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MarkFL
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Here are the questions:
I have posted a link there to this topic so the OP can see my work.
I have posted a link there to this topic so the OP can see my work.
Duncan G's Questions on Implicit Diff. & Related Rates
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In summary, we used implicit differentiation to find the second derivative of an implicitly defined curve, and then used it to solve for points where the tangent line has a given slope. We then used parametric equations to solve for the rate of change of the distance between two moving objects, and finally used similar triangles and implicit differentiation to solve for the rate of change of the position of a shadow.
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MarkFL
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MHB
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Hello Duncan G,
1.) We are given the implicitly defined curve:
\(\displaystyle x^{\frac{1}{2}}+y^{\frac{1}{2}}=1\)
Implcitly differentiating with respect to $x$, we find:
\(\displaystyle \frac{1}{2}x^{-\frac{1}{2}}+\frac{1}{2}y^{-\frac{1}{2}}\frac{dy}{dx}=0\)
Solving for \(\displaystyle \frac{dy}{dx}\), we obtain:
\(\displaystyle \frac{dy}{dx}=-\frac{y^{\frac{1}{2}}}{x^{\frac{1}{2}}}\)
Implicitly differentiating again with respect to $x$, we get:
\(\displaystyle \frac{d}{dx}\left(\frac{dy}{dx} \right)=-\frac{x^{\frac{1}{2}}\frac{1}{2}y^{-\frac{1}{2}}\frac{dy}{dx}-y^{\frac{1}{2}}\frac{1}{2}x^{-\frac{1}{2}}}{\left(x^{\frac{1}{2}} \right)^2}\)
Using the first derivative, and simplifying, we get:
\(\displaystyle \frac{d^2y}{dx^2}=\frac{1+\frac{y^{\frac{1}{2}}}{x^{\frac{1}{2}}}}{2x}\)
Multiplying by \(\displaystyle \frac{x^{\frac{1}{2}}}{x^{\frac{1}{2}}}\), we have:
\(\displaystyle \frac{d^2y}{dx^2}=\frac{x^{\frac{1}{2}}+y^{\frac{1}{2}}}{2x^{\frac{3}{2}}}\)
Using the fact that \(\displaystyle x^{\frac{1}{2}}+y^{\frac{1}{2}}=1\), we then get:
\(\displaystyle \frac{d^2y}{dx^2}=\frac{1}{2x^{\frac{3}{2}}}\)
We can check this result by explicitly differentiating. Let's write the original relation as:
\(\displaystyle y=\left(1-x^{\frac{1}{2}} \right)^2\)
\(\displaystyle \frac{dy}{dx}=2\left(1-x^{\frac{1}{2}} \right)\left(-\frac{1}{2}x^{-\frac{1}{2}} \right)=1-x^{-\frac{1}{2}}\)
\(\displaystyle \frac{d^2y}{dx^2}=\frac{1}{2}x^{-\frac{3}{2}}=\frac{1}{2x^{\frac{3}{2}}}\)
2.) We are given the implicit relation:
\(\displaystyle x^2y^2+xy=2\)
and asked to find the points where the slope of the tangent line is $-1$. So, implcitly differentiating with respect to $x$, we find:
\(\displaystyle x^2\cdot2y\frac{dy}{dx}+2xy^2+x\frac{dy}{dx}+y=0\)
Moving all terms not having \(\displaystyle \frac{dy}{dx}\) as a factor to the right, we get:
\(\displaystyle x^2\cdot2y\frac{dy}{dx}+x\frac{dy}{dx}=-2xy^2-y\)
Factoring both sides:
\(\displaystyle \left(x(2xy+1) \right)\frac{dy}{dx}=-y(2xy+1)\)
Hence:
\(\displaystyle \frac{dy}{dx}=-\frac{y}{x}\)
Equating this slope to $-1$, we obtain:
\(\displaystyle y=x\)
Substituting for $y$ in the original equation, and writing in standard form, we may solve for $x$:
\(\displaystyle x^4+x^2-2=0\)
Factor:
\(\displaystyle \left(x^2+2 \right)\left(x^2-1 \right)=0\)
The first factor has imaginary roots, so we are left with:
\(\displaystyle x=\pm1\)
Now, since $y=x$, the two points are then:
\(\displaystyle (-1,-1),\,(1,1)\)
Here is a plot of the curve ant its two tangent lines with slope $-1$:
View attachment 1501
3.) Let's orient our coordinate axes such that ship A is at the origin at noon, and so ship B is at $(150,0)$. Thus, we may describe the position of each ship parametrically as follows:
Ship A:
\(\displaystyle x=35t\)
\(\displaystyle y=0\)
Ship B:
\(\displaystyle x=150\)
\(\displaystyle y=25t\)
If we let $D$ be the distance between the ships at time $t$, we may use the distance formula to write:
\(\displaystyle D^2(t)=(35t-150)^2+(25t)^2=50\left(37t^2-210t+450 \right)\)
Implicitly differentiating with respect to $t$, we find:
\(\displaystyle 2D(t)D'(t)=50(74t-210)=100(37t-105)\)
Hence:
\(\displaystyle D'(t)=\frac{50(37t-105)}{D(t)}=\frac{\sqrt{50}(37t-105)}{\sqrt{37t^2-210t+450}}\)
At 4:00 pm, we have $t=4$, and so we find:
\(\displaystyle D'(4)=\frac{\sqrt{50}(37(4)-105)}{\sqrt{37(4)^2-210(4)+450}}=\frac{215}{\sqrt{101}}\,\frac{\text{km}}{\text{hr}}\)
Thus, at 4:00 pm the distance between the ships is increasing at a rate of about 21.3933 kph.
4.) Please refer to the following diagram:
View attachment 1502
$P$ is the height of the pole, $M$ is the height of the man, $D$ is the man's distance from the pole, and $x$ is the length of the shadow.
By similarity, we find:
\(\displaystyle \frac{D+x}{P}=\frac{x}{M}\)
\(\displaystyle DM+Mx=Px\)
\(\displaystyle x=\frac{DM}{P-M}\)
Observing that $x$ and $D$ are changing with time, we find by implcitly differentiating with respect to time $t$, we have:
\(\displaystyle \frac{dx}{dt}=\frac{M}{P-M}\frac{dD}{dt}\)
Also observing that movement of the tip of the man's shadow is the result not only of the shadow growing but also of the man's movement, we find that we must add these rates of change, so that the rate of change of the point $T$ of the tip of the shadow is given by:
\(\displaystyle \frac{dT}{dt}=\frac{dD}{dt}+\frac{dx}{dt}=\frac{dD}{dt}\left(1+\frac{M}{P-M} \right)=\frac{P}{P-M}\frac{dD}{dt}\)
Now, using the given data (did you notice we do not need $D$?):
\(\displaystyle P=15\text{ ft},\,M=6\text{ ft},\,\frac{dD}{dt}=5\,\frac{\text{ft}}{\text{s}}\)
We find:
\(\displaystyle \frac{dT}{dt}=\frac{15}{15-6}\cdot5\,\frac{\text{ft}}{\text{s}}=\frac{25}{3}\,\frac{\text{ft}}{\text{s}}\)
1.) We are given the implicitly defined curve:
\(\displaystyle x^{\frac{1}{2}}+y^{\frac{1}{2}}=1\)
Implcitly differentiating with respect to $x$, we find:
\(\displaystyle \frac{1}{2}x^{-\frac{1}{2}}+\frac{1}{2}y^{-\frac{1}{2}}\frac{dy}{dx}=0\)
Solving for \(\displaystyle \frac{dy}{dx}\), we obtain:
\(\displaystyle \frac{dy}{dx}=-\frac{y^{\frac{1}{2}}}{x^{\frac{1}{2}}}\)
Implicitly differentiating again with respect to $x$, we get:
\(\displaystyle \frac{d}{dx}\left(\frac{dy}{dx} \right)=-\frac{x^{\frac{1}{2}}\frac{1}{2}y^{-\frac{1}{2}}\frac{dy}{dx}-y^{\frac{1}{2}}\frac{1}{2}x^{-\frac{1}{2}}}{\left(x^{\frac{1}{2}} \right)^2}\)
Using the first derivative, and simplifying, we get:
\(\displaystyle \frac{d^2y}{dx^2}=\frac{1+\frac{y^{\frac{1}{2}}}{x^{\frac{1}{2}}}}{2x}\)
Multiplying by \(\displaystyle \frac{x^{\frac{1}{2}}}{x^{\frac{1}{2}}}\), we have:
\(\displaystyle \frac{d^2y}{dx^2}=\frac{x^{\frac{1}{2}}+y^{\frac{1}{2}}}{2x^{\frac{3}{2}}}\)
Using the fact that \(\displaystyle x^{\frac{1}{2}}+y^{\frac{1}{2}}=1\), we then get:
\(\displaystyle \frac{d^2y}{dx^2}=\frac{1}{2x^{\frac{3}{2}}}\)
We can check this result by explicitly differentiating. Let's write the original relation as:
\(\displaystyle y=\left(1-x^{\frac{1}{2}} \right)^2\)
\(\displaystyle \frac{dy}{dx}=2\left(1-x^{\frac{1}{2}} \right)\left(-\frac{1}{2}x^{-\frac{1}{2}} \right)=1-x^{-\frac{1}{2}}\)
\(\displaystyle \frac{d^2y}{dx^2}=\frac{1}{2}x^{-\frac{3}{2}}=\frac{1}{2x^{\frac{3}{2}}}\)
2.) We are given the implicit relation:
\(\displaystyle x^2y^2+xy=2\)
and asked to find the points where the slope of the tangent line is $-1$. So, implcitly differentiating with respect to $x$, we find:
\(\displaystyle x^2\cdot2y\frac{dy}{dx}+2xy^2+x\frac{dy}{dx}+y=0\)
Moving all terms not having \(\displaystyle \frac{dy}{dx}\) as a factor to the right, we get:
\(\displaystyle x^2\cdot2y\frac{dy}{dx}+x\frac{dy}{dx}=-2xy^2-y\)
Factoring both sides:
\(\displaystyle \left(x(2xy+1) \right)\frac{dy}{dx}=-y(2xy+1)\)
Hence:
\(\displaystyle \frac{dy}{dx}=-\frac{y}{x}\)
Equating this slope to $-1$, we obtain:
\(\displaystyle y=x\)
Substituting for $y$ in the original equation, and writing in standard form, we may solve for $x$:
\(\displaystyle x^4+x^2-2=0\)
Factor:
\(\displaystyle \left(x^2+2 \right)\left(x^2-1 \right)=0\)
The first factor has imaginary roots, so we are left with:
\(\displaystyle x=\pm1\)
Now, since $y=x$, the two points are then:
\(\displaystyle (-1,-1),\,(1,1)\)
Here is a plot of the curve ant its two tangent lines with slope $-1$:
View attachment 1501
3.) Let's orient our coordinate axes such that ship A is at the origin at noon, and so ship B is at $(150,0)$. Thus, we may describe the position of each ship parametrically as follows:
Ship A:
\(\displaystyle x=35t\)
\(\displaystyle y=0\)
Ship B:
\(\displaystyle x=150\)
\(\displaystyle y=25t\)
If we let $D$ be the distance between the ships at time $t$, we may use the distance formula to write:
\(\displaystyle D^2(t)=(35t-150)^2+(25t)^2=50\left(37t^2-210t+450 \right)\)
Implicitly differentiating with respect to $t$, we find:
\(\displaystyle 2D(t)D'(t)=50(74t-210)=100(37t-105)\)
Hence:
\(\displaystyle D'(t)=\frac{50(37t-105)}{D(t)}=\frac{\sqrt{50}(37t-105)}{\sqrt{37t^2-210t+450}}\)
At 4:00 pm, we have $t=4$, and so we find:
\(\displaystyle D'(4)=\frac{\sqrt{50}(37(4)-105)}{\sqrt{37(4)^2-210(4)+450}}=\frac{215}{\sqrt{101}}\,\frac{\text{km}}{\text{hr}}\)
Thus, at 4:00 pm the distance between the ships is increasing at a rate of about 21.3933 kph.
4.) Please refer to the following diagram:
View attachment 1502
$P$ is the height of the pole, $M$ is the height of the man, $D$ is the man's distance from the pole, and $x$ is the length of the shadow.
By similarity, we find:
\(\displaystyle \frac{D+x}{P}=\frac{x}{M}\)
\(\displaystyle DM+Mx=Px\)
\(\displaystyle x=\frac{DM}{P-M}\)
Observing that $x$ and $D$ are changing with time, we find by implcitly differentiating with respect to time $t$, we have:
\(\displaystyle \frac{dx}{dt}=\frac{M}{P-M}\frac{dD}{dt}\)
Also observing that movement of the tip of the man's shadow is the result not only of the shadow growing but also of the man's movement, we find that we must add these rates of change, so that the rate of change of the point $T$ of the tip of the shadow is given by:
\(\displaystyle \frac{dT}{dt}=\frac{dD}{dt}+\frac{dx}{dt}=\frac{dD}{dt}\left(1+\frac{M}{P-M} \right)=\frac{P}{P-M}\frac{dD}{dt}\)
Now, using the given data (did you notice we do not need $D$?):
\(\displaystyle P=15\text{ ft},\,M=6\text{ ft},\,\frac{dD}{dt}=5\,\frac{\text{ft}}{\text{s}}\)
We find:
\(\displaystyle \frac{dT}{dt}=\frac{15}{15-6}\cdot5\,\frac{\text{ft}}{\text{s}}=\frac{25}{3}\,\frac{\text{ft}}{\text{s}}\)
Attachments
1. What is implicit differentiation?
Implicit differentiation is a mathematical method used to find the derivative of a function that is not written in the form of y = f(x). Instead, the function is written in terms of both x and y, making it difficult to take the derivative using traditional methods. Implicit differentiation involves treating y as a function of x and using the chain rule to find the derivative.
2. How is implicit differentiation related to related rates?
Implicit differentiation is often used in problems involving related rates because it allows us to find the rate of change of a variable that is not explicitly written in the problem. This is useful in real-world situations where multiple variables are changing simultaneously and we need to find the rate of change of one variable with respect to another.
3. What is the chain rule and how is it used in implicit differentiation?
The chain rule is a calculus rule that is used to find the derivative of composite functions. In implicit differentiation, we use the chain rule to find the derivative of y with respect to x, treating y as a function of x. This allows us to find the derivative of a function that is not explicitly written in terms of x.
4. Can implicit differentiation be used to find higher order derivatives?
Yes, implicit differentiation can be used to find higher order derivatives, such as the second derivative or third derivative. This involves taking the derivative of the original function using implicit differentiation, and then using implicit differentiation again to find the derivative of the derivative.
5. In what types of problems is implicit differentiation commonly used?
Implicit differentiation is commonly used in problems involving curves and equations that cannot be easily solved for y in terms of x. This includes problems involving related rates, optimization, and implicit equations. It is also commonly used in physics and engineering problems involving changing variables.
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