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

$6.42

step-by-step explanation:

step-by-step explanation:

4  

25

5258y

54.

t5

5r

54

5856

542

5458

86

.8558.65

565

565

55898

786

8/95

5896

8656265

556478