This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: HW09 – gilbert – (55035) 1 This printout should have 18 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find an equation for the plane passing through the origin that is parallel to the tangent plane to the graph of z = f (x, y) = 4x 2 − 2y 2 − 3x + y at the point (1, 1, f (1, 1)). 1. z + 5x + 3y − 8 = 0 2. z − 5x + 3y = 0 correct 3. z − 5x − 3y = 0 4. z + 5x − 3y − 2 = 0 5. z + 5x + 3y = 0 6. z − 5x − 3y + 8 = 0 Explanation: Parallel planes have parallel normals. On the other hand, the tangent plane to the graph of z = f (x, y) at the point (a, b, f (a, b)) has normal n = h−f x (a, b), −f y (a, b), 1 O . But when f (x, y) = 4x 2 − 2y 2 − 3x + y we see that f x = 8x − 3 , f y = −4y + 1 , and so when a = 1, b = 1, n = h−5, 3, 1 O . Thus an equation for the plane through the origin with normal parallel to n is hx, y, z O·n = hx, y, z O·h−5, 3, 1 O = 0 , which after evaluation becomes z − 5x + 3y = . keywords: 002 10.0 points Find an equation for the tangent plane to the graph of p f (x, y) = 5 − x 2 + 3y 2 at the point P (2, 1, f (2, 1)). 1. 3x + 2y − 2z − 5 = 2. 2x − 3y + 2z − 5 = 0 correct 3. 2x − 3y − 2z + 3 = 0 4. 3x − 2y + 2z + 3 = 0 5. 2x + 3y + 2z − 11 = 0 6. 3x + 2y − 2z − 11 = 0 Explanation: The equation of the tangent plane to the graph of z = f (x, y) at the point P (a, b, f (a, b)) is given by z = f (a, b) + ∂f ∂x ₃ (a, b) (x − a) + ∂f ∂y ₃ (a, b) (y − b) . ₃ ₃ ₃ ₃ Now when f (x, y) = p 5 − x 2 + 3y 2 , we see that ∂f x ∂x = − p 5 − x 2 + 3y 2 , while ∂f 3y ∂y = p 5 − x 2 + 3y 2 . HW09 – gilbert – (55035) 2 Thus at P , But the partial derivatives of f (2, 1) = 2 , while ∂f ∂x ₃ (2, 1) = − 1 , ∂f ∂y ₃ (2, 1) = 2 3 . ₃ ₃ ₃ ₃ So at P the tangent plane has equation z = 2 − (x − 2) + 3 2 (y − 1) , which after rearrangement becomes 2x − 3y + 2z − 5 = 0 . f (x, y) = tan −1 (x + 4y) are f x = 1 , f y = 4 . 1 + (x + 4y) 2 1 + (x + 4y) 2 It follows that ∂f ∂x ₃ (1, 0) = 2 1 ₃ and ₃ ∂f ∂y ₃ (1, 0) = 2 . ₃ Consequently, ₃ keywords: tangent plane, partial derivative, radical function, square root function, 003 10.0 points Find the linearization, L(x, y), of the func tion f (x, y) = tan −1 (x + 4y) at the point (1, 0) . π 1 2. L(x, y) = x + 2y + 4 + 2 3. L(x, y) = 1 2 x + 2y − 1 2 4. L(x, y) = x + 2y + π 4 − 1 2 5. L(x, y) = x + 2y − 1 2 6. L(x, y) = 1 2 x + 2y + π 4 − 1 2 correct Explanation: The linearization of f = f (x, y) at a point (a, b) is given by L(x, y) = f (a, b) + (x − a) ∂f ₃ +(y − b) ∂f ₃ ....
View
Full Document
 Spring '07
 Sadler
 Chain Rule, Derivative, dt

Click to edit the document details