Find an equation of the tangent to the curve x =5+lnt, y=t2+5 at the point (5,6) by both eliminating the parameter and without eliminating the parameter.

ANSWER

EXPLANATION

Part a)

Eliminating the parameter:

The parametric equation is

From the first equation we make t the subject to get;

We put it into the second equation.

We differentiate to get;

At x=5,

The slope of the tangent is 2.

The equation of the tangent through

(5,6) is given by

Without eliminating the parameter,

At x=5,

This implies that,

The slope of the tangent is 2.

The equation of the tangent through

(5,6) is given by

area = length x width

length = 2 x width

32 = 2 x width x width

32 = 2 x width^2

36 = width ^2  

width = 6

length = 12

area is 32

perimeter = 2 x 12 + 2 x 6 = 24 + 12 = 36

step-by-step explanation:

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so: