velocity of the computer during the third second?

    • Calculate the position at $t = 2$ and $t = 3$:
      $s(2) = -16(2)^2 + 32(2) + 48 = -64 + 64 + 48 = 48$
      $s(3) = -16(3)^2 + 32(3) + 48 = -144 + 96 + 48 = 0$
    • Calculate the average velocity:
      $v_{avg} = \frac{s(3) - s(2)}{3 - 2} = \frac{0 - 48}{1} = -48$
    • The average velocity during the third second is -48 ft/sec.

  • d. What is the height of the computer when the velocity is zero?

    • Set $v(t) = 0$ and solve for $t$:
      $0 = -32t + 32$
      $32t = 32$
      $t = 1$
    • Calculate the height at $t = 1$:
      $s(1) = -16(1)^2 + 32(1) + 48 = -16 + 32 + 48 = 64$
    • The height of the computer when the velocity is zero is 64 feet.

  • e. After how many seconds was the computer's velocity -24 ft/sec?

    • Set $v(t) = -24$ and solve for $t$:
      $-24 = -32t + 32$
      $-56 = -32t$
      $t = \frac{-56}{-32} = \frac{7}{4} = 1.75$
    • The computer's velocity was -24 ft/sec after 1.75 seconds.

Problem 8: True or False - IVT and Zeros

• Statement: If $f(-2) = -10$ and $f(3) = 6$, then $f(x)$ has at least one zero between -2 and 3.
• Answer: False.
• Justification: The Intermediate Value Theorem (IVT) states that if a function $f(x)$ is continuous on a closed interval $[a, b]$ and $k$ is any number between $f(a)$ and $f(b)$, then there exists at least one number $c$ in the interval $(a, b)$ such that $f(c) = k$. For $f(x)$ to have a zero between -2 and 3, $f(x)$ must be continuous on the interval $[-2, 3]$. However, we do not know if $f(x)$ is continuous. Therefore, we cannot conclude that $f(x)$ has at least one zero between -2 and 3.

Problem 9: Average Rate of Change of Secant

• Objective: Find the average rate of change of $f(x) = \sec(x)$ on the interval $[\frac{\pi}{6}, \frac{5\pi}{6}]$.
• Step 1: Calculate $f(\frac{\pi}{6})$ and $f(\frac{5\pi}{6})$:
  • Recall that $\sec(x) = \frac{1}{\cos(x)}$:
  • $f(\frac{\pi}{6}) = \sec(\frac{\pi}{6}) = \frac{1}{\cos(\frac{\pi}{6})} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$
  • $f(\frac{5\pi}{6}) = \sec(\frac{5\pi}{6}) = \frac{1}{\cos(\frac{5\pi}{6})} = \frac{1}{-\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}} = -\frac{2\sqrt{3}}{3}$
• Step 2: Calculate the average rate of change:
  • $\frac{f(\frac{5\pi}{6}) - f(\frac{\pi}{6})}{\frac{5\pi}{6} - \frac{\pi}{6}} = \frac{-\frac{2\sqrt{3}}{3} - \frac{2\sqrt{3}}{3}}{\frac{4\pi}{6}} = \frac{-\frac{4\sqrt{3}}{3}}{\frac{2\pi}{3}} = -\frac{4\sqrt{3}}{3} \cdot \frac{3}{2\pi} = -\frac{2\sqrt{3}}{\pi}$
  • The average rate of change is $-\frac{2\sqrt{3}}{\pi}$.

Problem 10: Horizontal Tangent Line

• Objective: At what point does $f(x) = 3x^2 - 9x + 1$ have a horizontal tangent line?
• Step 1: Find the derivative of $f(x)$.
  • Using the limit definition of the derivative:
    $m = \lim<em>{h \to 0} \frac{3(x+h)^2 - 9(x+h) + 1 - (3x^2 - 9x + 1)}{h}$$m = \lim</em>{h \to 0} \frac{3(x^2 + 2xh + h^2) - 9x - 9h + 1 - 3x^2 + 9x - 1}{h}$
    $m = \lim<em>{h \to 0} \frac{3x^2 + 6xh + 3h^2 - 9x - 9h + 1 - 3x^2 + 9x - 1}{h}$$m = \lim</em>{h \to 0} \frac{6xh + 3h^2 - 9h}{h}$
    $m = \lim_{h \to 0} 6x + 3h - 9$
  • Taking the limit as $h$ approaches 0:
    $m = 6x - 9$
• Step 2: Find where the derivative is equal to zero (horizontal tangent line).
  • Set $6x - 9 = 0$ and solve for $x$:
    $6x = 9$
    $x = \frac{9}{6} = \frac{3}{2}$
• Step 3: Find the y-coordinate at $x = \frac{3}{2}$.
  • $f(\frac{3}{2}) = 3(\frac{3}{2})^2 - 9(\frac{3}{2}) + 1 = 3(\frac{9}{4}) - \frac{27}{2} + 1 = \frac{27}{4} - \frac{54}{4} + \frac{4}{4} = \frac{27 - 54 + 4}{4} = \frac{-23}{4}$
• Step 4: State the point.
  • The point where $f(x)$ has a horizontal tangent line is $(\frac{3}{2}, -\frac{23}{4})$.