A right tringle has a base of 10cm10\operatorname{\mathrm{cm}}10cm and a height of 12cm12\operatorname{\mathrm{cm}}12cm. The height of the right triangle is decreasing at a rate of 0.4cms0.4\operatorname{\frac{\mathrm{cm}}{s}}0.4scm, at what rate is the area of the triangle decreasing?

− 2 c m 2 s ⁡ -2\operatorname{\frac{cm^2}{s}}^{} − 2 s c m 2

A right tringle has a base of 10 c m 10\operatorname{\mathrm{cm}} 10 cm and a height of 12 c m 12\operatorname{\mathrm{cm}} 12 cm . The height of the right triangle is decreasing at a rate of 0.4 c m s 0.4\operatorname{\frac{\mathrm{cm}}{s}} 0.4 s cm , at what rate is the area of the triangle decreasing?

A right triangle has a base of 10 centimeters and a height of 12 centimeters. The height of the triangle is decreasing at this rate right here, and we're asked at what rate is the area of the triangle decreasing. Okay. So let's get into this. Now notice in this problem that we have a right triangle. So what I'm going to do is I'm going to draw this diagram right here. Anytime you have a situation where you're dealing with these kinds of changing shapes, it's good to start with a diagram. So we have a right triangle, and we're told that the base of this triangle is 10 centimeters and the height of this triangle is 12 centimeters. So this is the situation. Now what we're also told about this triangle is the rate at which one of the sides is decreasing, and specifically we're told the rate at which the height is decreasing. And so we can see that that's 0.4 centimeters per second. Now there's something we need to recall. When dealing with situations where you have a decreasing rate, you want to make that rate negative. So this 0.4 centimeters per second is actually going to be a negative 0.4 centimeters per second. So going over to this diagram right here, this height right here is going to be decreasing at a rate of negative 0.4 centimeters per second. This is what we end up getting. Now what we're asked to do in this problem is find the rate at which the area of the triangle is decreasing. And if we do our calculations right, we should get a negative result for our area or or a rate for our area because that's gonna be decreasing as well. So if I think about it, the best equation that I can think of for this entire situation is going to be the area of a triangle, which is 1 half base times height. I think that makes the most sense since we have a base, we have a height, we have the rate at which our height is decreasing. I think the time derivative is gonna give us all of that. When we do these problems, whenever you guess an equation and you go through the derivative process, the variables you get are going to tell you whether or not you got the right equation, but I'm pretty sure it's gonna line up with this. So before I take this derivative though, let's actually see what variables we can glean from the triangle here. Well, I can I can see we have a base of 10 centimeters? I can see that we have a height of 12 centimeters. I can see we have a rate for the height, which I can see is d h over d t, the rate at which the height changes as negative 0.4 centimeters per second. But what we're looking for is the rate at which the area changes. So that's gonna be what we're ultimately trying to solve for in this problem. So now let's take our time derivative to see what we get. If I take the time derivative of this entire equation right here, we'll first have the time derivative of the area. That's gonna give us d a over d t. Now next, I need to take the derivative of this right side, and this is the tricky part. But what you wanna do is notice something here. There's a product happening, and we know the product rule for derivatives. The product rule says if we take the derivative, and I'm gonna do it with respect to t, of 2 functions being multiplied, that's going to be f primeg plus fgprime. So it's just derivative of first time second plus first times derivative of second. So let's go ahead and apply that up here. So what I can do is factor this one half out because we're just going to focus on the base and the height here. So we won't have the derivative of the base, which is gonna be d b over d t times the height plus the base b times the derivative of the height d h over d t. Now we can go ahead and just leave it like this. We don't have to do anything more here since just b and h were by themselves. We factor this one half out. We just get these two rates with respect to time. So this is what our equation would look like. So now what we need to do is plug in what we have. Well, there's a problem here. We don't have the changing base with respect to time. But notice something, that's not a rate in this problem. The base is set at 10 centimeters. It's only the height that's decreasing. So since there is no rate for which the base changes, the derivative of the base, it's constant. So that whole thing's just going to be 0. So we're going to have 0 multiplied by the height of 12, but that doesn't matter. The whole thing's gonna go to 0 anyways. So this whole term is just gonna disappear. So all we're going to have is this plus the base right here, which I can see the base is 10 centimeters times the changing height with respect to time, which I can see right here is negative 0.4 centimeters. And this whole thing is going to allow us to solve for the changing area d a over d t, which is exactly what we're looking for in this problem. Right? Since this was the rate that we're looking for, this was already on one side of the equation. So we just had to plug in these variables like we see right here. Now doing this, we'll have 1 half and then this will turn one away. So it's just gonna be 1 half of 10 times negative 0.4. So we have this being multiplied out right here. And what you can go ahead and do is put this onto a calculator. Or you can if you know this by hand, you should know that 10 times 0.4 is just going to give you 4, then divide that by 2, you're going to get negative 2. So the DA over DT, the changing rate with respect to of the area with respect to time is going to be negative 2. And then the area is in square centimeters, and that's going to be divided by the time of seconds. We need to deal with centimeters and seconds, and we know an area is square centimeters. So this right here would be our changing rate over time. Hope you found this video helpful.

A right tringle has a base of 10 c m ⁡ 10\operatorname{\mathrm{cm}} 10 cm and a height of 12 c m ⁡ 12\operatorname{\mathrm{cm}} 12 cm . The height of the right triangle is decreasing at a rate of 0.4 c m s ⁡ 0.4\operatorname{\frac{\mathrm{cm}}{s}} 0.4 s cm , at what rate is the area of the triangle decreasing?

− 2 c m 2 s ⁡ -2\operatorname{\frac{cm^2}{s}}^{} − 2 s c m 2

− 2 . 4 c m 2 s ⁡ -2\operatorname{.4\frac{cm^2}{s}}^{} − 2 .4 s c m 2

− 1 c m 2 s ⁡ -1\operatorname{\frac{cm^2}{s}}^{} − 1 s c m 2

− 4 c m 2 s ⁡ -4\operatorname{\frac{cm^2}{s}}^{} − 4 s c m 2

4. Applications of Derivatives

Related Rates

Calculus

4. Applications of Derivatives

Related Rates

Struggling with Calculus?

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Multiple Choice

A right tringle has a base of $10\operatorname{\mathrm{cm}}$ and a height of $12\operatorname{\mathrm{cm}}$ . The height of the right triangle is decreasing at a rate of $0.4\operatorname{\frac{\mathrm{cm}}{s}}$ , at what rate is the area of the triangle decreasing?

$-2\operatorname{.4\frac{cm^2}{s}}^{}$

$-1\operatorname{\frac{cm^2}{s}}^{}$

$-2\operatorname{\frac{cm^2}{s}}^{}$

$-4\operatorname{\frac{cm^2}{s}}^{}$

Verified step by step guidance

First, recall the formula for the area of a triangle: A = (1/2) * base * height. In this case, the base is constant at 10 cm, and the height is changing over time.

To find the rate at which the area is decreasing, we need to differentiate the area with respect to time. This involves using the chain rule since the height is a function of time.

Let h(t) be the height of the triangle at time t. The area A(t) can be expressed as A(t) = (1/2) * 10 * h(t). Differentiate A(t) with respect to t: dA/dt = (1/2) * 10 * dh/dt.

We are given that the height is decreasing at a rate of 0.4 cm/s, which means dh/dt = -0.4 cm/s. Substitute this value into the differentiated area formula: dA/dt = (1/2) * 10 * (-0.4).

Simplify the expression to find the rate at which the area is decreasing. This will give you the final answer in cm²/s.

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