Let $z = x ^ { 2 } - y ^ { 2 }$ where x = 3r - s, y = r + 2s. When r = 1, s = 1, $z _ { r }$ is

A) $- 1$
B) $\frac { 17 } { 6 }$
C)4
D)6
E) $\frac { 11 } { 6 }$

Question

Let $z = x ^ { 2 } - y ^ { 2 }$ where x = 3r - s, y = r + 2s. When r = 1, s = 1, $z _ { r }$ is

A) $- 1$
B) $\frac { 17 } { 6 }$
C)4
D)6
E) $\frac { 11 } { 6 }$

Quizplus · Accepted Answer

$z_r$ represents the partial derivative of $z$ with respect to $r$. To find $z_r$, we first substitute $x = 3r - s$ and $y = r + 2s$ into $z = x^2 - y^2$, then differentiate $z$ with respect to $r$. After substitution and differentiation, when $r = 1$ and $s = 1$, $z_r = 6$.

Let z=x2−y2z = x ^ { 2 } - y ^ { 2 }z=x2−y2

Let $z = x ^ { 2 } - y ^ { 2 }$