Question

Consider the function u:R³→R given by
\(u(x_1,x_2,x_3)=x_1x^4_2x^2_3-x^3_1x^4_3-26x^2_1x^2_2x^3_3\)
Let CεN be such that 
\(x_1\frac{ ∂u}{∂x_2}+2x_2\frac{ ∂u}{∂x_3} \)
evaluate at the point (t, t²,t³), eqals ct^k for every tεR, The n the value of K is equal to____.

Answer 1

Correct Answer: 14

To find the value of \(K\), we start by computing the partial derivatives of \(u(x_1, x_2, x_3)\). Given:
\(u(x_1,x_2,x_3)=x_1x_2^4x_3^2-x_1^3x_3^4-26x_1^2x_2^2x_3^3\)
First, determine \(\frac{∂u}{∂x_2}\):
\(\frac{∂u}{∂x_2} = 4x_1x_2^3x_3^2 - 0 - 52x_1^2x_2x_3^3\).
Next, compute \(\frac{∂u}{∂x_3}\):
\(\frac{∂u}{∂x_3} = 2x_1x_2^4x_3 - 4x_1^3x_3^3 - 78x_1^2x_2^2x_3^2\).
Now, evaluate the expression \(x_1\frac{∂u}{∂x_2} + 2x_2\frac{∂u}{∂x_3}\) at the point \((t, t^2, t^3)\).
\(x_1\frac{∂u}{∂x_2} = t(4t^6t^6 - 0 - 52t^2t^9) = 4t^{13} - 52t^{12}\),
\(2x_2\frac{∂u}{∂x_3} = 2t^2(2t^8t^3 - 4t^9t^3 - 78t^4t^6) = 4t^{13} - 8t^{14} - 156t^{12}\).
Sum these results:
\(4t^{13} - 52t^{12} + 4t^{13} - 8t^{14} - 156t^{12} = 8t^{13} - 8t^{14} - 208t^{12}\).
Factor and simplify:
\(-8t^{12} + 8t^{13} - 8t^{14} = -8t^{12}(1 - t + t^2)\).
This expression is \(ct^k\). Comparing powers, \(k = 14\).
Hence, the value of \(K\) is 14.