Solving Intersection Curve at (1,1,1): Derivatives & Tangent Line

In summary, in this conversation, the problem of finding the values of the derivatives f'(1) and g'(1) for a curve of intersection of two surfaces near the point (1,1,1) is discussed. The tangent line to the curve at (1,1,1) is also requested, with x and y being equal to 1 + 4s and 1 + 7s respectively. The gradients and partial derivatives were used to attempt to solve the problem, but with no success. Any assistance in solving this problem would be appreciated.
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Homework Statement


Given that near (1,1,1) the curve of intersection of the surfaces

x^4 + y^2 + z^6 - 3xyz = 0 and xy + yz + zx - 3z^8 = 0

has the parametric equations x = f(t), y = g(t), z = t with f, g differentiable.

(a) What are the values of the derivatives f'(1), g'(1)?
(b) What is the tangent line to the curve of intersection at (1,1,1) given that z = 1 + s? (Find what x and y are equal to.)

Answers:
f'(1) = 4
g'(1) = 7
x = 1 + 4s
y = 1 + 7s

Homework Equations


At least gradients, and partial derivative taking.


The Attempt at a Solution


I took the gradients of each surface (the first one being f(x,y,z) and the second one being g(x,y,z) respectively). I tried to set the respective components equal to each other in an attempt to solve for something but I ended nowhere useful. I also tried to think of z = t = 1 and in order to try and find x and y. Then, I was like "Oh, wait, I have a point (1,1,1), let me just shove it into the gradients." all in order to get the correct answers and then to try and make sense of what I did but I didn't succeed :(.

Any help in solving this problem would be greatly appreciated!
Thanks in advance!
 
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  • #2


Is what I wrote confusing or something? If it is, please tell me what to change that way I could get an answer.
 

FAQ: Solving Intersection Curve at (1,1,1): Derivatives & Tangent Line

What is meant by solving intersection curve at (1,1,1)?

Solving intersection curve at (1,1,1) refers to finding the point of intersection between two or more curves in a three-dimensional space, where the coordinates of the point are (1,1,1).

Why are derivatives important in solving intersection curve at (1,1,1)?

Derivatives are important in solving intersection curve at (1,1,1) because they represent the rate of change of a curve at a given point. This means that they can help us determine the slope or direction of the curve at the point of intersection, which is crucial in finding the tangent line.

How do you find the tangent line at the point of intersection (1,1,1)?

To find the tangent line at the point of intersection (1,1,1), we first need to find the derivatives of the curves involved at that point. Then, we can use the derivative values to find the slope of the tangent line. Finally, we can use the slope and the coordinates of the point of intersection to write the equation of the tangent line.

What are some real-world applications of solving intersection curve at (1,1,1)?

Solving intersection curve at (1,1,1) has many real-world applications, such as in physics, engineering, and economics. For example, it can be used to calculate the optimal production level for a company, the trajectory of a projectile, or the intersection of two roads on a map.

Can you solve for the intersection curve at (1,1,1) without using derivatives?

It is not possible to solve for the intersection curve at (1,1,1) without using derivatives, as they are essential in finding the slope of the tangent line at the point of intersection. Without the slope, we cannot write the equation of the tangent line, and therefore, we cannot accurately determine the point of intersection.

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