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Sloan [31]
3 years ago
7

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 e

liminating the parameter.
Mathematics
1 answer:
svet-max [94.6K]3 years ago
5 0

ANSWER

y = 2x -4

EXPLANATION

Part a)

Eliminating the parameter:

The parametric equation is

x = 5 +  ln(t)

y =  {t}^{2}  + 5

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

x - 5 =  ln(t)

t =  {e}^{x - 5}

We put it into the second equation.

y =  { ({e}^{x - 5}) }^{2}  + 5

y =  { ({e}^{2(x - 5)}) }  + 5

We differentiate to get;

\frac{dy}{dx}  = 2 {e}^{2(x - 5)}

At x=5,

\frac{dy}{dx}  = 2 {e}^{2(5 - 5)}

\frac{dy}{dx}  = 2 {e}^{0}  = 2

The slope of the tangent is 2.

The equation of the tangent through

(5,6) is given by

y-y_1=m(x-x_1)

y - 6 = 2(x - 5)

y = 2x - 10 + 6

y = 2x -4

Without eliminating the parameter,

\frac{dy}{dx}  =  \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} }

\frac{dy}{dx}  =  \frac{ 2t}{  \frac{1}{t} }

\frac{dy}{dx}  =  2 {t}^{2}

At x=5,

5 = 5 +  ln(t)

ln(t)  = 0

t =  {e}^{0}  = 1

This implies that,

\frac{dy}{dx}  =  2 {(1)}^{2}  = 2

The slope of the tangent is 2.

The equation of the tangent through

(5,6) is given by

y-y_1=m(x-x_1)

y - 6 = 2(x - 5) =

y = 2x -4

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