Derivative of s with respect to t when y is constant

  • MHB
  • Thread starter karush
  • Start date
In summary, when $y$ is constant, the derivative of $s$ with respect to $t$ can be found using the chain rule as $\frac{ds}{dt}=\frac{-6x\d{x}{t}+2y\frac{dy}{dt}}{\sqrt{6x^2 +2 y^2 }}$. Since $y$ is constant, \d{y}{t}=0 and therefore the term $2y\d{y}{t}$ in the numerator vanishes. Additionally, the denominator can be simplified to $\sqrt{6x^2}$, making the final result $\frac{ds}{dt}=\frac{-6x\d{x}{t}}{\sqrt{6x^
  • #1
karush
Gold Member
MHB
3,269
5
Let $x$ and $y$ be differentiable functions of $t$, and let
$$s=\sqrt{6x^2+2 y^2 }$$ be a function of $x$ and $y$.
A. How is ds/dt related to dx/dt if y is constant?
 
Last edited:
Physics news on Phys.org
  • #2
What would you get if you apply the chain rule to find \(\displaystyle \d{s}{t}\)?
 
  • #3
$$\frac{ds}{dt}=\frac{-6x\d{x}{t}+2y\frac{dy}{dt}}{\sqrt{6x^2 +2 y^2 }}$$

I got ? on this
 
  • #4
karush said:
$$\frac{ds}{dt}=\frac{-6x\d{x}{t}+2y\frac{dy}{dt}}{\sqrt{6x^2 +2 y^2 }}$$

I got ? on this

Where did the negative sign in front of the 6 come from?

If $y$ is a constant, then what is \(\displaystyle \d{y}{t}\)? What is the denominator equal to?
 
  • #5
A constant would go to zero

$\frac{ds}{dt}=\frac{6x\d{x}{t}+(0)}{\sqrt{6x^2 +2 y^2 }}$
 
Last edited:
  • #6
Yes, we would have \(\displaystyle \d{y}{t}=0\)...how about that denominator? Can you replace it with something simpler?
 
  • #7
Does the $2y^2 = 0\ \ $ ?
 
  • #8
karush said:
Does the $2y^2 = 0\ \ $ ?

No, but since it is being multiplied by \(\displaystyle \d{y}{t}=0\), the entire term does vanish.
 
  • #9
How does that happen?
 
  • #10
karush said:
How does that happen?

I'm sorry, I meant the term:

\(\displaystyle 2y\d{y}{t}\)

in the numerator vanishes to zero. The denominator of your result can be simplified in another way, if you look at the problem statement, before you did any differentiating.
 

FAQ: Derivative of s with respect to t when y is constant

What is the definition of ds/dt?

The notation ds/dt represents the rate of change of a variable s with respect to time t. It is also known as the derivative of s with respect to t.

What does dx/dt represent?

dx/dt is the notation for the rate of change of a variable x with respect to time t. It is also known as the derivative of x with respect to t.

How is ds/dt related to dx/dt?

ds/dt and dx/dt are related by the chain rule, which states that the derivative of a composite function is equal to the product of the derivatives of its individual functions. In this case, ds/dt is equal to (ds/dx) * (dx/dt).

What is the physical interpretation of ds/dt and dx/dt?

ds/dt represents the instantaneous velocity of an object along a curve at a specific point in time, while dx/dt represents the instantaneous velocity of an object along a straight line at a specific point in time.

Can ds/dt and dx/dt have different values for the same object?

Yes, ds/dt and dx/dt can have different values for the same object. This is because the object can be moving along a curve and a straight line simultaneously, and the rates of change for these two movements may be different.

Similar threads

I Partial derivatives of the function f(x,y)
Replies
6
Views
2K
I Partial Derivative of Convolution
Replies
2
Views
2K
I Is My Calculus of Variations Approach Correct?
Replies
12
Views
2K
I Second derivative, chain rules and order of operations
Replies
6
Views
3K
I Derivative of a function is equal to zero
Replies
4
Views
2K
I Lipschitz continuity of vector-valued function
Replies
3
Views
2K
I How to manipulate functions that are not explicitly given?
Replies
16
Views
1K
A Non solvable integral? (dx/dt)^2 dt
Replies
3
Views
2K
I Different Substitution for Solving an Integral
Replies
6
Views
660
MHB How to find F(x,y) for $f(x,y)= h(y)\hat{i} + g(x)\hat{j}?$
Replies
11
Views
2K

Hot Threads

Recent Insights

Back
Top