If x2+y2+z2=9, dx / dt = 5, and dy /d1 = 4, find dz/ dt when (x, y, z) = (2, 2, 1).

x2+y2+z2=9Differentiate both sides with respect to td(x2+y2+z2)dt=d(9)dtd(x2)dx+d(y2)dt+d(z2)dt=0Use chain rule as shown below:d(x2)dx×dxdt+d(y2)dy×dydt+d(z2)dz×dzdt=0(2x)×dxdt+(2y)×dydt+(2z)×dzdt=0Divide both sides by 2(x)×dxdt+(y)×dydt+(z)×dzdt=0Substitute x=2, y=2, z=1 and dxdt=5,dydt=4(2)×(5)+(2)×(4)+(1)×dzdt=010+8+dzdt=0dzdt=−18Result:dzdt=−18

Get the solution to "If x2+y2+z2=9, dx / dt = 5, and..."

Grilletta1m

Open question

2022-08-19

x^{2} + y^{2} + z^{2} = 9

, dx / dt = 5, and dy /d1 = 4, find dz/ dt when (x, y, z) = (2, 2, 1).

Answer & Explanation

micelarnyiz

Beginner2022-08-20Added 8 answers

x^{2} + y^{2} + z^{2} = 9

Differentiate both sides with respect to t

\frac{d (x^{2} + y^{2} + z^{2})}{d t} = \frac{d (9)}{d t}

\frac{d (x^{2})}{d x} + \frac{d (y^{2})}{d t} + \frac{d (z^{2})}{d t} = 0

Use chain rule as shown below:

\frac{d (x^{2})}{d x} \times \frac{d x}{d t} + \frac{d (y^{2})}{d y} \times \frac{d y}{d t} + \frac{d (z^{2})}{d z} \times \frac{d z}{d t} = 0

(2 x) \times \frac{d x}{d t} + (2 y) \times \frac{d y}{d t} + (2 z) \times \frac{d z}{d t} = 0

Divide both sides by 2

(x) \times \frac{d x}{d t} + (y) \times \frac{d y}{d t} + (z) \times \frac{d z}{d t} = 0

Substitute x=2, y=2, z=1 and

\frac{d x}{d t} = 5, \frac{d y}{d t} = 4

(2) \times (5) + (2) \times (4) + (1) \times \frac{d z}{d t} = 0

10 + 8 + \frac{d z}{d t} = 0

\frac{d z}{d t} = - 18

Result:

\frac{d z}{d t} = - 18

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Calculus 2

Didn't find what you were looking for?