Question

How do you find the equation of the tangent line to the curve `y=xsqrtx` that is parallel to the line `y=1+3x`?

Answer 1

The equation is `y = 3x- 4`.

Explanation:

A parallel line means equal slope. The slope of the curve at `x =a` is given by evaluating `f'(a)`, where `f'(x)` is the derivative.

We need find the derivative, therefore.

First of all, the function can be rewritten as `y= x(x)^(1/2) = x^(1 + 1/2) = x^(3/2)`.

We can now use the power rule.

`dy/dx = 3/2x^(1/2)`

We know the slope of the line `y = 1 + 3x` is `3`, so we set `dy/dx` to `3` and solve for `x`.

`3 = 3/2x^(1/2)`

`3/(3/2) = x^(1/2)`

`2 = x^(1/2)`

`x = 4`

We now use this point to determine the corresponding y-coordinate.

`y= 4sqrt(4) = 4(2) = 8`

We now know the slope and the point of contact . The equation is given by:

`y- y_1 = m(x- x_1)`

`y - 8 = 3(x- 4)`

`y- 8 = 3x - 12`

`y= 3x - 4`

Hopefully this helps!