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Unformatted text preview: Math 23
B. Dodson
Week 3b Homework:
14.4 tangent plane, diﬀerentials
14.5 chain rule Problem 14.4.3:
Find the tangent plane to the surface
the z = f (x, y ) = 4 − x2 − 2y 2 at (1, −1, 1). Solution:
We recall that the tangent plane at (a, b, f (a, b))
is the plane with equation
z − f (a, b) = fx (a, b)(x − a) + fy (a, b)(y − b).
1 We compute fx = 1 (4 − x2 − 2y 2 )− 2 (−2x),
2
1 1
and fy = 2 (4 − x2 − 2y 2 )− 2 (−4y ),
1 so fx (1, −1) = −(4 − 1 − 2)− 2 = −1,
1 fy (1, −1) = 2(4 − 1 − 2)− 2 = 2, giving
z − 1 = −(x − 1) + 2(y + 1), or −x + 2y − z + 2 = 0,
the plane through (1, −1, 1) with normal
< fx , fy , −1 >=< −1, 2, −1 > .
As an example of diﬀerential (linear) approximation,
we solve #19, 14.4. 2 Week 4 Homework: 14.5 chain rule
Problem 14.5.3:
Find f ′ (t) when f (x, y ) = sin x cos y, with
√
x = x(t) = πt, y = y (t) = t.
Solution:
We have f ′ (t) = fx x′ + fy y ′ , with fx and fy
evaluated at (x, y ) = (x(t), y (t)). Here
fx = cos x cos y, fy = sin x(− sin y ), so
√
x = x(t) = πt, y = y (t) = t gives
√
√
√
fx (πt, t) = cos(πt) cos( t), fy = sin(πt)(− sin t).
1 Then x′ = π, y ′ = 1 t− 2 , so
2
√
√
1
f ′ (t) = π cos(πt) cos( t) + 1 t− 2 sin(πt)(− sin t).
2
We note that this calculation gives vector structure
to what is otherwise a straight calculus 1 calculation,
√
d
(sin(πt) cos( t)),
dt
using the product rule and the 1variable chainrule. The vector formulation is as a dot product, of a
vector < fx , fy >, with < x′ , y ′ >=< x′ (t), y ′ (t) >,
where < fx , fy > is evaluated at (x, y ) = (x(t), y (t)).
As a preview, the ﬁrst vector is called the gradient of f .
We also solved #13 and #23 of section 14.5. ...
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 Spring '06
 YUKICH
 Calculus, Chain Rule, Derivative, Gradient, 2 Week, 14.5 Chain

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