Find the equation for the plane tangent to each surface

Juan Hewlett 2021-12-20 Answered

Find the equation for the plane tangent to each surface z=f(x,y) at the indicated point.
(a)z=x3+y32xy,at the point(1,2,5)
(b)z=(cos(x))(cos(y)),at the point(0,π2,0)
(c)z=(cos(x))(sin(y)),at the point(0,π2,1)

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Expert Answer

Jeremy Merritt
Answered 2021-12-21 Author has 31 answers

Part (a): We have:
Surface z=x3+y32xy,at the point(1,2,5)
(x0,y0,z0)=(1,2,5)
Used formula in equation:
An equation of the tangent plane to the surface z=f(x,y) at the point P(x0,y0,z0) is:
zz0=fx(x0,y0)(xx0)+fy(x0,y0)(yy0)
Differentiating partially equation of surface with respect to x and y:

fx=zx=3x22y

zx=3(1)22(2)=1

fy=zx=3y22x

zx=3(2)22(1)=10

Now, put the valuesin formula:
zz0=fx(x0,y0)(xx0)+fy(x0,y0)(yy0)
z(5)=(1)(x1)+(10)(y2)
x10y+z+24=0

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temnimam2
Answered 2021-12-22 Author has 36 answers

Part (b):Given:
Surface z=cosx×cosy,at the point(0,π2,0)(x0,y0,z0)=(0,π2,0)
Differentiating partially equation of surface with respect to x and y:
fx=zx=(sinx)(cosy)
zx=(sin0)(cosπ2)=0
fy=zx=(cosx)(siny)
zx=(cos0)(sinπ2)=1
Now, put the values in formula:
zz0=fx(x0,y0)(xx0)+fy(x0,y0)
z(0)=(0)(x0)+(1)(yπ2)
y+zπ2=0

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nick1337
Answered 2021-12-27 Author has 575 answers

Part (c):Given:
Surface z=(cosx)(cosy), at the point (0,π2,1)
(x0,y0,z0)=(0,π2,1)
Differentiating partially equation of surface with respect to x and y:
fx=zx=(sinx)(siny)
zx=(sin0)(sinπ2)=0
fyzx=(cosx)(cosy)
zx=(cos0)(cosπ2)=0
Now, put the values in formula:
zz0=fx(x0,y0)(x0,y0)+fy(x0,y0)(yy0)
z(1)=(0)(x0)+(0)(yπ2)
z1=0

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