Introduction to Related Rates - AP Calculus BC
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What is the derivative of $C=2\text{π}r$ with respect to time?
What is the derivative of $C=2\text{π}r$ with respect to time?
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$\frac{dC}{dt} = 2\text{π}\frac{dr}{dt}$. Differentiate circumference with respect to time.
$\frac{dC}{dt} = 2\text{π}\frac{dr}{dt}$. Differentiate circumference with respect to time.
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Convert $C=2\pi r$ to a related rates form.
Convert $C=2\pi r$ to a related rates form.
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$\frac{dC}{dt}=2\pi\frac{dr}{dt}$. Differentiate circumference formula with respect to time.
$\frac{dC}{dt}=2\pi\frac{dr}{dt}$. Differentiate circumference formula with respect to time.
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How do you express $\frac{dC}{dt}$ for a circle with $C=2\pi r$?
How do you express $\frac{dC}{dt}$ for a circle with $C=2\pi r$?
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$\frac{dC}{dt} = 2\pi \frac{dr}{dt}$. Direct differentiation of circumference formula.
$\frac{dC}{dt} = 2\pi \frac{dr}{dt}$. Direct differentiation of circumference formula.
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Identify the related rates formula for $s = \frac{1}{2}at^2$.
Identify the related rates formula for $s = \frac{1}{2}at^2$.
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$\frac{ds}{dt} = at$. Differentiate quadratic position to get velocity formula.
$\frac{ds}{dt} = at$. Differentiate quadratic position to get velocity formula.
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Which mathematical process is essential in related rates?
Which mathematical process is essential in related rates?
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Differentiation with respect to time. Taking derivatives converts static equations to rate relationships.
Differentiation with respect to time. Taking derivatives converts static equations to rate relationships.
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What is the derivative of $A=\frac{1}{2}bh$ in related rates context?
What is the derivative of $A=\frac{1}{2}bh$ in related rates context?
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$\frac{dA}{dt} = \frac{1}{2}(b\frac{dh}{dt} + h\frac{db}{dt})$. Product rule for triangle area in related rates context.
$\frac{dA}{dt} = \frac{1}{2}(b\frac{dh}{dt} + h\frac{db}{dt})$. Product rule for triangle area in related rates context.
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Differentiate $V = \text{π}r^2h$ to find $\frac{dV}{dt}$.
Differentiate $V = \text{π}r^2h$ to find $\frac{dV}{dt}$.
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$\frac{dV}{dt} = \text{π}(2rh\frac{dr}{dt} + r^2\frac{dh}{dt})$. Product rule application to find cylinder volume rate.
$\frac{dV}{dt} = \text{π}(2rh\frac{dr}{dt} + r^2\frac{dh}{dt})$. Product rule application to find cylinder volume rate.
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What is the relationship between $\frac{dV}{dt}$ and $\frac{dr}{dt}$ for a sphere?
What is the relationship between $\frac{dV}{dt}$ and $\frac{dr}{dt}$ for a sphere?
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$\frac{dV}{dt} = 4\text{π}r^2\frac{dr}{dt}$. Surface area formula differentiated for sphere problems.
$\frac{dV}{dt} = 4\text{π}r^2\frac{dr}{dt}$. Surface area formula differentiated for sphere problems.
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Differentiate $V = \frac{4}{3}\text{π}r^3$ with respect to time.
Differentiate $V = \frac{4}{3}\text{π}r^3$ with respect to time.
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$\frac{dV}{dt} = 4\text{π}r^2\frac{dr}{dt}$. Chain rule applied to sphere volume formula.
$\frac{dV}{dt} = 4\text{π}r^2\frac{dr}{dt}$. Chain rule applied to sphere volume formula.
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How do you express $\frac{dV}{dt}$ for a cone, $V = \frac{1}{3}\text{π}r^2h$?
How do you express $\frac{dV}{dt}$ for a cone, $V = \frac{1}{3}\text{π}r^2h$?
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$\frac{dV}{dt} = \frac{1}{3}\text{π}(2rh\frac{dr}{dt} + r^2\frac{dh}{dt})$. Product rule applied to cone volume formula.
$\frac{dV}{dt} = \frac{1}{3}\text{π}(2rh\frac{dr}{dt} + r^2\frac{dh}{dt})$. Product rule applied to cone volume formula.
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Differentiate $s = ut + \frac{1}{2}at^2$ with respect to time.
Differentiate $s = ut + \frac{1}{2}at^2$ with respect to time.
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$\frac{ds}{dt} = u + at$. Derivative of position gives velocity in kinematics.
$\frac{ds}{dt} = u + at$. Derivative of position gives velocity in kinematics.
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Differentiate $A = lw$ with respect to time.
Differentiate $A = lw$ with respect to time.
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$\frac{dA}{dt} = l\frac{dw}{dt} + w\frac{dl}{dt}$. Product rule: each variable's rate times the other variable.
$\frac{dA}{dt} = l\frac{dw}{dt} + w\frac{dl}{dt}$. Product rule: each variable's rate times the other variable.
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How do you express $\frac{ds}{dt}$ for $s=ut+\frac{1}{2}at^2$?
How do you express $\frac{ds}{dt}$ for $s=ut+\frac{1}{2}at^2$?
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$\frac{ds}{dt} = u + at$. Velocity formula from differentiating position equation.
$\frac{ds}{dt} = u + at$. Velocity formula from differentiating position equation.
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Find $\frac{dV}{dt}$ for $V=\frac{4}{3}\text{π}r^3$, $r=5$, $\frac{dr}{dt}=0.3$.
Find $\frac{dV}{dt}$ for $V=\frac{4}{3}\text{π}r^3$, $r=5$, $\frac{dr}{dt}=0.3$.
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$\frac{dV}{dt} = 30\text{π}$. Calculate using $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} = 4\pi \cdot 25 \cdot 0.3$.
$\frac{dV}{dt} = 30\text{π}$. Calculate using $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} = 4\pi \cdot 25 \cdot 0.3$.
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What is $\frac{dA}{dt}$ when $A=\pi r^2$ and $r=5$, $\frac{dr}{dt}=0.3$?
What is $\frac{dA}{dt}$ when $A=\pi r^2$ and $r=5$, $\frac{dr}{dt}=0.3$?
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$\frac{dA}{dt} = 3\pi$. Substitute values into $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.
$\frac{dA}{dt} = 3\pi$. Substitute values into $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.
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What is the related rate derivative for $s = \text{π}r^2$?
What is the related rate derivative for $s = \text{π}r^2$?
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$\frac{ds}{dt} = 2\text{π}r\frac{dr}{dt}$. Differentiate area formula $s = \pi r^2$ with respect to time.
$\frac{ds}{dt} = 2\text{π}r\frac{dr}{dt}$. Differentiate area formula $s = \pi r^2$ with respect to time.
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What is the related rate for $A=\frac{1}{2}ab$?
What is the related rate for $A=\frac{1}{2}ab$?
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$\frac{dA}{dt} = \frac{1}{2}(a\frac{db}{dt} + b\frac{da}{dt})$. Product rule applied to triangle area formula.
$\frac{dA}{dt} = \frac{1}{2}(a\frac{db}{dt} + b\frac{da}{dt})$. Product rule applied to triangle area formula.
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What equation relates volume and radius for a cylinder?
What equation relates volume and radius for a cylinder?
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$V = \text{π}r^2h$. Basic cylinder volume relating radius and height.
$V = \text{π}r^2h$. Basic cylinder volume relating radius and height.
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Identify the relationship between $\frac{dA}{dt}$ and $\frac{da}{dt}$ for $A=a^2$.
Identify the relationship between $\frac{dA}{dt}$ and $\frac{da}{dt}$ for $A=a^2$.
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$\frac{dA}{dt} = 2a \times \frac{da}{dt}$. Power rule applied to area formula gives linear relationship.
$\frac{dA}{dt} = 2a \times \frac{da}{dt}$. Power rule applied to area formula gives linear relationship.
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When differentiating $x^2 + y^2 = r^2$, what is $\frac{dy}{dt}$?
When differentiating $x^2 + y^2 = r^2$, what is $\frac{dy}{dt}$?
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$\frac{dy}{dt} = \frac{-x}{y} \times \frac{dx}{dt}$. From implicit differentiation: $2x + 2y\frac{dy}{dt} = 0$.
$\frac{dy}{dt} = \frac{-x}{y} \times \frac{dx}{dt}$. From implicit differentiation: $2x + 2y\frac{dy}{dt} = 0$.
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What is the chain rule for related rates in implicit differentiation?
What is the chain rule for related rates in implicit differentiation?
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Use $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$. Standard chain rule application for composite functions.
Use $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$. Standard chain rule application for composite functions.
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What is the first step in solving a related rates problem?
What is the first step in solving a related rates problem?
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Identify all given information and the rate to be found. Essential setup before writing equations and differentiating.
Identify all given information and the rate to be found. Essential setup before writing equations and differentiating.
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State the chain rule used in related rates.
State the chain rule used in related rates.
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If $y=f(u)$ and $u=g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$. Links derivatives of composite functions for rate calculations.
If $y=f(u)$ and $u=g(x)$, then $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$. Links derivatives of composite functions for rate calculations.
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What is the definition of a related rates problem?
What is the definition of a related rates problem?
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A problem involving rates of change of related variables. Variables change together; find how one rate affects another.
A problem involving rates of change of related variables. Variables change together; find how one rate affects another.
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Differentiate $V = \text{π}r^2h$ with respect to time.
Differentiate $V = \text{π}r^2h$ with respect to time.
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$\frac{dV}{dt} = \text{π}(2rh\frac{dr}{dt} + r^2\frac{dh}{dt})$. Product rule applied to cylinder volume formula.
$\frac{dV}{dt} = \text{π}(2rh\frac{dr}{dt} + r^2\frac{dh}{dt})$. Product rule applied to cylinder volume formula.
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What is the general method for solving related rates?
What is the general method for solving related rates?
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Differentiate the relation between variables with respect to time. Core strategy: relate variables, then differentiate both sides.
Differentiate the relation between variables with respect to time. Core strategy: relate variables, then differentiate both sides.
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Which equation relates the rates of a circle's area and radius?
Which equation relates the rates of a circle's area and radius?
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Use $A = \text{π}r^2$ and differentiate with respect to time. Differentiating gives $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.
Use $A = \text{π}r^2$ and differentiate with respect to time. Differentiating gives $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.
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Differentiate $A = \frac{1}{2}bh$ with respect to time.
Differentiate $A = \frac{1}{2}bh$ with respect to time.
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$\frac{dA}{dt} = \frac{1}{2}(b\frac{dh}{dt} + h\frac{db}{dt})$. Product rule for triangle area with two variables.
$\frac{dA}{dt} = \frac{1}{2}(b\frac{dh}{dt} + h\frac{db}{dt})$. Product rule for triangle area with two variables.
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Find $\frac{dv}{dt}$ for a sphere with $V=\frac{4}{3}\text{π}r^3$, $r=7$, $\frac{dr}{dt}=0.2$.
Find $\frac{dv}{dt}$ for a sphere with $V=\frac{4}{3}\text{π}r^3$, $r=7$, $\frac{dr}{dt}=0.2$.
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$\frac{dv}{dt} = 117.6\text{π}$ cm³/s. Apply $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$ with given values.
$\frac{dv}{dt} = 117.6\text{π}$ cm³/s. Apply $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$ with given values.
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Find $\frac{dV}{dt}$ for a spherical balloon with $r=10$, $\frac{dr}{dt}=0.5$ cm/s.
Find $\frac{dV}{dt}$ for a spherical balloon with $r=10$, $\frac{dr}{dt}=0.5$ cm/s.
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$\frac{dV}{dt} = 200\text{π}$ cm³/s. Substitute $r=10$, $\frac{dr}{dt}=0.5$ into sphere rate formula.
$\frac{dV}{dt} = 200\text{π}$ cm³/s. Substitute $r=10$, $\frac{dr}{dt}=0.5$ into sphere rate formula.
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