What is the mean absolute deviation and standard deviation of 2,4,6,9,14
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_______________
dy
Find —— for an implicit function:
dx
cos(xy) = 3x + 1.
First, differentiate implicitly both sides with respect to x. Keep in mind that y is not just a variable, but it is also a function of x, so you have to use the chain rule there:
![\mathsf{\dfrac{d}{dx}\big[cos(xy)\big]=\dfrac{d}{dx}(3x+1)}\\\\\\ \mathsf{-\,sin(xy)\cdot \dfrac{d}{dx}(xy)=\dfrac{d}{dx}(3x)+\dfrac{d}{dx}(1)}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdfrac%7Bd%7D%7Bdx%7D%5Cbig%5Bcos%28xy%29%5Cbig%5D%3D%5Cdfrac%7Bd%7D%7Bdx%7D%283x%2B1%29%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B-%5C%2Csin%28xy%29%5Ccdot%20%5Cdfrac%7Bd%7D%7Bdx%7D%28xy%29%3D%5Cdfrac%7Bd%7D%7Bdx%7D%283x%29%2B%5Cdfrac%7Bd%7D%7Bdx%7D%281%29%7D)
Apply the product rule to differentiate that term at the left-hand side:
![\mathsf{-\,sin(xy)\cdot \left[\dfrac{d}{dx}(x)\cdot y+x\cdot \dfrac{dy}{dx}\right]=3+0}\\\\\\ \mathsf{-\,sin(xy)\cdot \left[1\cdot y+x\cdot \dfrac{dy}{dx}\right]=3}\\\\\\ \mathsf{-\,sin(xy)\cdot \left[y+x\cdot \dfrac{dy}{dx}\right]=3}](https://tex.z-dn.net/?f=%5Cmathsf%7B-%5C%2Csin%28xy%29%5Ccdot%20%5Cleft%5B%5Cdfrac%7Bd%7D%7Bdx%7D%28x%29%5Ccdot%20y%2Bx%5Ccdot%20%5Cdfrac%7Bdy%7D%7Bdx%7D%5Cright%5D%3D3%2B0%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B-%5C%2Csin%28xy%29%5Ccdot%20%5Cleft%5B1%5Ccdot%20y%2Bx%5Ccdot%20%5Cdfrac%7Bdy%7D%7Bdx%7D%5Cright%5D%3D3%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B-%5C%2Csin%28xy%29%5Ccdot%20%5Cleft%5By%2Bx%5Ccdot%20%5Cdfrac%7Bdy%7D%7Bdx%7D%5Cright%5D%3D3%7D)
Now, multiply out the terms to get rid of the brackets at the left-hand
dy
side, and then isolate —— :
dx
and there it is.
I hope this helps. =)
Tags: <span><em>implicit function derivative implicit differentiation chain product rule differential integral calculus</em>
</span>
_______________
dy
Find —— for an implicit function:
dx
cos(xy) = 3x + 1.
First, differentiate implicitly both sides with respect to x. Keep in mind that y is not just a variable, but it is also a function of x, so you have to use the chain rule there:
Apply the product rule to differentiate that term at the left-hand side:
Now, multiply out the terms to get rid of the brackets at the left-hand
dy
side, and then isolate —— :
dx
and there it is.
I hope this helps. =)
Tags: <span><em>implicit function derivative implicit differentiation chain product rule differential integral calculus</em>
</span>
<h2>a. What is your equation?</h2>
This is a problem of projectile motion. A projectile is an object you throw with an initial velocity and whose trajectory is determined by the effect of gravitational acceleration. The general equation in this case is described as:
Where:
So:
Finally, the equation is:
The rocket will reach the maximum height at the vertex of the parabola described by the equation . Therefore, our goal is to find
at this point. In math, a parabola is described by the quadratic function:
So the x-coordinate of the vertex can be calculated as:
From our equation:
So:
So the rocket will take its maximum value after 1.99 seconds.
<h2>c. What is the maximum height the rocket will reach?</h2>
From the previous solution, we know that after 1.99 seconds, the rocket will reach its maximum, so it is obvious that the maximum height is given by . Thus, we can find this as follows:
So the maximum height the rocket will reach is 66.68ft
<h2>d. How long is the rocket in the air?</h2>
The rocket is in the air until it hits the ground. This can be found setting , so:
We can't have negative value of time, so the only correct option is and rounding to the nearest hundredth we have definitively: