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W1erm Fuewoug Angwie s LarCalc11 6,509XPEHodAak ToEt Teca .85point movino along the graphgiven function such that dxldt is centimetens per secono Find Dyldt for the...

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W1erm Fuewoug Angwie s LarCalc11 6,509XPEHodAak ToEt Teca .85point movino along the graphgiven function such that dxldt is centimetens per secono Find Dyldt for the given values of x-5+2(a) * dyldtcm/s4Cdyldt =cnsecdyldtcm secNeed Help?Rade Iuech /Miacv Jc0 Ji pD'nia LiCak1126.011.FlotadeE Yo.r Jaxc-5iThe radius r of circle increasinorte Of = centimeters per minute Find the rate change of Lhe area when31 centimeters.cm? /minNeed Help?Raae IFjalfinttatSubmi: AnswerSave ProgressPractice Enotn

W1erm Fuewoug Angwie s LarCalc11 6,509XP EHod Aak ToEt Teca .85 point movino along the graph given function such that dxldt is centimetens per secono Find Dyldt for the given values of x- 5+2 (a) * dyldt cm/s4C dyldt = cnsec dyldt cm sec Need Help? Rade I uech / Miacv Jc 0 Ji pD'nia LiCak1126.011. Flotad eE Yo.r Jaxc-5i The radius r of circle increasino rte Of = centimeters per minute Find the rate change of Lhe area when 31 centimeters. cm? /min Need Help? Raae I Fjalfinttat Submi: Answer Save Progress Practice Enotne _ Versior



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In parts (a)-(d), let $A$ be the area of a circle of radius $r,$ and assume that $r$ increases with the time $t$ (a) Draw a picture of the circle with the labels $A$ and $r$ placed appropriately. (b) Write an equation that relates $A$ and $r .$ (c) Use the equation in part (b) to find an equation that relates $d A / d t$ and $d r / d t$ (d) At a certain instant the radius is $5 \mathrm{cm}$ and increasing at the rate of $2 \mathrm{cm} / \mathrm{s}$. How fast is the area increasing at that instant?

I love related rates problems. This is another one. They're not a big deal on this problem. They tell me that I have a circle with area A. And radius R. And the radius is changing. The radius is increasing so the area is too. So part A says draw a picture of a circle with area and radius R. Part B is write an equation describing the relationship between A and R. We learned in earlier school A equals pi R square. See says, take the derivative of the equation and be with respect to T. I get D A. D. T. The derivative of aid to the first is A. To the zero times the derivative what's inside D. A. D. T. The derivative of the right sound is pie the derivative of two R squared with respect to T. Is to our D. R. D. T. I'm not taking the derivative with respect to our but with respect to T. It's usually written D A D. T. Is equal to pi R D. R. DT. This is the circumference times a teeny tiny change in our, so the area is changing. You can kind of think about this as a really small rectangle that's stretched out and being added to the outside. It's the circumference times a teeny thickness D. R. D. T. See circumference times drd t. Okay so I start with a picture. I read an equation relating the variables they tell me I differentiated implicit implicitly with respect to R. And then I plug in values For part d. They said at the point in time when our is 5cm D. R. DT Is two cm per second meaning the radius is changing at that rate. They want me to find D. A. D. T. At this point I'm just plugging in numbers two times pi times five times two. If I look at the units multiplied centimeters times centimeters per second. So I get centimeters squared per second which is a change in area with time. So I end up with five times two times 2 20 pie centimetres squared per second. They may use a calculator to get a Number value for that. It will be 62.8 and change Because two pi is two times 3.14 mm 4159. That would be 318 maybe. I don't know how many Places after the decimal. They want to carry that one. In calculus we carry three or more places. So that's 20 times pi. I'm checking with my calculator 62.83185. So that could round anywhere you wanted it. Hope that helps. Have a nice day.

Okay so first we need to right down the area of the circle. The area of the circle. Yes. Okay so a echoes tie times R. Squared. So here are it's a radius. So now let's find let's differentiating with respect to T. So I have the A. Over D. T. Oh actually echoes five times to our time's D. R. Over DT. Okay so we know that D. A. Over D. T. Actually echoes two pi R. Time's D. R. Over DT. Sending me their square per minute. Yeah. Yeah. Okay since we know that the our our oh D. P. Mhm. Yes three cm for a second. Okay mm. So this is a gable in the text of the problem so we can write T. A. Over D. T. Yes six pi R. 30 m squared per second per minute. Okay. Uh huh. Yeah yeah for part a local part A. So what are yeah Find a rental change of the area where our echo six cm top. Mhm. The radio change off the area is Yeah the A. Over D. T. You go 65 times six which of course 36 times pi centimetres squared per minute For part B. Okay. What are equals 24 centimetres. The region change of area is yeah. Yeah. Mhm. D. O. R. D. T. He calls 65 times 24 Thank God this is 144 pies cm squared per minute. Okay

Here are is the radius of a circle and is increased that the rate of four centimeters per minute. Given that our is 37 sentimentalists, we would like to find the rate of change of the area of the circle. We know that the area of a circle is the people to pi. Times are square here, isa function offthe e So ace, also functional. Think we can definition respect to Tebow sides of this equation and we have t a d d equal pies the constants that we get five times the derivative of r squared which is to our so again our is a function of tea. So we multiply with the David DeVeaux are with respect to the d. P. R. Did we know the radius is 37? So here we can plugging the value for our and we are also given the information that are is increased that the rate of four centimeters per minute This means that they are Davey is equal to four sentiment it's permanent Going back to our immigration. We can plug in the value off the Arditti and find the value of P a deity which is the rate of change of the area. So here we will have to multiply hi times two times 37 centimeters times or centimeters per minute and multiplying two times 37 times before we get to 96 and part time spy. And here's sentiment is time. Centimeters gives a centimeters squared over for a minute, so the area changes with this range.

In problem 12 We have a circle and we need to find the rate of change of the area of that circle at two different lengths of the radius are also given the rate of change of the radius with respect to time. So in order to solve for this, we're gonna have to relate the area through the radius and then take that implicit derivative and probably using the chain rule. And then we're gonna end up plugging these values for our How do we relate the area and the radius of a circle? Well, it's quite easy. A is going to be equal to pi r squared. That's just the equation of the area of a circle. Now, if we take through derivative of that with respect to time, well, pie is just constant. And for our, that's just going to be the our rules that we have two pi r We took the derivative of this with respect are now we need to take the derivative of our with respect to time. We use the chain rule there. Now that we have this, we know that the derivative Arth respect time is just four or centimeters for second or per minute if you attacked. And that means that our rate of change here is going to be eight. I are now. We can just evaluate this at two different points. So when we're dealing with our case and a going to go ahead and evaluate this at eight, that's going to be equal to eight times pi times eight, which is just going to be 64 pi. And then that's going to be, Of course, we can't forget our units in our final answer. Don't make sure you write that as centimeters per minute and then for B, it's going to be the exact same thing where it's playing in a different number. So for being yeah, it was a derivative of a with respect to time or evaluating it. At 32 that's going to be eight times pi Times 32 which is going to be equal to number eight times 32 is going to be 256 pie, and you wanna write on our units or we have centimeters per minute


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