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

answer: solve 5\cdot 2^x=2405⋅2  

x

=2405, dot, 2, start superscript, x, end superscript, equals, 240.

solution

to solve for xxx, we must first isolate the exponential part. to do this, divide both sides by 555 as shown below. we do not multiply the 555 and the 222 as this goes against the order of operations!

\begin{aligned} 5\cdot 2^x& =240 2^x& =48 \\ \end{aligned}  

5⋅2  

x

2  

x

​    

=240

=48

​

step-by-step explanation: