Related Rates and The Implicit

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Find  $\frac{dv}{dt}\ if\ v=t+\frac{8}{t}$  

•  $1-\frac{8}{t^2}$  

•  $1-\frac{8}{t}$  

•  $t-\frac{8}{t^2}$  

•  $1+\frac{8}{t^2}$  

Tags

CCSS.HSA.APR.D.6

CCSS.HSA.APR.D.7

CCSS.HSA.SSE.A.2

CCSS.HSA.SSE.B.3

• A

• B

• C

• D

Tags

CCSS.HSA.SSE.A.2

CCSS.HSN.RN.A.2

Implicit Differentiation is essentially using the __________rule, with y² = (f(x))² as one example

 $\frac{d}{dt}\left[\sin\theta\right]\ =$  

•  $\left(\cos\ \theta\right)\ \cdot\frac{d\theta}{dt}$  

•  $\left(-\cos\ \theta\right)\ \cdot\frac{d\theta}{dt}$  

•  $\sin\left(\frac{d\theta}{dt}\right)$  

• all of these

 $\frac{d}{dx}\left[\left(2y^3\ +\ x\right)^5\right]\ =\$  

•  $5\left(6y^2y'\ +\ 1\ \right)^4$  

•  $5\left(2y^3+x\right)^4\left(6y^2\ +\ 1\right)$  

•  $5\left(2y^3\ +\ x\right)^4\left(6y^2\ \frac{dy}{dx}\ +1\right)$  

• none of these

A water tank, shaped like an inverted circular cone, has a base radius of 6 ft and a height of 9 ft. The valve is opened and the water begins to decrease at a rate of 2 ft³/sec. When asked to find how fast the height of the water is changing when the water is 2 ft deep, what formula will we differentiate with respect to time?

• The Pythagorean Theorem

• The surface area of a cone

• The volume of a cone

• The height of a balloon

A water tank, shaped like an inverted circular cone, has a base radius of 6 ft and a height of 9 ft. The valve is opened and the water begins to decrease at a rate of 2 ft³/sec. Which of these are correct statements? Check all that are true.

• $\frac{r}{h}=\frac{6}{9}$

• $\frac{h}{r}=\frac{9}{6}$

• $r=\frac{2}{3}h$
  

• $h=\frac{3}{2}r$
  

A water tank, shaped like an inverted circular cone, has a base radius of 6 ft and a height of 9 ft. The tank is completely full and needs to be drained. **The valve is opened and the water begins to decrease at a rate of 2 ft³/sec.**  Which of these expressions represents the ** information?

•  $\frac{dV}{dt}=\ \frac{2ft^3}{\sec}$  

•  $\frac{dV}{dt}=-\frac{2ft^3}{\min}$  
  

•  $\frac{dV}{dt}=-\frac{2ft^3}{\sec}$  
  

• none of these

Find dy/dx by Implicit Differentiation 
x + 4y = 1

• -1/4
• -1 
• 4
• 1