Find The Equation Of The Tangent Plane To The Grap

TheBob7

From: US
Posts: 25
Votes: 75
Find the equation of the tangent plane to the graph of z = e ^x ln y at the point (3,1,0). Show work.

 

Answer No.1

MisterNewton

From: US
Posts: 42
Votes: 84
Refer this example and solve in the same approach: Find an equation for the tangent plane to the graph of z = sqrt (60 - x^2 - 2y^2) at the point (3, 5, 1) Solution: z=sqrt(60-x^2-2y^2) at (Xo,Yo,Zo)=(3,5,1) Tangent plane equation: fx(Xo,Yo)*(x-Xo)+fy(Xo,Yo)*(y-Yo) - (z-Zo)=0 Where fx(Xo,Yo) means the partial derivative of f w.r.t. x, evluated at (Xo,Yo) ***Note: All derivative shown below should be in partial derivative (since I cannot type the notation of partial derivative in my computer) dz/dx = (1/2)*[sqrt(60-x^2-2y^2)] ^ ( -1/2 )*(-2x) dz/dy = (1/2)*[sqrt(60-x^2-2y^2)] ^ (-1/2) * (-4y) dz/dx (3,5)= -3 dz/dy(3,5)= -10 Hence, the tangent plane equation is: (-3)* (x-3) + (-10 )*(y-5) -(z-1) =0 3x+10y+z=60

Answer No.2

JonnyMaster9

From: US
Posts: 7
Votes: 14
The point (3,1,0) is on the surface determined by z=e^x ln(y), because 0=(e^3)*ln(1). Let (x0,y0,z0)=(3,1,0). The equation of a tangent plane at the point (x0,y0,z0) on the graph of z=f(x,y) is: z-z0=(?f/?x)0*(x-x0)+(?f/?y)0*(y-y0), where (?f/?x)0 and (?f/?y)0 are the partial derivatives of f(x,y) at (x0,y0,z0). Here z=e^x ln(y) ==> (?f/?x)=e^x ln(y) and (?f/?y)=(e^x)/y (?f/?x)0=e^3 ln(1)=0 and (?f/?y)0=(e^3)/1=e^3 The equation of the tangent plane to the graph of z=e^x ln(y) at the point (3,1,0) is: z=(e^3)*(y-1) or (e^3)y-z-e^3=0 The tangent plane is parallel to x axis.

Answer No.3

BBMaster

From: US
Posts: 20
Votes: 80
It may help you a lot. Thanks point (x,y,z) --->(2, pi/6, 9) z = 18 cos(xy) ?z / ?x = fx = -18 y sin(xy) ?z / ?y = fy = -18 x sin(xy) The equation of the tangent plane: fx(x-2)+fy(y-pi/6)-(z-9)=0 -18Ysin(xy)(x-2) -18Xsin(xy)(y-pi/6) - (z-9)=0 where xy = pi/3 , X=2, and Y =pi/6

Answer No.4

Anonymous

From: -
Posts: -
Votes: -
So the vector perpendicular to the plane lies on the graph z=e^x *ln(y) we already have a point (3,1,0) let another point be (1,1,0)[it satisfies the equation z=exp(x)*ln(y)] then the vector perpendicular to the plane is (3-1,1-1,0-0)=(2,0,0) so the equation of plane becomes 2*(x-3)+0*(y-0)+0*(z-0)=0 ==>x=3

Answer No.5

GamaIntegral3

From: US
Posts: 4
Votes: 4
So the vector perpendicular to the plane lies on the graph z=e^x *ln(y) we already have a point (3,1,0) let another point be (1,1,0)[it satisfies the equation z=exp(x)*ln(y)] then the vector perpendicular to the plane is (3-1,1-1,0-0)=(2,0,0) so the equation of plane becomes 2*(x-3)+0*(y-0)+0*(z-0)=0 ==>x=3

Answer No.6

Anonymous

From: -
Posts: -
Votes: -
Given that z=e^xlny or,point (x,y,z) -->(2, pi/6, 9) z = 18 cos(xy) The equation of the tangent plane: fx(x-2)+fy(y-pi/6)-(z-9)=0 -18Ysin(xy)(x-2) -18Xsin(xy)(y-pi/6) - (z-9)=0 where xy = pi/3 , X=2, and Y =pi/6

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