Partial derivaives
𝜕𝑧 𝜕𝑥
𝜕𝑧 𝜕𝑦
(1,2)=−8,
(1,2)=−4
2
Therefore, the tangent line at (1,2) for the curve 𝑧(𝑥, 𝑦) = 10 − 4𝑥 2 −𝑦 2 for the plane 𝑥 =1 has a slope of −8 , while for the plane 𝑦 =2 has a slope of −4 . Thus, the direction vector of the line tangent to the curve at (1,2) for the plane 𝑥 =1 is ( 0 1 −4 ) because as the y value increases by 1, the z value decreases by 4, and the x value is kept constant as we are moving along the plane 𝑥 =1 . For the same reasons, the direction vector of the line for the plane 𝑦 =
1 0 −8
2 is (
) .
Below we calculate the value of 𝑧(1,2) to find the position vector so that we can put together our vector equations.
𝑧(1,2) = 10 − 4(1) 2 −(2) 2
𝑧(1,2)=2
Thus, below are the vector equations for tangent lines to the curve for the plane 𝑥 =1 and 𝑦 =2 .
1 0 −8
1 2 2
0 1 −4
1 2 2
𝑟 𝑥=1 =(
)+𝜆(
) ,
𝑟 𝑦=2 =(
)+𝜇(
)
1+𝜇 2 2−8𝜇
1 2+𝜆 2−4𝜆
𝑟 𝑥=1 =(
),
𝑟 𝑦=2 =(
)
Using the two direction vectors, we can also find the plane tangent to the curve at ( 1 2 2
) like so. First, we
use the cross product to find the direction vector perpendicular to the two direction vectors.
1 0 −8
0 1 −4
−8 −4 −1
(
)×(
)=(
)
2 Dawkins (2003).
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