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Sagot :
Answer:
[tex]\displaystyle \frac{dh}{dt} = \frac{1}{20 \pi} \ cm/s[/tex]
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
- Left to Right
Algebra I
- Equality Properties
Geometry
- Volume of a Cone: [tex]\displaystyle V = \frac{1}{3} \pi r^2h[/tex]
Calculus
Derivatives
Derivative Notation
Differentiating with respect to time
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Step-by-step explanation:
Step 1: Define
[tex]\displaystyle \frac{r}{h} = \frac{2}{3} \\\frac{dV}{dt} = 5 \ cm^3/s\\h = 15 \ cm[/tex]
Step 2: Rewrite Cone Volume Formula
Find the volume of the cone with respect to height.
- Define ratio: [tex]\displaystyle \frac{r}{h} = \frac{2}{3}[/tex]
- Isolate r: [tex]\displaystyle r = \frac{2}{3} h[/tex]
- Substitute in r [VC]: [tex]\displaystyle V = \frac{1}{3} \pi (\frac{2}{3}h)^2h[/tex]
- Exponents: [tex]\displaystyle V = \frac{1}{3} \pi (\frac{4}{9}h^2)h[/tex]
- Multiply: [tex]\displaystyle V = \frac{4}{27} \pi h^3[/tex]
Step 3: Differentiate
- Basic Power Rule: [tex]\displaystyle \frac{dV}{dt} = \frac{4}{27} \pi \cdot 3 \cdot h^{3-1} \cdot \frac{dh}{dt}[/tex]
- Simplify: [tex]\displaystyle \frac{dV}{dt} = \frac{4}{9} \pi h^{2} \frac{dh}{dt}[/tex]
Step 4: Find Height Rate
Find dh/dt.
- Substitute in known variables: [tex]\displaystyle 5 \ cm^3/s = \frac{4}{9} \pi (15 \ cm)^{2} \frac{dh}{dt}[/tex]
- Isolate dh/dt: [tex]\displaystyle \frac{5 \ cm^3/s}{\frac{4}{9} \pi (15 \ cm)^{2} } = \frac{dh}{dt}[/tex]
- Rewrite: [tex]\displaystyle \frac{dh}{dt} = \frac{5 \ cm^3/s}{\frac{4}{9} \pi (15 \ cm)^{2} }[/tex]
- Evaluate Exponents: [tex]\displaystyle \frac{dh}{dt} = \frac{5 \ cm^3/s}{\frac{4}{9} \pi (225 \ cm^2) }[/tex]
- Evaluate Multiplication: [tex]\displaystyle \frac{dh}{dt} = \frac{5 \ cm^3/s}{100 \pi cm^2 }[/tex]
- Simplify: [tex]\displaystyle \frac{dh}{dt} = \frac{1}{20 \pi} \ cm/s[/tex]
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