Can some one teach me how to solve this problem in calculus????

Question

Hi, I need someone to teach me how to solve this problem:

 

Find an equation of the tangent plane to the surface 

ƒ(x,y) = tan(x²y) 

 

at the point P where x= 1, y=π(pi)/4 . State the equation in simplified exact form (no decimals) 

 

 

--- and Find the linear approximation L (x,y) for this function at P. compare L(x,y) and ƒ(x,y) for x=1.1, y=π(pi)/8.

State your answers in decimal form accurate to 4 places. 

 

 

Thank you,, 

Answer 1

The way to solve this type of equation is to work with the gradient operator (grad).

The grad operator involves the partial derivatives of the function  f =  tan(x²y).

The grad operator creates a vector function with  i component equal to the partial with respect to x  etc.

 

grad f  =  2 xy /cos²(x²y) i^→  + x² / cos(x² y)  j^→  

 

Evaluation of this at  the point P  where x = 1  and y =  Π/4    results in

 

grad =  (Π/2)√2   i^→  +  √2  j^→  

 

The general equation for a plane in three dimensions is   a x + b y + c z  =1

for some constants a, b,c   ( called direction cosines)

Any vector perpendicular to this plane is parallel to    N^→ =  a i^→ + b j^→  + c k^→

 

The needed conditions  is for   grad to be parallel to  N^→    This is because the gradient is perpendicular to the tangent plane.    This condition leads to a set of three equations:

 

λ a =   (Π/2) √2   

 

λ b =  √2 

 

λ c =  0

 

for some value of a constant λ

 

The equation of the plane can be rewritten (without loss of generality) as

 

λ a x  +  λ b y  + λ c z = λ

substituting from the set of three equations with x = 1  and y =  Π/4   yields

 

(Π/2) √2 + (Π/4) √2  =  λ     so  λ =  (3Π/4) √2

 

The final equation of the line tangent at the point P is thus

 (Π/2) x + y  =  (3Π/4)

 

The linear estimate asked for is obtained using a generalization of a Taylor series

 

f ~  f(x=1, y = Π/4)  + grad • [   ( 1.1- 1) i^→  + ( (Π/8 ) -(Π/4) ) j^→  ]

 

This evaluates to  1 + .22214 - .55536 =   0.6669