Question #138348

A student was asked to find the equation of the tangent plane to the surface z=x^4−y^5 at the point (x,y)=(4,1). The student's answer was z=255+4x^3 (x−4)−(5y^4)(y−1).

(a) At a glance, how do you know this is wrong. What mistakes did the student make? Select all that apply.

A. The answer is not a linear function.
B. The partial derivatives were not evaluated a the point.
C. The 255 should not be in the answer.
D. The (x - 4) and (y - 1) should be x and y.
E. All of the above

(b) Find the correct equation for the tangent plane.
z=

Expert's answer

(a) At a glance this is wrong because of:

A. The answer is not a linear function;

B. The partial derivatives were not evaluated at the point.


(b) Common equation for the tangent plane:

Fx′(x0,y0,z0)(x−x0)+Fy′(x0,y0,z0)(y−y0)+F_x'(x_0,y_0,z_0)(x-x_0)+F_y'(x_0,y_0,z_0)(y-y_0)+

+Fz′(x0,y0,z0)(z−z0)=0+F_z'(x_0,y_0,z_0)(z-z_0)=0 , then:

z0=x04−y05=(4)4−(1)5=256−1=255;z_0=x^4_{0}-y^5_{0}=(4)^4-(1)^5=256-1=255;

Find the equation of the tangent plane to the surface z=x4−y5z=x^4-y^5 at the point (x0,y0,z0)=(4,1,255)(x_0,y_0,z_0)=(4,1,255):

F(x,y,z)=−z+x4−y5=0F(x,y,z)=-z+x^4-y^5=0 ;

Fx′=4x3F_x'=4x^3 ; Fx′(x0,y0,z0)=4⋅43=256F_x'(x_0,y_0,z_0)=4\cdot4^3=256 ;

Fy′=−5y4F_y'=-5y^4 ; Fy′(x0,y0,z0)=−5⋅14=−5F_y'(x_0,y_0,z_0)=-5\cdot1^4=-5 ;

Fz′=−1F_z'=-1 ; Fz′(x0,y0,z0)=−1F_z'(x_0,y_0,z_0)=-1 ;

256(x−x0)−5(y−y0)−(z−z0)=0256(x-x_0)-5(y-y_0)-(z-z_0)=0 ;

256(x−4)−5(y−1)−(z−255)=0256(x-4)-5(y-1)-(z-255)=0 ;

z=255+256(x−4)−5(y−1)z=255+256(x-4)-5(y-1).

Answer: (a)A, B;

(b) z=255+256(x−4)−5(y−1)z=255+256(x-4)-5(y-1).


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