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In this video, we’re going to learn how we can use derivatives to find the relation between rates of two or more quantities in related rates problems.
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In related rates problems, the idea is to compute the rate of change of one quantity in terms of the rate of change of another quantity, which is sometimes more easily measured.
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We’ll look at how to use differentiation and the various rules for differentiation to achieve this before considering how these techniques can help us in more complicated examples.
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So, it’s important that you have a sound understanding of the rules for differentiation before watching this video.
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For related rates problem, it’s usually best to jump straight into a problem and see how it all works.
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The radius of a circle is increasing at a rate of three millimetres per second.
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Find the rate of change of the area of the circle when the radius of the circle is 15 millimetres.
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The first thing we should always do in related rates questions is to identify what information we’ve been given and what we’re trying to find.
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We are told that the radius of the circle increases at a rate of three millimetres per second.
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We know that the rate of change is essentially the derivative of some function with respect to time.
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So, we can say that d𝑟 by d𝑡, the derivative of our radius with respect to time, is equal to three.
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We’re being asked to find the rate of change of area of the circle.
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That’s the derivative of the area with respect to time.
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It’s d𝐴 by d𝑡.
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And specifically, we want to find the rate of change of the area of the circle when the radius is equal to 15, when 𝑟 is equal to 15.
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So, how do we link these three pieces of information?
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Here, we have to delve into the recesses of our brain and recall the formula for the area of a circle.
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It’s 𝜋 times the radius squared.
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So, the derivative of 𝐴 with respect to time is the same as the derivative of 𝜋𝑟 squared with respect to time.
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Now we can’t differentiate 𝜋𝑟 squared with respect to time using the usual rules for differentiating a polynomial.
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But we know that the radius is essentially a function in time.
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So, actually, we see that we need to differentiate a function of a function, or a composite function.
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And we’ll use the chain rule to do this.
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The chain rule says that if 𝑦 is a function in 𝑢 and 𝑢 is a function in 𝑥, then d𝑦 by d𝑥 is equal to d𝑦 by d𝑢 times d𝑢 by d𝑥.
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We can change this a little and say that, well, the derivative of our area with respect to time can be found by multiplying d𝐴 by d𝑟, the derivative of the area with respect to its radius, by the derivative of the radius with respect to time, d𝑟 by d𝑡.
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We saw that 𝐴 is equal to 𝜋𝑟 squared.
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So, we can say that the derivative of 𝐴 with respect to 𝑟 is two times 𝜋𝑟.
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It’s two 𝜋𝑟.
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We also saw that the derivative of our radius with respect to time, d𝑟 by d𝑡, is three.
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So, we can say that d𝐴 by d𝑡 is equal to two 𝜋𝑟 times three.
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And you can see we found a formula for the rate of change of the area of the circle in terms of 𝑟.
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We simplify it, and we see that d𝐴 by d𝑡 is equal to six 𝜋𝑟.
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And there is one piece of information we haven’t yet used.
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We were trying to find the rate of change of the area of the circle when the radius was equal to 15.
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So, we’re going to substitute 𝑟 equals 15 into the formula we created.
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That gives us six times 𝜋 times 15, which is equal to 90𝜋.
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And it’s sensible to refer back to the question when thinking about our units.
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The radius was clearly in millimetres and the time was in seconds, so our area is going to be square millimetres.
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And the rate of change of the area of the circle, when the radius is equal to 15 millimetres, is 90𝜋 square millimetres per second.
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In this example, we’ve seen how the chain rule can help us to answer related rates problems.
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We’ll now consider another example of this type.
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A spherical balloon leaks helium at a rate of 48 cubic centimetres per second.
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What is the rate of change of its surface area when its radius is 41 centimetres?
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The first thing we should do in related rates questions such as this is to identify what information we’ve been given and what we’re trying to find.
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We’re told that the balloon is in the shape of a sphere.
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And it’s leaking helium at a rate of 48 cubic centimetres per second.
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Now another way of thinking about this is that the volume of the balloon is decreasing at a rate of 48 cubic centimetres per second.
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This tells us two things.
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Firstly, volume is a function of time and the rate of change, which is simply the derivative with respect of 𝑡, is negative 48.
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And it’s negative because the balloon is leaking, the volume is decreasing.
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So, we can say that d𝑣 by d𝑡 is equal to negative 48.
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We’re trying to find the rate of change of the surface area.
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Let’s call that 𝐴.
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And we’re trying to find the derivative of our area with respect to time.
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We also know that there’s a point at which the radius of the balloon is equal to 41 centimetres.
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Now it’s going to be sensible to set up an equation for the surface area of our sphere.
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The standard formula for the surface area of a sphere with a radius 𝑟 is four 𝜋𝑟 squared.
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So, the derivative of 𝐴 with respect to time is equal to the derivative of four 𝜋𝑟 squared with respect to time.
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𝐴 is a function in 𝑟.
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But actually, we know that the radius is going to be a function in 𝑡.
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And this means, to differentiate 𝐴 with respect to time, we’re going to use the chain rule.
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This says that if 𝑦 is a function in 𝑢 and 𝑢 is a function in 𝑥, then d𝑦 by d𝑥 is equal to d𝑦 by d𝑢 times d𝑢 by d𝑥.
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Let’s change this a little to match our question.
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And we see d𝐴 by d𝑡 must be equal to d𝐴 by d𝑟, the derivative of the area with respect to the radius, multiplied by d𝑟 by d𝑡, that’s the derivative of 𝑟 with respect to 𝑡.
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The surface area of the sphere is four 𝜋𝑟 squared.
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So, we can differentiate that 𝐴 with respect to 𝑟 as two times four 𝜋𝑟, which is eight 𝜋𝑟.
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But what about d𝑟 by d𝑡?
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We haven’t actually been given a value for d𝑟 by d𝑡.
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So, we’re going to use d𝑣 by d𝑡 to help us.
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We’ll also use the fact that the volume of the sphere with a radius 𝑟 is four-thirds 𝜋𝑟 cubed.
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So, we can say that the derivative of four-thirds of 𝜋𝑟 cubed with respect to 𝑡 is equal to negative 48.
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The constant multiple rule tells us that we can take the multiple of four-thirds outside of the derivative and focus on differentiating the function itself.
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So, we have four-thirds times the derivative of 𝑟 cubed with respect to 𝑡 equals negative 48.
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We then divide through by four-thirds, and we see that d by d𝑡 of 𝑟 cubed is negative 36.
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But we need d𝑟 by d𝑡 not d of 𝑟 cubed by d𝑡.
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But since 𝑟 is a function of 𝑡, we can use the chain rule here too.
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This time we can say that the derivative of 𝑟 cubed with the respect to 𝑡 is equal to the derivative of 𝑟 cubed with respect to 𝑟 times the derivative of 𝑟 with respect to 𝑡.
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And we see that the derivative of 𝑟 cubed with respect to 𝑟 is three 𝑟 squared.
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So, we have three 𝑟 squared times d𝑟 by d𝑡 equals negative 36.
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We can divide both sides of this equation by three 𝑟 squared.
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And we now have d𝑟 by d𝑡.
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It’s negative 12 over 𝑟 squared.
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Let’s substitute this back into our expression for d𝐴 by d𝑡.
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And we get eight 𝜋𝑟 times negative 12 over 𝑟 squared.
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We can simplify by this common factor of 𝑟.
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And we can now see that d𝐴 by d𝑡, the formula for the rate of change of surface area with respect to time, is given as negative 96𝜋 over 𝑟.
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We can now evaluate this at the point where the radius of our balloon is equal to 41 centimetres.
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That’s negative 96𝜋 over 41.
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And since this is the rate of change of the surface area, and, in this case, surface area will be measured in centimetres squared, our answer is negative 96𝜋 over 41 square centimetres per second.
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In this example, we needed to apply the chain rule twice.
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In our next example, we’ll see how the other rules for differentiation can help us with related rates problems.
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The length of a rectangle is increasing at a rate of 15 centimetres per second and its width at a rate of 13 centimetres per second.
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Determine the rate at which the area of the rectangle increases when the length of the rectangle is 25 centimetres and its width is 12 centimetres.
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Remember, we should always try to identify what information we’ve been given and what we’re being asked to find.
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We know that the length of the rectangle is increasing at a rate of 15 centimetres per second.
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This tells us two things.
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Firstly, length is a function of time.
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It also tells us though that the rate of change, which is simply the derivative with respect to time, is 15.
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So, we can say that d𝑙, where 𝑙 is the length, by d𝑡 is equal to 15.
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Similarly, we’re told the width of the rectangle is increasing at a rate of 13 centimetres per second.
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So, width is also a function of time.
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And since the rate of change is the derivative with respect to time, d𝑤, where 𝑤 is the width, by d𝑡 is equal to 13.
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Now we’re looking to find the rate at which the area is increasing.
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So, we use two pieces of information.
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Firstly, we know that the area of a rectangle is given by its width multiplied by its length.
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But we’re trying to find the rate of change of the area, that’s d𝐴 by d𝑡.
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Let’s say area is 𝑤𝑙, width times length, then we can say that d𝐴 by d𝑡 is the derivative of 𝑤 times 𝑙 with respect to time.
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We’re trying to find the derivative of a product of two functions of 𝑡.
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So, we can use the product rule to evaluate the rate of change of the area of our rectangle.
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This says that if 𝑢 and 𝑣 are differentiable functions, then the derivative of 𝑢 times 𝑣 with respect to 𝑥 is equal to 𝑢 times d𝑣 by d𝑥 plus 𝑣 times d𝑢 by d𝑥.
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And what does this mean for the derivative of the area with respect to time?
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Well, we can say it’s 𝑤 times d𝑙 by d𝑡 plus 𝑙 times d𝑤 by d𝑡.
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And actually, we have all of these things.
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We want to find the rate at which the area of the rectangle increases when its width is 12.
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So, 𝑤 times d𝑙 by d𝑡 is 12 times 15.
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And we want to find the rate of change of the area when the length is 25.
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So, 𝑙 times d𝑤 by d𝑡 is 25 times 13.
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12 multiplied by 15 plus 25 multiplied by 13 is 505.
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Since the width and length of the rectangle are measured in centimetres, and the rate of change of these measurements is centimetres per second, we know that the rate at which the area of the rectangle increases when the length of the rectangle is 25 centimetres and its width is 12 centimetres is 505 square centimetres per second.
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In our next example, we’ll look at how implicit differentiation can be used in related rates problems.
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A particle is moving along the curve six 𝑦 squared plus two 𝑥 squared minus two 𝑥 plus five 𝑦 minus 13 equals zero.
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If the rate of change of its 𝑥-coordinate with respect to time as it passes through the point negative one, three is two, find the rate of change of its 𝑦-coordinate with respect to time at the same point.
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Let’s have a look at what we know and what we’re trying to find out.
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The equation six 𝑦 squared plus two 𝑥 squared minus two 𝑥 plus five 𝑦 minus 13 equals zero describes the movement of the particle.
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We’re told that the rate of change of its 𝑥-coordinate is two, so 𝑥 is a function in time.
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And we know that 𝑦 itself is a function of 𝑥.
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So, we’re going to use implicit differentiation to differentiate our entire equation with respect to 𝑡.
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Now, of course, we differentiate both sides with respect to time but the derivative of zero is zero.
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So, we have d by d𝑡 of six 𝑦 squared plus two 𝑥 squared minus two 𝑥 plus five 𝑦 minus 13 equals zero.
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At this stage, it might be sensible to separate the derivative.
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And then, we use implicit differentiation, which is essentially a special case of the chain rule, to differentiate each term.
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The derivative of six 𝑦 squared with respect to 𝑡 is equal to the derivative of six 𝑦 squared with respect to 𝑦 times the derivative of 𝑦 with respect to 𝑡.
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We can differentiate six 𝑦 squared as normal.
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It’s two times six 𝑦, which is 12𝑦.
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We then differentiate two 𝑥 squared with respect to 𝑡.
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This time it’s the derivative of two 𝑥 squared with respect to 𝑥 times the derivative of 𝑥 with respect to 𝑡.
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The derivative of two 𝑥 squared is two times two 𝑥, which is four 𝑥.
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So, d by d𝑡 of two 𝑥 squared is four 𝑥 times d𝑥 by d𝑡.
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We’ll repeat this process with the derivative of negative two 𝑥 with respect to 𝑡.
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It’s the derivative of negative two 𝑥 with respect to 𝑥.
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That’s negative two times d𝑥 by d𝑡.
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We then see that d by d𝑡 of five 𝑦 is five times d𝑦 by d𝑡.
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And that’s because the derivative of five 𝑦 with respect to 𝑦 is just five.
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And, of course, the derivative of our constant negative 13 is zero.
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And now we can evaluate.
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We know that the rate of change of its 𝑥-coordinate with respect to time is two.
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So, we can say that d𝑥 by d𝑡 is equal to two.
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We also know it passes through the point negative one, three, so we’ll set 𝑥 to be negative one and 𝑦 to be equal to three.
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And our equation becomes 12 times three times d𝑦 by d𝑡 plus four times negative one times two plus negative two times two plus five times d𝑦 by d𝑡.
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We can simplify.
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And we end up with 41 d𝑦 by d𝑡 minus 12 equals zero.
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And then, we add 12 to both sides.
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So, 41 d𝑦 by d𝑡 is equal to 12.
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And that means that d𝑦 by d𝑡 is equal to 12 over 41.
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And this means that the rate of change of the 𝑦-coordinate with respect to time at negative one, three is 12, 41.
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In our final example, we’re going to see how we need to be careful when it comes to considering the units of a question and how differentiation isn’t always our first port of call when it comes to answering related rates questions.
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Given that a rocket of mass 26 metric tons is burning fuel at a constant rate of 80 kilograms per second, find the mass of the rocket 25 seconds after takeoff?
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Let’s begin by identifying what we know and what we’re trying to find.
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We know that the rocket is burning fuel at a rate of 80 kilograms per second.
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In other word, assuming the rocket isn’t losing mass from anywhere else, which is a safe assumption to make, we know that the rocket is losing 80 kilograms of its mass per second.
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We also know that the rocket initially, in other words, when 𝑡 is equal to zero, has a mass of 26 metric tons.
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One metric ton is equal to 1000 kilograms, so the starting mass of the rocket must be equal to 26000 kilograms.
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To find the total mass of the rocket 25 seconds after takeoff, we can find the total amount of fuel lost in this time.
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We know it loses 80 kilograms per second.
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So, after 25 seconds, it will have lost 80 times 25 kilograms.
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That’s 2000 kilograms in 25 seconds.
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The new mass of the rocket after 25 seconds will, therefore, be 26000 minus 2000.
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That’s 24000 kilograms, which is 24 metric tons.
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And so, the mass of the rocket after 25 seconds is 24 metric tons.
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In this video, we’ve seen how we can use derivatives to find the relation between the rates of two or more quantities in related rates problems.
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We saw how we can use the chain rule and other rules for differentiation to help us solve related rates problems.
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We saw that we do need to be careful though, as derivatives aren’t always the most efficient method to answer these questions.