Solving: Finding $\frac{dT}{dt}$ for a Swimming Duck

[mathmari]

Hey!

Suppose that a duck swims in a circle $x=\cos t$, $y=\sin t$ and that the temperature of the water is given from the formula $T=x^2e^y-xy^3$. Find $\frac{dT}{dt}$:

1. using the chain rule
2. expressing $T$ as a function of $t$ and differentiating

1. At the chain rule do we not have $\frac{\partial}{\partial{t}}$ ?? Why are we looking for $\frac{dT}{dt}$ ?? (Wondering)
2. $$T(t)=x^2(t)e^{y(t)}-x(t)y^3(t)=\cos^2 t e^{\sin t}-\cos t \sin^3 t$$
   $$\frac{dT}{dt}=2 \cos t (-\sin t)e^{\sin t}+\cos^2 te^{\sin t} (\cos t)-(-\sin t)\sin^3 t-\cos t 3 \sin^2 t (\cos t)= -2 \cos t \sin t e^{\sin t}+\cos^3 t e^{\sin t}+\sin^4 t -3 \cos^2 t \sin^2 t$$
   
   Is this correct?? (Wondering) Could I improve somehing?? (Wondering)

 

[MarkFL]

For part 1, I would begin with:

\(\displaystyle T=x^2e^y-xy^3\)

Now, differentiating with respect to $t$, we find:

\(\displaystyle \d{T}{t}=x^2e^y\d{y}{t}+2x\d{x}{t}e^y-2xy^2\d{y}{t}-\d{x}{t}y^3\)

\(\displaystyle \d{T}{t}=xe^y\left(x\d{y}{t}+2\d{x}{t}\right)-y^2\left(2x\d{y}{t}+\d{x}{t}y\right)\)

Now, using the following:

\(\displaystyle \d{x}{t}=-y\)

\(\displaystyle \d{y}{t}=x\)

we may write:

\(\displaystyle \d{T}{t}=xe^y\left(x^2-2y\right)-y^2\left(2x^2-y^2\right)\)

See if this matches the result you obtained for part 2. :D

 

[I like Serena]

Hey!

Suppose that a duck swims in a circle $x=\cos t$, $y=\sin t$ and that the temperature of the water is given from the formula $T=x^2e^y-xy^3$. Find $\frac{dT}{\dt}$:

1. using the chain rule
2. expressing $T$ as a function of $t$ and differentiating

• At the chain rule do we not have $\frac{\partial}{\partial{t}}$ ?? Why are we looking for $\frac{dT}{dt}$ ?? (Wondering)

Hi mathmari! (Smile)

The first is a partial derivative, which is used when there are multiple variables, but we differentiate with respect to only one of them.

The second is a total derivative with respect to $t$, meaning that we apply the chain rule for any variable that depends on $t$. Or else that we expand $T$ making all dependencies on $t$ explicit, and differentiating only then.

That means:
$$\pd T x = 2xe^y-y^3$$
$$\pd T t = 0$$
$$\d T t = \pd T t + \pd T x \d x t + \pd T y \d y t$$

You're supposed to evaluate the last expression. (Thinking)

• $$T(t)=x^2(t)e^{y(t)}-x(t)y^3(t)=\cos^2 t e^{\sin t}-\cos t \sin^3 t$$
  $$\frac{dT}{dt}=2 \cos t (-\sin t)e^{\sin t}+\cos^2 te^{\sin t} (\cos t)-(-\sin t)\sin^3 t-\cos t 3 \sin^2 t (\cos t)= -2 \cos t \sin t e^{\sin t}+\cos^3 t e^{\sin t}+\sin^4 t -3 \cos^2 t \sin^2 t$$
  
  Is this correct?? (Wondering) Could I improve somehing?? (Wondering)

Completely correct! (Nod)

 

[mathmari]

The first is a partial derivative, which is used when there are multiple variables, but we differentiate with respect to only one of them.

The second is a total derivative with respect to $t$, meaning that we apply the chain rule for any variable that depends on $t$. Or else that we expand $T$ making all dependencies on $t$ explicit, and differentiating only then.

This mean that we have $$T(t)=x^2(t)e^{y(t)}-x(t)y^3(t)$$ so $T$ depends only on $t$, right ?? (Wondering)

Or does the way I wrote it mean that we write $T$ as a function of $t$, as we should do at the second part??

 

[I like Serena]

This mean that we have $$T(t)=x^2(t)e^{y(t)}-x(t)y^3(t)$$ so $T$ depends only on $t$, right ?? (Wondering)

Or does the way I wrote it mean that we write $T$ as a function of $t$, as we should do at the second part??

You can still go either way. You have only made the dependencies explicit.
That is, you can still first take a partial derivative with respect to x and apply the chain rule.
Or else, you can substitute the expression for x(t) and differentiate then, as you did in the second part. (Wasntme)

 

[mathmari]

So, we have to use the following version of the chain rule:

We suppose that $c: \mathbb{R} \rightarrow \mathbb{R}^3$ and $f: \mathbb{R}^3 \rightarrow \mathbb{R}$. Let $h(t)=f(c(t))=f(x(t),y(t),z(t))$, where $c(t)=(x(t),y(t),z(t))$. Then $$\frac{dh}{dt}=\frac{\partial{f}}{\partial{x}}\frac{dx}{dt}+\frac{\partial{f}}{\partial{y}}\frac{dy}{dt}+\frac{\partial{f}}{\partial{z}}\frac{dz}{dt}$$

right ?? (Wondering) Do I have to write $T$ as a composition of two functions, for example $T(t)=f(c(t))=f(x(t),y(t))=x^2(t)e^{y(t)}-x(t)y^3(t)$, where $c(t)=(x(t),y(t))$ ?? Or isn't it necessary ?? (Wondering)

 

[I like Serena]

So, we have to use he following version of the chain rule:

We suppose that $c: \mathbb{R} \rightarrow \mathbb{R}^3$ and $f: \mathbb{R}^3 \rightarrow \mathbb{R}$. Let $h(t)=f(c(t))=f(x(t),y(t),z(t))$, where $c(t)=(x(t),y(t),z(t))$. Then $$\frac{dh}{dt}=\frac{\partial{f}}{\partial{x}}\frac{dx}{dt}+\frac{\partial{f}}{\partial{y}}\frac{dy}{dt}+\frac{\partial{f}}{\partial{z}}\frac{dz}{dt}$$

right ?? (Wondering)

Yes. (Nod)

Do I have to write $T$ as a composition of two functions, for example $T(t)=f(c(t))=f(x(t),y(t))=x^2(t)e^{y(t)}-x(t)y^3(t)$, where $c(t)=(x(t),y(t))$ ?? Or isn't it necessary ?? (Wondering)

I don't think that is necessary. I think you're only supposed to show that you know what you're doing. Getting stuck into cosmetic notational issues isn't part of that. (Wasntme)

 

[mathmari]

So, do we have to write

$$\frac{dT}{dt}=\frac{\partial{T}}{\partial{x}}\frac{dx}{dt}+\frac{\partial{T}}{\partial{y}}\frac{dy}{dt}$$

?? (Wondering)

Or is it a little confusing that we write $T(t)=x^2(t)e^{y(t)}-x(t)y^3(t)$ and then we take $\frac{\partial{T}}{\partial{x}}$ ?? (Wondering)

 

[I like Serena]

So, do we have to write

$$\frac{dT}{dt}=\frac{\partial{T}}{\partial{x}}\frac{dx}{dt}+\frac{\partial{T}}{\partial{y}}\frac{dy}{dt}$$

?? (Wondering)

Yup. (Nod)

Or is it a little confusing that we write $T(t)=x^2(t)e^{y(t)}-x(t)y^3(t)$ and then we take $\frac{\partial{T}}{\partial{x}}$ ?? (Wondering)

That is confusing.
Formally, we can't take $\frac{\partial{T}}{\partial{x}}$ from $T(t)$, because $T(t)$ is not a function of x. (Worried)Actually, we're dealing with 2 distinct functions: $T_1(x,y)$ and $T_2(t)=T_1(x(t),y(t))$.
As you can see, they have different domains.
So we can talk about $\pd {T_1} x$, but not about $\pd {T_2} x$. (Doh)
When writing $\d T t$, this means $T_2'(t)$, which is equal to $\pd {T_1} x x'(t) + \pd {T_1} y y'(t)$ due to the chain rule. In practice these distinctions are often not made: we write $x$ instead of $x(t)$ unless we want to emphasize that $x$ is a function of $t$. And we write $T$ to mean any of $T(x,y)$, $T(t)$, or $T(t,x,y)$. (Malthe)