Math 221 Quiz 4 Spring 2013 Several Variable Calculus 1. Suppose π§ = π₯ ππ¦ . Find an equation of the plane tangent to this surface at the point (1, 0, 1). Solution. This problem is Exercise 36 on page 790. One equation for the tangent plane is the following: π§−1= ππ§ ππ§ (1, 0, 1)(π₯ − 1) + (1, 0, 1)(π¦ − 0). ππ₯ ππ¦ ππ§ ππ§ ππ§ ππ§ = ππ¦ , so (1, 0, 1) = 1, while = π₯ ππ¦ , so (1, 0, 1) = 1. ππ₯ ππ₯ ππ¦ ππ¦ Substituting this information into the preceding equation yields the following result: π§ − 1 = (π₯ − 1) + π¦, or π§ = π₯ + π¦. Now An alternative method is to rewrite the equation of the surface as follows: π§ − π₯ ππ¦ = 0. This equation displays the surface as a level surface, and the gradient vector β¨−ππ¦ , −π₯ ππ¦ , 1β© is normal to the surface. At the specified point, the gradient vector becomes β¨−1, −1, 1β©, so an equation for the tangent plane is the following: −π₯ − π¦ + π§ = π, where π = −1 − 0 + 1 = 0. This answer again says that π§ = π₯ + π¦. 2. If π§ = π (π₯, π¦) and π₯ = π(π , π‘) and π¦ = β(π , π‘), then π§ can be viewed as a function of π and π‘. Suppose at a certain point ππ ππ ππ ππ πβ πβ = 2, = 3, = 5, = 7, = −1, = −4. ππ₯ ππ¦ ππ ππ‘ ππ ππ‘ Use this information to determine the value of ππ§ . ππ‘ Solution. One method is to say that ππ§ ππ§ ππ₯ ππ§ ππ¦ ππ ππ ππ πβ = + = + = 2 × 7 + 3 × (−4) = 2. ππ‘ ππ₯ ππ‘ ππ¦ ππ‘ ππ₯ ππ‘ ππ¦ ππ‘ Alternatively, you could argue with differentials as follows: dπ§ = February 7, 2013 ππ ππ dπ₯ + dπ¦ = 2 dπ₯ + 3 dπ¦. ππ₯ ππ¦ Page 1 of 2 Dr. Boas Math 221 Quiz 4 Spring 2013 Several Variable Calculus Similarly ππ ππ dπ + dπ‘ = 5 dπ + 7 dπ‘, ππ ππ‘ πβ πβ dπ¦ = dπ + dπ‘ = − dπ − 4 dπ‘. ππ ππ‘ dπ₯ = Substituting the expressions for dπ₯ and dπ¦ into the expression for dπ§ shows that dπ§ = (10 dπ + 14 dπ‘) + (−3 dπ − 12 dπ‘) = 7 dπ + 2 dπ‘. ππ§ ππ§ ππ§ ππ§ Since dπ§ = dπ + dπ‘, it follows that = 7 and = 2. ππ ππ‘ ππ ππ‘ 3. Suppose π (π₯, π¦, π§) = π§ ππ₯π¦ . Find the direction in which π (π₯, π¦, π§) increases most rapidly at the point (0, 1, 2). Solution. This problem is Exercise 56 on page 791. The gradient vector points in the direction in which the function increases most rapidly. Now β© β¨ ππ ππ ππ , , = β¨π¦π§ ππ₯π¦ , π₯π§ ππ₯π¦ , ππ₯π¦ β© , ∇π = ππ₯ ππ¦ ππ§ so ∇π (0, 1, 2) = β¨2,β¨ 0, 1β©. The vector β© β¨2, 0, 1β© is an acceptable answer, and 2 1 so is the unit vector √ , 0, √ . 5 5 February 7, 2013 Page 2 of 2 Dr. Boas