Solving for variable when the integral is known

[enott312]

I have this question posed to me for homework.

A parabolic prism is being filled with water. The top of the prism is 30cm wide and is 20cm deep. What is the depth of the water when the prism is half full.

I have found the area of the face to be 110 cm² , by finding the equation of the parabola and then the integral between -15 and +15.

Given that the volume of a prism is Ah, to be half full would be found with (A/2)h.

This is where I get stuck.

Help?

 

[Deleted member 4993]

I have this question posed to me for homework.

A parabolic prism is being filled with water. The top of the prism is 30cm wide and is 20cm deep. What is the depth of the water when the prism is half full.

I have found the area of the face to be 110 cm² , by finding the equation of the parabola and then the integral between -15 and +15.

Given that the volume of a prism is Ah, to be half full would be found with (A/2)h.

This is where I get stuck.

Help?

What is the equation of the parabola?

 

[Harry_the_cat]

I have this question posed to me for homework.

A parabolic prism is being filled with water. The top of the prism is 30cm wide and is 20cm deep. What is the depth of the water when the prism is half full.

I have found the area of the face to be 110 cm² , by finding the equation of the parabola and then the integral between -15 and +15.

Given that the volume of a prism is Ah, to be half full would be found with (A/2)h.

This is where I get stuck.

Help?

How did you get 110cm^2? That is incorrect. What equation did you get?

 

[enott312]

I have this question posed to me for homework.

A parabolic prism is being filled with water. The top of the prism is 30cm wide and is 20cm deep. What is the depth of the water when the prism is half full.

I have found the area of the face to be 110 cm² , by finding the equation of the parabola and then the integral between -15 and +15.

Given that the volume of a prism is Ah, to be half full would be found with (A/2)h.

This is where I get stuck.

Help?

Sorry, was looking at the wrong number.
The area of the face is 400cm² and the equation of the parabola is y=0.08889x² or more precisely y=(4/45)x²

 

[stapel]

The area of the face is 400cm² and the equation of the parabola is y=0.08889x² or more precisely y=(4/45)x²

I will guess that you set up your flat face as a parabola with its vertex at the origin, so the points (x, y) = (-15, 20) and (x, y) = (15, 20) are the upper "corner" points of the flat face. Plugging one of these points into "y = ax²", you found a = 4/45, so the modelling equation for the flat faces is y = (4/45)x².

I will guess that you found the area of a flat face by setting up an integral along the lines of the following:

. . . . .\(\displaystyle \displaystyle \begin{align} 2\, \int_0^{15}\, \left(\, 20\, -\, \dfrac{4}{45}\, x^2\, \right)\, dx\, &=\, 2\, \left(\, 20x\, -\, \dfrac{4}{135}\, x^3\, \right)\, \bigg|_0^{15}

\\ \\ &=\, 2\, \Big[\, (300\, -\, 100)\, -\, (0\, -\, 0)\, \Big]

\\ \\ &=\, 2\, (200)

\\ \\ &=\, 400 \end{align}\)

I will guess that you are using "A" to stand for "the area of a flat face, or the cross-sectional area" and "h" to stand for "the length, perpendicular to the flat faces".

You found the area A by integrating between the top ("20") and the bottom ("y = (4/45)x²"). In this case, you need to find the top.

Looking at the graph, you know that the top is some y-value; the y-value is given by some x-value, x₀, where integrating from x = 0 to x = x₀ (and then multiplying by 2, of course) will give you half the value of the total area A. So set up the integral, and work backwards:

. . . . .\(\displaystyle \displaystyle 2\, \int_0^{x_0}\, \left(\, \dfrac{4}{45}x_0^2\, -\, \dfrac{4}{45}x^2\, \right)\, dx\, =\, 200\)

Don't forget that x₀ is a constant!

 

[enott312]

I will guess that you set up your flat face as a parabola with its vertex at the origin, so the points (x, y) = (-15, 20) and (x, y) = (15, 20) are the upper "corner" points of the flat face. Plugging one of these points into "y = ax²", you found a = 4/45, so the modelling equation for the flat faces is y = (4/45)x².

I will guess that you found the area of a flat face by setting up an integral along the lines of the following:

. . . . .\(\displaystyle \displaystyle \begin{align} 2\, \int_0^{15}\, \left(\, 20\, -\, \dfrac{4}{45}\, x^2\, \right)\, dx\, &=\, 2\, \left(\, 20x\, -\, \dfrac{4}{135}\, x^3\, \right)\, \bigg|_0^{15}

\\ \\ &=\, 2\, \Big[\, (300\, -\, 100)\, -\, (0\, -\, 0)\, \Big]

\\ \\ &=\, 2\, (200)

\\ \\ &=\, 400 \end{align}\)

I will guess that you are using "A" to stand for "the area of a flat face, or the cross-sectional area" and "h" to stand for "the length, perpendicular to the flat faces".

You found the area A by integrating between the top ("20") and the bottom ("y = (4/45)x²"). In this case, you need to find the top.

Looking at the graph, you know that the top is some y-value; the y-value is given by some x-value, x₀, where integrating from x = 0 to x = x₀ (and then multiplying by 2, of course) will give you half the value of the total area A. So set up the integral, and work backwards:

. . . . .\(\displaystyle \displaystyle 2\, \int_0^{x_0}\, \left(\, \dfrac{4}{45}x_0^2\, -\, \dfrac{4}{45}x^2\, \right)\, dx\, =\, 200\)

Don't forget that x₀ is a constant!

Thanks.

I as eventually able to solve this using guess/check method, I knew that it would be slightly more than half way, so I guessed y=12.5cm, solved for x and then solved the integral and multiplied by 2. The final answer was 11.9cm. However this took 5 attempts and I had to do this in an exam it would take too long.

I don't know how to use your integral to find x as there are two variables x₀ and x. I have only learned to solve with one. Can you please show how to solve with yours?

I was told to find the inverse function to y=(4/45)x², which is y^-1=SQRT(45x/4), integrate this and then solve where the integral = 100. Then find the equivalent value. Again, this method takes too long.

 

[stapel]

...So set up the integral, and work backwards:

. . . . .\(\displaystyle \displaystyle 2\, \int_0^{x_0}\, \left(\, \dfrac{4}{45}x_0^2\, -\, \dfrac{4}{45}x^2\, \right)\, dx\, =\, 200\)

Don't forget that x₀ is a constant!

I don't know how to use your integral to find x as there are two variables x₀ and x.

There aren't "two variables"; there's a variable and a constant. You can rename the x0 as "a" or "n" or whatever, if you like, but then you still integrate in the usual way.

Can you please show how to solve with yours?

To integrate a constant, use the regular rule:

. . . . .\(\displaystyle \displaystyle \int_a^b\, 1\, dx\, =\, 1\, \cdot\, x\,\bigg|_a^b\, =\, x\, \bigg|_a^b\, =\, b\, -\, a\)

. . . . .\(\displaystyle \displaystyle \int_a^b\, a\, dx\, =\, a\, \cdot\, x\,\bigg|_a^b\, =\, ax\, \bigg|_a^b\, =\, ab\, -\, a^2\)

. . . . .\(\displaystyle \displaystyle \int_a^b\, b\, dx\, =\, b\, \cdot\, x\,\bigg|_a^b\, =\, bx\, \bigg|_a^b\, =\, b^2\, -\, ab\)

. . . . .\(\displaystyle \displaystyle \int_a^b\, c\, dx\, =\, c\, \cdot\, x\,\bigg|_a^b\, =\, cx\, \bigg|_a^b\, =\, bc\, -\, ac\)

. . . . .\(\displaystyle \displaystyle \int_a^b\, x_0\, dx\, =\, x_0\, \cdot\, x\,\bigg|_a^b\, =\, x_0\, x\, \bigg|_a^b\, =\, b\,x_0\, -\, a\,x_0\)

I was told to find the inverse function to y=(4/45)x², which is y^-1=SQRT(45x/4), integrate this and then solve where the integral = 100. Then find the equivalent value. Again, this method takes too long.

But if this is the method that your instructor requires that you use (though I'm not sure why), then you should probably use it, even if it's "hard" or messy.

 

[enott312]

There aren't "two variables"; there's a variable and a constant. You can rename the x0 as "a" or "n" or whatever, if you like, but then you still integrate in the usual way.


To integrate a constant, use the regular rule:

. . . . .\(\displaystyle \displaystyle \int_a^b\, 1\, dx\, =\, 1\, \cdot\, x\,\bigg|_a^b\, =\, x\, \bigg|_a^b\, =\, b\, -\, a\)

. . . . .\(\displaystyle \displaystyle \int_a^b\, a\, dx\, =\, a\, \cdot\, x\,\bigg|_a^b\, =\, ax\, \bigg|_a^b\, =\, ab\, -\, a^2\)

. . . . .\(\displaystyle \displaystyle \int_a^b\, b\, dx\, =\, b\, \cdot\, x\,\bigg|_a^b\, =\, bx\, \bigg|_a^b\, =\, b^2\, -\, ab\)

. . . . .\(\displaystyle \displaystyle \int_a^b\, c\, dx\, =\, c\, \cdot\, x\,\bigg|_a^b\, =\, cx\, \bigg|_a^b\, =\, bc\, -\, ac\)

. . . . .\(\displaystyle \displaystyle \int_a^b\, x_0\, dx\, =\, x_0\, \cdot\, x\,\bigg|_a^b\, =\, x_0\, x\, \bigg|_a^b\, =\, b\,x_0\, -\, a\,x_0\)


But if this is the method that your instructor requires that you use (though I'm not sure why), then you should probably use it, even if it's "hard" or messy.

Thanks for you help. I feel really stupid because when I used my Ti Nspire calculator it solved the problem for me. Once I have the answer, I worked backwards and it makes total sense.