Calculus Parametric Equations

Question

Find the tangent line(s) to the parametric curve x = t⁶-4t⁴ and y = t² at (0,4). Write your tangent line(s) as a cartesian equation. If there is only one tangent line, enter 'DNE' into the second box below.

y =

y =

Answer 1

dy/dx=(dy/dt)/(dx/dt) = 2t/(6t⁵-16t³) = 2/(6t⁴-16t²).

The point (0,4) occurs when t = 2 or -2. (because t⁶-4t⁴= 0 => t⁴(t²-4) = 0 => t = 0, t=2, t = -2). and t²=4 => t = 2, t = -2.

dy/dx|t=2 = 2/(96-64) = 2/32 = 1/16. (Same for a value of t = -2).

So slope of tangent line is 1/16, and point of tangency is (0,4).

y - 4 = 1/16(x-0)

y = x/16 + 4

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Answer 2

Convert parametric equations to cartesian

y = t²

√y = t

x = t⁶-4t⁴

x = (√y)⁶-4(√y)⁴

x = y³-4y²

Find dy/dx with implicit differentiation

1 = (3y²-8y)(dy/dx)

dy/dx = 1/(3y²-8y)

At (0,4):

dy/dx = 1/(3(4)²-8(4)) = 1/(3*16 - 32) = 1/(48-32) = 1/16

Hence, the slope of the tangent line will be 1/16 at (0,4). Now, we can use point-slope form to get the equation of our tangent line:

y-y₀ = m(x-x₁)

y-4 = (1/16)(x-0)

y-4 = (1/16)x

y = (1/16)x+4

Thus, the tangent line as a cartesian equation is y = (1/16)x+4

Hope this helped!