Semantron 24 Summer 2024

Partial derivaives

We define the vector point ( 𝑥 𝑦 𝑧

𝑥 𝑦 𝑧

) as a point on the plane which is tangent to the curve at ( 1 2 2 ) . (

)−

( 1 2 2 ) is then a direction vector parallel to the plane. We know that if two direction vectors are perpendicular to each other, their dot product equals 0 due to the formula cos(𝜃)= 𝑎⃗ ∙𝑏⃗ |𝑎||𝑏| . This gives:

𝑥 𝑦 𝑧

1 2 2

−8 −4 −1

((

)−(

))∙ (

)=0

𝑥 𝑦 𝑧

−8 −4 −1

1 2 2

−8 −4 −1

(

) ∙ (

)=(

) ∙ (

)

−8𝑥−4𝑦−𝑧=−8−8−2

Thus, the equation of the plane that is tangent to the curve 𝑧(𝑥, 𝑦) = 10 − 4𝑥 2 −𝑦 2 at the point ( 1 2 2 ) is 8𝑥+4𝑦+𝑧=18 .

Utilization of chain rule for partial derivatives

Likewise, we can use the chain rule for partial derivatives to simplify difficult questions. Take the functions below for example, it is a function of x and y but those are both respectively functions of s and t. How would you find 𝜕𝑓 𝜕𝑡 ?

𝑥 = 𝑡 3 −𝑠 2 ,

𝑦 = √2𝑠 + 𝑡 2

𝑓(𝑥, 𝑦) = tan(2𝑥) ln(3𝑦) ,

Take a look at the general situation in which a change in the function can be observed. A change in the value of the function can come about from either a change along the x axis or a change along the y axis or both. This gives the line below:

130

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