Find the points on the curve where the tangent is horizontal or vertical. You may want to use a graph from a calculator or computer to check your work. (If an answer does not exist, enter DNE.) $x = t^3 - 3t, \quad y = t^2 - 6$

• Horizontal tangent
  $(x, y) = \boxed{?}$

• Vertical tangent (smaller $x$-value)
  $(x, y) = \boxed{?}$

• Vertical tangent (larger $x$-value)
  $(x, y) = \boxed{?}$

Neetesh Kumar | January 3, 2025

Calculus Homework Help

This is the solution to Math 1c
Assignment: 10.2 Question Number 8
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Step-by-step solution:

Step 1: Conditions for horizontal and vertical tangents

The slope of the tangent line is given by:

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

1. Horizontal tangent:

   $\frac{dy}{dx} = 0$, which occurs when $\frac{dy}{dt} = 0$.

2. Vertical tangent:

   $\frac{dy}{dx}$ is undefined, which occurs when $\frac{dx}{dt} = 0$.

Step 2: Compute $\frac{dy}{dt}$ and $\frac{dx}{dt}$

Differentiate $y = t^2 - 6$:

$\frac{dy}{dt} = 2t$

Differentiate $x = t^3 - 3t$:

$\frac{dx}{dt} = 3t^2 - 3$

Step 3: Solve for horizontal tangents

For horizontal tangents, set $\frac{dy}{dt} = 0$:

$2t = 0 \implies t = 0$

Substitute $t = 0$ into $x = t^3 - 3t$ and $y = t^2 - 6$:

• At $t = 0$:

  $x = 0^3 - 3(0) = 0, \quad y = 0^2 - 6 = -6$

The horizontal tangent is at:

$(x, y) = \boxed{(0, -6)}$

Step 4: Solve for vertical tangents

For vertical tangents, set $\frac{dx}{dt} = 0$:

$3t^2 - 3 = 0 \implies t^2 = 1 \implies t = \pm 1$

Case 1: $t = -1$

Substitute $t = -1$ into $x = t^3 - 3t$ and $y = t^2 - 6$:

• At $t = -1$:

  $x = (-1)^3 - 3(-1) = -1 + 3 = 2, \quad y = (-1)^2 - 6 = 1 - 6 = -5$

The point is:

$(x, y) = (2, -5)$

Case 2: $t = 1$

Substitute $t = 1$ into $x = t^3 - 3t$ and $y = t^2 - 6$:

• At $t = 1$:

  $x = (1)^3 - 3(1) = 1 - 3 = -2, \quad y = (1)^2 - 6 = 1 - 6 = -5$

The point is:

$(x, y) = (-2, -5)$

Final Answers:

• Horizontal tangent:
  $(x, y) = \boxed{(0, -6)}$

• Vertical tangent (smaller $x$-value):
  $(x, y) = \boxed{(-2, -5)}$

• Vertical tangent (larger $x$-value):
  $(x, y) = \boxed{(2, -5)}$

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