Question

The equation of the tangent plane to h(x,y)=yln(x) at the point (1,5) is
Select one:
a. z=1/5x+1/5y

b. z=x+y


c. z=1/4x+1/4y


d. z=4(x−1)


e. z=5(x+1)


f. z=5(x−1)

Answer 1

Let us rewrite our equation as $z= h(x,y)=y\ln x.$ Point $(x_0,y_0) = (1,5)$ . We may write the equation of a tangent plane in form (see https://openstax.org/books/calculus-volume-3/pages/4-4-tangent-planes-and-linear-approximations)

$z=h(x_0,y_0) + h_x(x_0,y_0)(x-x_0) + h_y(x_0,y_0)(y-y_0).$

Let us determine the derivatives:

$h_x(x,y) = y\cdot \dfrac1x, \;\; h_y(x,y) = 1\cdot\ln x.$ Therefore, $h_x(x_0,y_0) = 5\cdot\dfrac11 = 5, \;\; h_y(x_0,y_0) = 1\cdot\ln 1 = 0.$

$h(x_0,y_0) = 5\ln1 = 0.$

So the equation of tangent plane will be

$z =$ 0 + 5(x-1) + 0(y-5) = 5(x-1).

Therefore, the correct answer is f. z=5(x-1).