Furry saves you from Related Rates - AP Calculus - Cone Problem + 4 more problems

Added: 2026-05-12

[00:00:00]Hello everyone, Mathcat here.

[00:00:04]Oh no, your friend is falling out of the sky again. Um, so the distance he's falling from can be drawn as h and the distance how far you we how far you are away from your friend can be marked as a distance x.

[00:00:25]So, um, he's falling at a rate of 27.3 m/s squared, and you're trying to run at him to catch him at a rate of 4 m/s squared.

[00:00:40]So, do you notice the language I've kind of used? Rate. A rate is essentially just a derivative with respect to time.

[00:00:51]So if something is changing like the height this he's falling at a rate of 27.3 m/s squared it's just how fast the height is changing in respect to time and over here it's how fast the distance or the horizontal distance is changing in respect to time. So this is the this is the vertical distance. This is the horizontal distance. So we can denote that as the change in vertical height how fast that changes in respect to time. So if h is the height, the vertical component between you and your friend, this would be the rate of which the vertical distance is changing in respect to time.

[00:01:51]And same thing here, can you um create a rate similar to this for horizontal distance? I'll give you a moment.

[00:02:03]Well, I hopefully you tried that. Um, it would just be dx dt, the change in horizontal distance x in respect to time.

[00:02:15]So, another example would be if water is leaking out of the cone, the volume is changing, right? So you would write the rate as dv dt where v is the volume and in this form it says that the volume is changing how fast the volume is changing in respect to time.

[00:02:44]So a rate is a derivative with respect to time or yeah with respect to time.

[00:02:54]Now, now let's move on to very useful tools that you will use when you're trying to solve related rate problems.

[00:03:04]So, um hopefully you're familiar with the chain rule. Um but if not, let's do a quick review of the chain rule and implicit differentiation.

[00:03:17]So, let's start with implicit differentiation. Um let's say you just take the derivative of x to the 3r using the power rule. We know that this is 3 x², right? There's no extra term and that's what they want you to think.

[00:03:37]There is an implicit term. You multiply this by dx over in respect to x and this actually just cancels out to one.

[00:03:51]So that's why it's implicit. We don't write that. It's unnecessary.

[00:03:59]Now um let's say you take a derivative of the same thing but it's in respect to a different variable.

[00:04:09]dy x cubed.

[00:04:15]Well, that would be just the power rule.

[00:04:18]The same thing here.

[00:04:21]3x^ 2 * dx dy. Now, this this cannot cancel to one.

[00:04:33]Why does this appear? Because of the chain rule. If you apply let's say um in respect to y of actually no not respect to y let's do t of f of x.

[00:04:54]This would be the derivative the outside multiplied by the derivative of the inside.

[00:05:06]So you can kind of see that this form is taking shape in these equations and the same thing applies for the product rule and the quotient rule.

[00:05:24]So just as a quick reminder um let's say you have ddt of a product of u and v.

[00:05:40]This would be um do you remember the product rule?

[00:05:47]That would be f prime * g plus g prime * f. So let's um define f. f can be u f prime that would be du in respect to time.

[00:06:10]uh and then uh g can be v and g prime would be dv in respect to time because we're differentiating using respect to time like if this was a y then this would be a y this would be a y and then we would just write it out identify dudt E uh * G plus G prime times U and this would be your answer.

[00:07:00]Now let's move on to um some problems involving related rates.

[00:07:08]So before we even um begin with solving the problem, we need to list out everything we know. And also drawing a picture may help. So let's say we have a ladder that leans against the wall.

[00:07:34]Okay.

[00:07:35]Well, this ladder is 10 feet tall. So, let's immediately um write that out. We can label that as a height h. And I got that before I even continued reading the whole problem. So, a 10 ft ladder leans on a vertical wall.

[00:07:57]The base slides away uh let's write 10 ft. The base slides away at a rate of 2 feet per second.

[00:08:12]So, how fast is the top sliding down when the base is 6 ft away from the wall? So, if the base is 6 ft away from the wall, let's label that X. We know that x is 6 ft and the base slides away at 2 ft per second.

[00:08:39]That's very important. Um, can you try and find a rate in like the derivative notation to respect to time?

[00:08:48]I'll give you a moment.

[00:08:52]Hopefully, you tried that out. That would be dx dt is equal to 2 feet per second. So the x is changing in respect to time at 2 ft per second.

[00:09:10]How fast is the top sliding down when the base is 6 ft away from the wall?

[00:09:17]Well, for related rates, we want to try and find an equation that fully encaptures our situation.

[00:09:27]So, um, if you can see here, try and find a shape that we can use and an equation for that shape.

[00:09:40]Hopefully, you tried that. Um, I can see here that it kind of creates a triangle situation.

[00:09:48]Um, and if we just draw out that triangle, that would be right next to the drawing.

[00:10:10]This would be 6 feet and this would be 10 ft.

[00:10:18]And then there's this unknown here.

[00:10:21]Let's label the y.

[00:10:26]So we can use the Pythagorean theorem to create kind of an equation that will help us help us find the rate of change of how fast is the top sliding down. So you want to know when this like how fast it slides down. That would be we're trying to find the rate of the y like the rate of the vertical height and how it changes in respect to time. Right?

[00:11:07]So um we would want to yeah again the Pythagorean theorem we know that it's a^2 + b^2 is equal to c^2. Let's change it a bit due to our situation x^2 + y^2 and that is equal to the hypotenuse which would be this squared. So that would be 100.

[00:11:39]Now I want to emphasize that before we plug in any numbers, actually I think it's good practice if we just write it as h squ. Before we plug in any numbers, we need to make sure we finish all the algebraic steps like solving for the derivative in respect to time. So do not plug anything in until the end.

[00:12:06]It often makes all the steps a lot easier. And if you want to plug things in before the end, do not plug anything in before differentiation.

[00:12:17]So um we have this system here.

[00:12:22]Let's um find the derivative in respect to time.

[00:12:28]So um let's apply the derivative in respect to time across the whole system plus y^2 and that is equal to the derivative in respect to time of h squ.

[00:12:53]So um let's apply the derivative respect to t to all these um terms. So that would be 2x * dx dt + 2y * dy dt over d or sorry 2 H * DH DT.

[00:13:37]And I want you to notice something. What can we do about this side of the equation? dHdt that is just the that's saying that the rate of the ladder length changes in respect to time. But if we look at our situation, does the hypotenuse of the ladder change? Does the rate of or does this length ever change? Is there a rate that can be applied to that?

[00:14:10]Um this the length of the ladder uh I don't think it's possible for it to change for our situation. So the rate of which the height of the ladder is changing at is zero. So this can be immediately crossed out to zero.

[00:14:32]Now we have this side of the equation 2x * dxdt and so on. Um, what we're trying to find is how fast is the top sliding down when the base is 6 ft away from the wall.

[00:14:52]That would be um we're trying to find dy dt how fast the um top part is changing in respect to time. So how fast y changes in respect to time.

[00:15:10]So we can rewrite this equation 2x * dx dt.

[00:15:20]This is equal to 2y * dy dt.

[00:15:34]And one thing important to note is that um we need to include direction in our rates. So going up here, let's define that the positive x direction is to the right and the positive y direction is up. So we know that dydt must be negative at the end.

[00:16:00]Um, so now we can distribute the negative and um, let's try and get dydt by itself. And notice because we have all these other variables, it's just as simple as solving for this variable. So then we get 2x dx dt over say 2 y. Okay, I lied earlier but um we don't have y. So let's find that using the Pythagorean theorem.

[00:16:47]So um we have 6^2 + y^2 that is equal to 10^2 um 10^2 - 6^2 is equal to y^2.

[00:17:01]So to find y we need to take the root.

[00:17:04]If you're wondering what I'm doing here, this is just the Pythagorean theorem. So 100us 36, what is that?

[00:17:14]Uh 64. And that means the square t of 64 is equal to y. y is equal to 8. Now we have everything.

[00:17:26]So uh what is our x again? Our x is 6 ft and the change in x in respect to time is 2 ft per second. So rewriting this, plugging in everything after we're finished, this would be 2 * 6 * dxdt, which was 2, I believe.

[00:17:54]And this is over -2 y, and y was 8.

[00:18:06]And this is equal to the change in height in respect to time.

[00:18:12]Calculating this uh 2 * 2 * 6 that would be 24 over -16.

[00:18:27]And uh this roughly simplifies to -1.5 ft/s.

[00:18:39]And that would be the change in height in respect to time, which is what the question originally asks us, right?

[00:18:52]How fast is the top sliding down when the base is 6 feet away from the wall?

[00:18:59]So at 6 feet, that is when x is six and it's sliding at a rate of 2 ft per second and the height is 8 at that specific time.

[00:19:13]This is our final answer and we know it's correct because it's negative and it's sliding down and we defined negative as going down. Um we can make sure that our answer is correct by performing dimensional analysis.

[00:19:31]Um so it's just plugging in the units here to make sure they cancel out correctly. So 2x that would be a unit of feet.

[00:19:45]times feet per second um feet per second with that feet per second squared. No um over 2y which was a feet as well. These cancel and this returns our answer of dy dts per second and we know that it's correct.

[00:20:19]Now let's try another problem.

[00:20:26]A tank with the vertex down has height 10. So let's first draw the tank.

[00:20:38]put the vertex down and it has a height. Let's denote the height as h and a radius okay a radius of 5t.

[00:20:56]Water drains at 3 feet cubed per minute.

[00:21:01]How fast is the water level falling when h is equal to 6 feet?

[00:21:08]So we have our smaller height. So let's make this big height.

[00:21:16]And let's make this actually let's name it let's name it K big K.

[00:21:26]And this will be this can be little K.

[00:21:33]It's at 6 feet.

[00:21:39]So that's the height of the water and it's leaking out at a rate of 3 ft cubed per minute. So this would be the change in volume in respect to time. So writing out all of our variables, our radius would be 5 ft.

[00:22:04]Our height of the tank, this is big k would be 10 ft.

[00:22:13]And little k would be um 6 ft.

[00:22:22]What if I use another unit measurement to make it easier? Length or L.

[00:22:32]And we have the change in volume.

[00:22:37]Let's make it white.

[00:22:43]The change in volume in respect to time is 3 ft cubed per minute.

[00:22:51]Now see if you can solve this problem by yourself. I'll give you the formula for a cone and see if you can solve it by yourself.

[00:23:03]Here's the equation for the volume of a cone where this is the radius and this is the height of the cone.

[00:23:18]Given this we can actually try out with solving. So um we need to find what the rate is the rate that the um the rate of change of the height of the water when the height is six.

[00:23:41]So knowing that we can't immediately just start by using these variables because um if we started and plugged that in, it would assume that our water height is starting from up here and the rate at which the water is falling would be different or it wouldn't work. And then so what we need to do is we need to find the height and radius of the water when the height is 6 ft.

[00:24:15]We need to utilize something called similar triangles.

[00:24:20]So if we create a triangle up here, we can see another triangle where the water height is at.

[00:24:32]Let me draw it separately the diagram to make it clear.

[00:24:40]This is where the water height is at.

[00:24:44]Notice how these triangles are similar if I draw them separate.

[00:24:51]This triangle is essentially a um scaled up version. maybe by a factor of two, three, I'm not sure, but it's just scaled by a factor of this triangle.

[00:25:05]If we know the height and radius of one of the triangles, we can find the right the radius and height of another triangle because it's based off of a ratio.

[00:25:21]Like let's say um this triangle has a height of two and a radius of four. And let's say this bigger triangle just doubled in size. So that would be eight and four.

[00:25:40]If we take the ratio radius to height, height to radius because they depend on each other. We can take the ratio 8 over 4 which would be the radius over height is equal to the ratio of the radius over height. Here this is equivalent and also other way around works. Let's say you want to take the ratio of the height to radius instead. That would be 2 4 height to radius 4 8 and these are equivalent.

[00:26:25]So that's just a recap of similar triangles and we need to use it to solve for the radius at this specific height. We know that the um ratio of let's just do radius to height. The ratio of that is equal to 5 over 10.

[00:26:57]And this is equal to the ratio from the radius to the height is a half.

[00:27:06]So that means that the radius is approximately half of the height. So we can rewrite that as half of the height is equal to the radius.

[00:27:24]And then uh we can plug that in to this equation to get the formula in terms of just height. And we get rid of a variable which helps us out a lot. So we can rewrite this as V is 1/3 *<unk> and the radius has changed. It's half of the height. So that would be h / 2^ squar * height.

[00:27:56]So for this specific situation, this formula remains or this formula is consistent.

[00:28:05]Now if we simplify this this would be v = 1/3 power over h ^2 / 4 * h and this is equal to 1/3 pi r let's do 112th * h cubed and then Now this is the formula you want to find the derivative of in respect to time.

[00:28:45]Now applying the time derivative d v dt is equal to one.

[00:28:55]Uh let's take the constants out first and then now the derivative in respect to time of the height cubed.

[00:29:09]We can take out these constants because um of the power rule. We can do that.

[00:29:15]And then taking the derivative of this would equal 112 pi times 3 h^ 2 * the change in height in respect to time.

[00:29:42]Now if we look at all of our variables we have the height the radius um let's define that little radius this can be p little radius would be 3 ft we have everything this is our formula that we want to use so let's isolate the change in height in respect to Okay, that would be 12 over pi * dv dt and that is equal to I forgot to do something times 1 over 3 h. There we go. Now that is equal to dh dt and this is squared. So um canceling this out cancelelling this 3 and 12 we get 4 over pi times the change in volume in respect to time and that is equal to the change in height in respect to time and I forgot to include * 1 / h^2 So at that specific height at this height it is a height of L and L is 6 ft.

[00:31:25]L is 6 ft. So we can plug that in 4 pi * um 1 / 6 squared times the change in volume in respect to time which would be 3 cubed per minute and that is equal to dhdt t.

[00:31:59]Simplifying this down, it's just a matter if you can get the algebra correct. 12 over 36 * 1 /<unk> which is equal to uh/3 pi. Oh, uh something I forgot to Something I forgot to include earlier.

[00:32:30]Um it would be changing the volume would be negative because it's losing volume.

[00:32:36]That's important to note. Um so then our change in volume would be negative uh negative so then our answer would be 1/3 over pi is equal to the change in height in respect to time and this would be our final answer but make sure to include units feet cubed oh no just feet per Second, let's rearrange this. Um, our final answer would be 3 1 over 3 pi feet per second. And this would be the change in height intersected time.

[00:33:30]And that is the cone tank problem.

[00:33:41]Okay, now we have a trough that's leaking water. Same principles as the cone problem, but just different shape.

[00:33:50]So, drawing this out.

[00:33:53]Oh, is she in a different color?

[00:33:57]A trough.

[00:34:08]This is a trough. I hope you can trust me on that. And it is 4 ft wide at the top.

[00:34:19]So we can denote that as a width W and it is 3 ft deep. So that means it has a height of h.

[00:34:38]Water fills in at a rate of 2 ft cubed per minute. So we immediately know that the change in volume is positive.

[00:34:50]3 feet cubed per minute.

[00:34:55]Okay, I made an error. This is supposed to be two. I read the problem wrong. 2 feet cubed per minute. And we know that the water level is at one.

[00:35:11]Let's do blue water.

[00:35:16]So, the height of the little water, let's do um P the height of the water is 1 ft.

[00:35:33]Now, um defining all of our variables, the height is 3 ft and the uh the width is 4T.

[00:35:47]Can you find an equation that helps capture the shape of this water trough or this situation?

[00:35:57]It would be um what is the volume for a triangular prism?

[00:36:05]It's just V is equal to 12 time the width time the height time the length. Oh, I forgot to say that this trough has a length of L which is 10 ft.

[00:36:32]Okay. Now we have our equation and we're trying to find the change in height.

[00:36:44]So is there a relationship where we can get rid of W where we can get it in terms of the height? Well um we can use similar triangles again.

[00:36:58]So if we kind of zoom make a bigger picture this would be the trough at this specific view.

[00:37:13]There's a triangle that we can make and it's filled with water.

[00:37:22]We know the width is W.

[00:37:26]But for this triangle, for this section, it would be the width over two and then the height of h.

[00:37:38]So the width, let's try to find the height in terms of the width. We know that the width half of the width would be two over three.

[00:37:59]The height is equal to um and because we know the ratio now we can do the width over two over h. This is our relationship to get it in terms of the height. flip the variable three halves is equal to h over two and just multiply this to this side and we get for the width oh sorry I was blocking it um we'll get two/3 time oh sorry 3 times width / 2 is equal to h. So this would be 34s of the width is equal to height.

[00:39:02]Now that we have this relationship because we set up this ratio um we just I just plugged it in this in and then set it equal and we can solve for a variable. Um now we have the width in terms of the height.

[00:39:23]This would be 3/4 of the width is equal to the height.

[00:39:28]That would be V is equal to 12.

[00:39:38]Um, oh, uh, let's isolate. Let's move this to the other side. So, the width would be 4/3.

[00:39:49]We want to get rid of this width to get it in respect to the height. How fast is the water level rising? That relates to the height.

[00:40:02]H 4/3 * 4/3 times the height time the length.

[00:40:12]Now we have our equation in terms of height because L doesn't change. It's not a variable anymore.

[00:40:23]um that could just turn into 10.

[00:40:33]Simplifying this to get the volume in terms of height that would be um 46 or 40 6 * h 2.

[00:40:50]Now taking both sides in respect to time the change in volume in respect to time is equal to 46. This is a constant. We don't care about that.

[00:41:06]In respect to time of a squ the derivative respect to time which is equal to times uh let's keep this purple actually.

[00:41:20]So you can keep track maybe 2 h times the change in height in respect to time. Now this is what we want. Okay, we can just kind of like delete this and that would be our equation.

[00:41:43]Now we just need to do some algebra.

[00:41:48]640 * the change in volume in respect to time all divided by the height at that specific time is equal to the change in height in respect to time.

[00:42:06]And that can be simplified to 12 40 / 2 / 1 which is equal to 12 over 80 640 320 ft per minute and this would be the change in height and respect to time.

[00:42:32]Nice. That is the trough problem. Okay, you have a 20 ft street light.

[00:42:40]20 ft street light. Very bright.

[00:42:46]And it casts a shadow of a fiveT person walking away at 4 feet per second.

[00:43:00]Um, he is 5T.

[00:43:03]Let's label that.

[00:43:05]a height of D actually no height of Y and this can be a height of H and it's casting a shadow so I would assume the shadow be somewhere on the ground let's make the shadow green this line just representing the shadow um the length of the shadow can be How fast is the shadow length increasing? So we need to find the length of the shadow which is the rate of change of the length of shadow in respect to time.

[00:43:50]And let's focus on finding this first.

[00:43:53]So we can label the distance between the two um as x And notice, do you see a triangle and a similar triangle case? This triangle is just a smaller version of this triangle.

[00:44:14]Nice. We can um create the relationship that let's say oh we know the height is 20 feet and the height of the person is equal to 5 ft. So we can create a relationship a ratio through similar triangles.

[00:44:39]That would be let's do the height in respect to the length.

[00:44:47]So the height would be 20 feet. Let's start with a big triangle. 20 feet.

[00:44:56]And then the length x + s.

[00:45:02]Oops.

[00:45:08]And this is equal to um the ratio of the height to length of the smaller triangle.

[00:45:16]And we know he's 5 ft. So five over um s which is we don't know.

[00:45:28]So now we have this um formula here to get the denominator to the numerator numerator we can take the reciprocal which is s plus x over 20 is equal to s over five.

[00:46:00]All right.

[00:46:03]Now, um how fast is the shadows length increasing? That means we want we only care about S. So, let's get S in terms of X.

[00:46:16]expanding this that would be um s over 20 plus x over 20.

[00:46:38]This is equal to s over 5.

[00:46:44]And we can just move this to the other side.

[00:46:50]So we get x over 20 is equal to the length of the shadow over the height of the person minus the length of the shadow over 20 or the height of the lampost.

[00:47:13]And simplifying this out, we should get x over 20 of su 15 s over 100. And now um to isolate s we can do 5 * x / 15 is equal to s.

[00:47:54]So that is roughly/3 of x is equal to the length of the shadow. Now we can finally take the time derivative.

[00:48:08]The time derivative.

[00:48:10]So the change in x in respect to time * 1/3 is equal to the derivative of the length of the shadow is respect to time. I should make this red.

[00:48:31]Now um originally we were trying to solve for that and we have it in a combination of things we know dxdt that would be what it's moving the person is moving away at a rate of 4 feet per second right so that would mean that he is changing x in respect to time at a rate of 4 ft per So that' be 4 ft per second * 1/3 is equal to 4/3t per second and this would be equal to um the rate at which the length of the shadow is increasing.

[00:49:26]That answers the first part of the problem.

[00:49:29]But what is the second part of the problem? How fast is the tip of the shadow moving?

[00:49:35]Well, um the tip of the shadow can be modeled after x plus s cuz we want this point here, right? So x + s.

[00:49:51]And we just need to take the time derivative of that which is equal to dx dt plus d sdt.

[00:50:05]You just add up the rates together. 4/3t per second plus 4t/s.

[00:50:15]Um that would be equal to 163 ft per second that would be let me box both of my answers.

[00:50:33]So that is the rate at which this entire length is increasing. Not just the length of the shadow or the distance the sky is making. It's the length of the entire or the tip of the shadow.

[00:50:51]That is the street lamp problem.

[00:50:58]Now, at noon, a car is 60 miles north.

[00:51:05]So, let's um create our point.

[00:51:11]It's a box. He is 60 miles up north. The car can be green car A.

[00:51:21]It is 60 m north and traveling south at 30 miles hour.

[00:51:28]Um, let's label our axes.

[00:51:34]This is going to be blue. Y, and it's traveling east. Car B is at the intersection and it's traveling east 40 miles per hour.

[00:51:51]How fast is the distance changing at 100 p.m.? So, um this is our snapshot at uh 12 at noon 12:00 p.m. But after 1 hour, they are at different spots, right?

[00:52:17]So now maybe car P is here and car A is now at a distance uh let's say distance Y 2 this is Y1 this is at 1 PM so Y1 is equal to 60 mi north in the x direction y direction and car b is at zero but for now yeah um x1 is equal z and this can be x2 and at 100 p.m.

[00:53:09]It travels 30 mph because it takes an hour.

[00:53:14]It travels 30 mph. So now y 2 is equal to 30 mi away from the origin.

[00:53:25]Right?

[00:53:27]And x2 is now because it's traveling east at 40 miles hour. After 1 hour it travels 40 miles to the right.

[00:53:38]So that is 40 miles.

[00:53:45]And now um oh the change in height with respect to time. We know that based off of traveling south 30 miles hour. So dy in respect to time. Let's make that blue.

[00:54:09]that is equal to 30 miles hours but it's negative because uh let's define right positive up positive and down negative left negative and d x dt is equal to 40 mph in the positive direction.

[00:54:43]Is there a shape that we can use um to help us map out the situation?

[00:54:51]Yes, we can use a triangle.

[00:54:55]So the distance would be the hypotenuse between the two cars at 1 p.m. So y 2 is 30. We use the second situation because we only care about how fast is the distance changing at 1 p.m. So these values we can actually just get rid of them. They don't factor into our final uh calculation.

[00:55:33]And then x2 would be 40 miles away. And the hypotenuse would be the distance.

[00:55:42]Let's name that distance D.

[00:55:45]We can use the Pythagorean theorem. Now d^ squ is equal to actually just d is equal to the square t of 30^2 + 40^2 and that is equal to uh 250 I believe. No I like 50.

[00:56:12]So d is 50.

[00:56:17]And now let's write out our equation.

[00:56:22]We have x2 + y^2 is equal to d^2.

[00:56:30]And what we can do is now we can take the time derivative because we know x and y and we know the change with respect to anything and y and x do not depend on each other. So we can just simply take the time derivative of both sides.

[00:56:52]ddt time dt.

[00:56:56]So that would be d sorry 2x * dx dt plus 2y * dy dt and it is equal to 2d dd dt t and um we want to find again how fast is the distance changing and we have it here.

[00:57:42]This is how fast the distance distance is changing and we know all of these. So at 100 p.m.

[00:57:54]x distance is 40.

[00:58:00]40 * 2 is 80 * the change which is 40 mph in the positive direction plus 2 y uh y is y is 30 30 * 2 is 60 times the change in y in respect to time because the car is going down that is the negative change in y. So -30 and that is equal to uh let's isolate the change in distance.

[00:58:42]So we would divide this all by 2d and d the distance we found earlier at 100 p.m. is 50.

[00:58:53]So that would be 2 * 50 is 100.

[00:58:58]This would result in the change in distance in respect to time.

[00:59:03]Calculating this we get um 3,200 plus uh negative what 14 oh no negative 1800 divided by 100 and that is equal to And the change in distance in respect to time is miles per hour.

[00:59:41]This is your final answer.

[00:59:45]Made it to the end. If you're watching this before the AP calc test, good luck.

[00:59:51]Hello everyone. Thank you guys so much for watching. I really appreciate everyone who watches my videos through and through. And hopefully those who watch it fully can give me some feedback to better improve my teaching cuz I am not I wouldn't say I'm a teacher. Um but yeah, got all this wonderful fan art and I I will be looking into posting more calculus later. I wanted to post this before the calculus tests on Monday. AP calc AP calc test. Yeah. Um, but I will be looking to delve more into calculus as time goes on and um, I've been really busy with like scholarship applications and filling out a bunch of stuff. It's taking up a lot of my time. But hopefully this video helped. And I just know that I will eventually post um math videos covering every like section like like the organic chemistry tutor. It's a big inspiration to me. Um, and yeah, thank you so much for watching.