Answer to Question #138348 in Calculus for Promise Omiponle

Question

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=

Answer 1

(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:

$F_x'(x_0,y_0,z_0)(x-x_0)+F_y'(x_0,y_0,z_0)(y-y_0)+$

$+F_z'(x_0,y_0,z_0)(z-z_0)=0$ , then:

$z_0=x^4_{0}-y^5_{0}=(4)^4-(1)^5=256-1=255;$

Find the equation of the tangent plane to the surface $z=x^4-y^5$ at the point $(x_0,y_0,z_0)=(4,1,255)$:

$F(x,y,z)=-z+x^4-y^5=0$ ;

$F_x'=4x^3$ ; $F_x'(x_0,y_0,z_0)=4\cdot4^3=256$ ;

$F_y'=-5y^4$ ; $F_y'(x_0,y_0,z_0)=-5\cdot1^4=-5$ ;

$F_z'=-1$ ; $F_z'(x_0,y_0,z_0)=-1$ ;

$256(x-x_0)-5(y-y_0)-(z-z_0)=0$ ;

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

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

Answer: (a)A, B;

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