16. Parametric Equations & Polar Coordinates

Calculus with Parametric Curves

16. Parametric Equations & Polar Coordinates

Calculus with Parametric Curves

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Multiple Choice

Write the equation of the tangent line in cartesian coordinates for the given parameter $t$ .
$x equals 8 cosine t$ , $y equals 6 sine t$ , $t = \frac{π}{4}$

$y - 3 \sqrt{2} = - \frac{3}{4} (x - 4 \sqrt{2})$

$y - 3 \sqrt{2} = \frac{3}{4} (x - 4 \sqrt{2})$

$y - 4 \sqrt{2} = - \frac{3}{4} (x - 3 \sqrt{2})$

$y - 4 \sqrt{2} = \frac{3}{4} (x - 3 \sqrt{2})$

Verified step by step guidance

Step 1: Recall that the equation of a tangent line can be written in the form y - y₁ = m(x - x₁), where m is the slope of the tangent line, and (x₁, y₁) is the point of tangency. Here, the parametric equations are x = 8cos(t) and y = 6sin(t), and the parameter t is given as t = π/4.

Step 2: Compute the coordinates of the point of tangency (x₁, y₁) by substituting t = π/4 into the parametric equations. For x₁, substitute t = π/4 into x = 8cos(t). For y₁, substitute t = π/4 into y = 6sin(t). This gives the point (x₁, y₁).

Step 3: Find the slope of the tangent line, m, by calculating dy/dx. Since the equations are parametric, use the chain rule: dy/dx = (dy/dt) / (dx/dt). Compute dy/dt by differentiating y = 6sin(t) with respect to t, and compute dx/dt by differentiating x = 8cos(t) with respect to t.

Step 4: Substitute t = π/4 into the expressions for dy/dt and dx/dt to find the slope m = (dy/dt) / (dx/dt) at t = π/4.

Step 5: Write the equation of the tangent line using the point-slope form y - y₁ = m(x - x₁), where m is the slope found in Step 4 and (x₁, y₁) is the point found in Step 2.