Answer to Question #118267 in Calculus for Olivia

Question #118267
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)
1
Expert's answer
2020-05-28T18:00:22-0400

We have h(x,y)=yln(x)So, we get hx(x,y)=yxandhy(x,y)=ln(x).at (1,5), we gethx(1,5)=5 andhy(1,5)=ln(1)=0.As at(1,5) the equation ofthe tangent plane to h(x,y) isz=hx(1,5)(x1)+hy(1,5)(y5)So, we getz=5(x1).This implies that the correct answer isf.We \ have\ h(x,y)= y ln(x)\\ \text{So, we get }\\ h_x(x,y)=\frac{y}{x}\\ ​ and\\ h_y (x,y)=ln(x).\\ at\ (1,5), \ we \ get\\ h _x (1,5)=5 \ and\\ h _y (1,5)=ln(1)=0.\\ As \ at(1,5)\ the \ equation \ of \\ \text{the tangent plane to } h(x,y)\ is\\ z=h_x (1,5)(x−1)+h_y (1,5)(y−5)\\ \text{So, we get} \\ z=5(x−1).\\ \text{This implies that the correct answer is}\\ f.


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