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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

STEP 1

1. We are given the function F(r,s)=3r21s0.5+1s3 F(r, s) = 3r^2 \frac{1}{s^{0.5}} + \frac{1}{s^3} .
2. We need to find the equation of the tangent plane at the point (1,1) (1, 1) .

STEP 2

1. Calculate the partial derivatives of F(r,s) F(r, s) with respect to r r and s s .
2. Evaluate the partial derivatives at the point (1,1) (1, 1) .
3. Use the point and the evaluated partial derivatives to write the equation of the tangent plane.

STEP 3

Find the partial derivative of F(r,s) F(r, s) with respect to r r :
Fr(r,s)=∂∂r(3r21s0.5+1s3) F_r(r, s) = \frac{\partial}{\partial r} \left( 3r^2 \frac{1}{s^{0.5}} + \frac{1}{s^3} \right)
Fr(r,s)=∂∂r(3r21s0.5)+∂∂r(1s3) F_r(r, s) = \frac{\partial}{\partial r} \left( 3r^2 \frac{1}{s^{0.5}} \right) + \frac{\partial}{\partial r} \left( \frac{1}{s^3} \right)
Fr(r,s)=6r1s0.5+0 F_r(r, s) = 6r \frac{1}{s^{0.5}} + 0

STEP 4

Find the partial derivative of F(r,s) F(r, s) with respect to s s :
Fs(r,s)=∂∂s(3r21s0.5+1s3) F_s(r, s) = \frac{\partial}{\partial s} \left( 3r^2 \frac{1}{s^{0.5}} + \frac{1}{s^3} \right)
Fs(r,s)=3r2∂∂s(1s0.5)+∂∂s(1s3) F_s(r, s) = 3r^2 \frac{\partial}{\partial s} \left( \frac{1}{s^{0.5}} \right) + \frac{\partial}{\partial s} \left( \frac{1}{s^3} \right)
Fs(r,s)=3r2(−12s−1.5)−3s−4 F_s(r, s) = 3r^2 \left( -\frac{1}{2} s^{-1.5} \right) - 3s^{-4}
Fs(r,s)=−3r22s1.5−3s4 F_s(r, s) = -\frac{3r^2}{2s^{1.5}} - \frac{3}{s^4}

STEP 5

Evaluate Fr(r,s) F_r(r, s) at the point (1,1) (1, 1) :
Fr(1,1)=6(1)110.5=6 F_r(1, 1) = 6(1) \frac{1}{1^{0.5}} = 6

STEP 6

Evaluate Fs(r,s) F_s(r, s) at the point (1,1) (1, 1) :
Fs(1,1)=−3(1)22(1)1.5−314 F_s(1, 1) = -\frac{3(1)^2}{2(1)^{1.5}} - \frac{3}{1^4}
Fs(1,1)=−32−3=−92 F_s(1, 1) = -\frac{3}{2} - 3 = -\frac{9}{2}

STEP 7

Evaluate F(r,s) F(r, s) at the point (1,1) (1, 1) :
F(1,1)=3(1)2110.5+113 F(1, 1) = 3(1)^2 \frac{1}{1^{0.5}} + \frac{1}{1^3}
F(1,1)=3+1=4 F(1, 1) = 3 + 1 = 4

STEP 8

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)(r−1)+Fs(1,1)(s−1) z = F(1, 1) + F_r(1, 1)(r - 1) + F_s(1, 1)(s - 1)
z=4+6(r−1)−92(s−1) z = 4 + 6(r - 1) - \frac{9}{2}(s - 1)
z=4+6r−6−92s+92 z = 4 + 6r - 6 - \frac{9}{2}s + \frac{9}{2}
z=6r−92s+52 z = 6r - \frac{9}{2}s + \frac{5}{2}
The equation of the tangent plane is:
z=6r−92s+52 \boxed{z = 6r - \frac{9}{2}s + \frac{5}{2}}

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