Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the parameter. x = 6 + ln(t), y = t2 + 3, (6, 4)

Step-by-step explanation:

The given parametric equation is;

To eliminate the parameter we make the subject in one equation and put it inside the other.

We make the subject in because it is easier.

Or

We now substitute this into .

This gives us;

.

.

We have now eliminated the parameter.

The equation of the tangent at (6,4) is given by;

where the gradient function is given by;

We substitute into the gradient function to obtain the gradient.

The equation of the tangent becomes

We simplify to obtain

The given parametric equation is;

For

For

The slope is given by;

At the point, (6,4), we plug in any of the values into the parametric equation and find the corresponding value for .

Notice that

When ,

when ,

But the slope is the same when we plug in any of these values for t.

The equation of the tangent becomes

We simplify to obtain

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d

step-by-step explanation:

6mm^3 or .6cm^3

      216

      /   \

  36     6

  /   \     /

6     6   6

factor trees are nice.