Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter. X = t5 1, y = t6 t;

Question

2 years ago

11

t = −1

1 answer:

kifflom [539] · Answer 1 · 2023-04-18T06:48:34+03:00

The equation of the tangent to the curve at the point corresponding to the given value of the parameter will be x + y = c.

<h3>How to calculate the tangent of the parameter curves at a point?</h3>

The curves are given below.

x = t⁵ + 1 and y = t⁶ + t

Then differentiate the functions with respect to t, then we have

$\rm \dfrac{dx}{dt} = 5t^4\\$ ...1

and

$\rm \dfrac{dy}{dt} = 6t^5 + 1$ ...2

Divide equation 2 by equation 1, then we have

$\rm \dfrac{\dfrac{dy}{dt} }{\dfrac{dx}{dt} } = \dfrac{dy}{dx} = \dfrac{6t^5 + 1}{5t^4}$

Then the slope of the equation at t = −1, then we have

dy/dx = [6(-1)⁵ + 1]/[5(-1)⁴]

dy/dx = (-6+1)/5

dy/dx = -5/5

dy/dx = -1

Then the equation of the tangent line will be

y = -x + c

x + y = c

Where c is a constant.

More about the tangent of the parameter curves at a point link is given below.

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