Slope of tangent line to curves cut from surface

chetzread
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Homework Statement


find the slope of tangent line to curves cut from surface z = (3x^2) +(4y^2) - 6 by planes thru the point (1,1,1) and parallel to xz planes and yz planes ...

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The Attempt at a Solution


slope of tnagent that parallel to xz planes is dz/dy , while the slope of tangent that parallel to yz plane is dz /dz . Correct me if i am wrong
 
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To be parallel to the xz plane means that the y value is some constant number (y ≡ 1). So there is no dy. y is not involved in that slope except that its constant value (1) might be in the equation. Calculate dz/dx and plug in the 1s.
Similarly, to be parallel to the yz plane means that the x value is some constant number (x ≡ 1). So there is no dx. x is not involved in that slope except that its constant value (1) might be in the equation. Calculate dz/dy and plug in the 1s
 
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So
FactChecker said:
To be parallel to the xz plane means that the y value is some constant number (y ≡ 1). So there is no dy. y is not involved in that slope except that its constant value (1) might be in the equation. Calculate dz/dx and plug in the 1s.
Similarly, to be parallel to the yz plane means that the x value is some constant number (x ≡ 1). So there is no dx. x is not involved in that slope except that its constant value (1) might be in the equation. Calculate dz/dy and plug in the 1s
So , slope of tnagent that parallel to xz planes is dz/dx , while the slope of tangent thatparallel to yz plane is dz /dy . ?
 
chetzread said:
So

So , slope of tnagent that parallel to xz planes is dz/dx , while the slope of tangent thatparallel to yz plane is dz /dy . ?
yes
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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