Math  /  Calculus

QuestionFind an equation of the tangent plane at the given point: F(r,s)=3r21s0.5+1s3,\begin{array}{l} F(r, s)=3 r^{2} \frac{1}{s^{0.5}}+\frac{1}{s^{3}}, \end{array} Submit answer Next item

Studdy Solution
Use the point and the evaluated partial derivatives to write the equation of the tangent plane:
The equation of the tangent plane at (1,1) (1, 1) is given by:
z=F(1,1)+Fr(1,1)(r1)+Fs(1,1)(s1) z = F(1, 1) + F_r(1, 1)(r - 1) + F_s(1, 1)(s - 1)
z=4+6(r1)92(s1) z = 4 + 6(r - 1) - \frac{9}{2}(s - 1)
z=4+6r692s+92 z = 4 + 6r - 6 - \frac{9}{2}s + \frac{9}{2}
z=6r92s+52 z = 6r - \frac{9}{2}s + \frac{5}{2}
The equation of the tangent plane is:
z=6r92s+52 \boxed{z = 6r - \frac{9}{2}s + \frac{5}{2}}

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