EQUATION OF A LINE TANGENT TO THE CURVE WITH PARAMETRIC EQUATIONS

Problem 1 :

Find the equation of the line tangent to the curve with parametric equations

x = 3t

y = t² + 1

at t = 2.

Solution :

To find the value of the slope m, substutite t = 2 into  ᵈʸ⁄dₓ.

Substitute t = 2 into the given parametric equations to find the point of tangency (x₁, y₁).

(x₁, y₁) = (6, 5)

Equation of the line tangent to the curve :

y - y₁ = m(x - x₁)

Substitute (x₁, y₁) = (6, 5) and m = ⁴⁄₃.

Problem 2 :

Find the equation of the line tangent to the curve with parametric equations

x = 4cosθ

y = 9sinθ

at θ = ^π⁄₄.

Solution :

To find the value of the slope m, substutite θ = ^π⁄₄ into  ᵈʸ⁄dₓ.

Substitute θ = ^π⁄₄ into the given parametric equations to find the point of tangency (x₁, y₁).

(x₁, y₁) = (6, 5)

Equation of the line tangent to the curve :

y - y₁ = m(x - x₁)

Problem 3 :

Find the equation of the line tangent to the curve with parametric equations

x = t³ - 3t

y = t² - 5t

at t = 4.

Solution :

To find the value of the slope m, substutite t = 4 into  ᵈʸ⁄dₓ.

Substitute t = 4 into the given parametric equations to find the point of tangency (x₁, y₁).

(x₁, y₁) = (52, -4)

Equation of the line tangent to the curve :

y - y₁ = m(x - x₁)

Substitute (x₁, y₁) = (52, -4) and m = ¹⁄₁₅.

Problem 4 :

Find the equation of the line tangent to the curve with parametric equations

x = ln t

y = t²

at t = e.

Solution :

To find the value of the slope m, substutite t = e into  ᵈʸ⁄dₓ.

m = 2e²

Substitute t = 4 into the given parametric equations to find the point of tangency (x₁, y₁).

(x₁, y₁) = (1, e²)

Equation of the line tangent to the curve :

y - y₁ = m(x - x₁)

Substitute (x₁, y₁) = (1, e²) and m = 2e².

y - e² = 2e²(x - 1)

y - e² = 2e²x - 2e²

y = 2e²x - 2e² + e²

y = 2e²x - e²

Problem 5 :

Determine the values of t where the tangent lines to the curve with parametric equations

x = t³ - 3t

y = t² - 5t

are horizontal and where they are vertical.

Solution :

If the tangent line to the curve is horizontal, change on vertical axis with respect to t is zero. Since y  represents the vertical axis, we have

2t - 5 = 0

2t = 5

t = ⁵⁄₂

Therefore, the tangent line to the curve is horizontal, when

t = ⁵⁄₂

If the tangent line to the curve is vertical, change on horizontal axis with respect to t is zero. Since x represents the horizontal axis, we have

3t² - 3 = 0

3(t² - 1) = 0

t² - 1 = 0

t² = 1

Taking Square root on both sides,

t = ±1

Therefore, the tangent line to the curve is vertical, when

t = ±1

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