Answer to Question #117915 in Calculus for Olivia

Question #117915
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
2020-05-28T17:41:55-0400

ANSWER : f. z=5(x1)z=5(x-1)

EXPLANATION

The tangent plane to z=h(x,y) at the point (a,b) is zh(a,b)=hx(a,b) (xa)+hy(a,b)(yb)z-h(a,b)={ h }_{ x }^{ ' }(a,b)\ (x-a)+{ h }_{ y }^{ ' }(a,b)(y-b)

a=1,b=5, h(a,b) =0a=1,\quad b=5,\ h(a,b)\ =0\quad

hx(x,y)=yx,{ h }_{ x }^{ ' }(x,y)=\frac { y }{ x } ,\quad hx(1,5)=5hy(x,y)=lnx,hy(1,5)=0{ h }_{ x }^{ ' }(1,5)=5\quad { h }_{ y }^{ ' }(x,y)=\ln { x } , { h }_{ y }^{ ' }(1,5)=0 .

So, the tangent plane is z=5(x1)z=5(x-1)


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