Is anyone willing to do my ixl? I have loads of work to do and I need my ixl due tonight. It’s 8th grade math and I’d appreciate
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We are looking for the length of a curve, also known as the arc length. Before we get to the formula for arc length, it would help if we re-wrote the equation in y = form.
We are given:
We divide by 36 and take the root of both sides to obtain:
Note that the square root can be written as an exponent of 1/2 and so we can further simplify the above to obtain:
Let's leave that for the moment and look at the formula for arc length. The formula is
where ds is defined differently for equations in rectangular form (which is what we have), polar form or parametric form.
Rectangular form is an equation using x and y where one variable is defined in terms of the other. We have y in terms of x. For this, we define ds as follows:
As a note for a function x in terms of y simply switch each dx in the above to dy and vice versa.
As you can see from the formula we need to find dy/dx and square it. Let's do that now.
We can use the chain rule: bring down the 3/2, keep the parenthesis, raise it to the 3/2 - 1 and then take the derivative of what's inside (here
). More formally, we can let 
and then consider the derivative of 
. Either way, we obtain,
Looking at the formula for ds you see that dy/dx is squared so let's square the dy/dx we just found.
This means that in our case:
Recall, the formula for arc length:
Here, the limits of integration are given by 5 and 9 from the initial problem (the values of x over which we are computing the length of the curve). Putting it all together we have:
evaluated from 9 to 5 (I cannot seem to get the notation here but usually it is a straight line with the 9 up top and the 5 on the bottom -- just like the integral with the 9 and 5 but a straight line instead). This means we plug 9 into the expression and from that subtract what we get when we plug 5 into the expression.
That is,![[(\frac{1}{2}) ( \frac{9^3}{3}) -9]-([(\frac{1}{2}) ( \frac{5^3}{3}) -5]=( \frac{9^3}{6}-9)-( \frac{5^3}{6}-5})=\frac{290}{3}](https://tex.z-dn.net/?f=%5B%28%5Cfrac%7B1%7D%7B2%7D%29%20%28%20%5Cfrac%7B9%5E3%7D%7B3%7D%29%20-9%5D-%28%5B%28%5Cfrac%7B1%7D%7B2%7D%29%20%28%20%5Cfrac%7B5%5E3%7D%7B3%7D%29%20-5%5D%3D%28%20%5Cfrac%7B9%5E3%7D%7B6%7D-9%29-%28%20%5Cfrac%7B5%5E3%7D%7B6%7D-5%7D%29%3D%5Cfrac%7B290%7D%7B3%7D)
We are given:
We divide by 36 and take the root of both sides to obtain:
Note that the square root can be written as an exponent of 1/2 and so we can further simplify the above to obtain:
Let's leave that for the moment and look at the formula for arc length. The formula is
Rectangular form is an equation using x and y where one variable is defined in terms of the other. We have y in terms of x. For this, we define ds as follows:
As a note for a function x in terms of y simply switch each dx in the above to dy and vice versa.
As you can see from the formula we need to find dy/dx and square it. Let's do that now.
We can use the chain rule: bring down the 3/2, keep the parenthesis, raise it to the 3/2 - 1 and then take the derivative of what's inside (here
Looking at the formula for ds you see that dy/dx is squared so let's square the dy/dx we just found.
This means that in our case:
Recall, the formula for arc length:
Here, the limits of integration are given by 5 and 9 from the initial problem (the values of x over which we are computing the length of the curve). Putting it all together we have:
That is,