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Find the limit. $ \displaystyle \lim_{x\to 0} \f…

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Problem 38 Hard Difficulty

An object with weight $W$ is dragged along a horizontal plane by a force acting along a rope attached to the object. $\\$
If the rope makes an angle $\theta$ with the plane, then the magnitude of the force is $\\$
$$
F=\frac{\mu W}{\mu \sin \theta+\cos \theta}
$$ $\\$
where $\mu$ is a constant called the coefficient of friction. $\\$
(a) Find the rate of change of $F$ with respect to $\theta$. $\\$
(b) When is this rate of change equal to $0 ?$$\\$
(c) If $W=50$ Ib and $\mu=0.6$, draw the graph of $F$ as a function of $\theta$ and use it to locate the value of $\theta$ for which $d F / d \theta=0 .$ Is the value consistent with your answer to part (b)?

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Differentiate with respect to t. y = a cos(t) + t^2 sin(t)

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Video Transcript

All right here is another really cool application problem. A box is being pulled along a horizontal floor, but the force is at an angle and we're told that the force has this equation here in green is equal to mu or mu as a coefficient of friction mu times w, which is the weight all over m Sin of theta, plus cosine of theta, and so we're going to work with this equation. First thing we are asked to do is find in part a the derivative with respect to theta. So what i'm going to do before i take the derivative is go ahead and rewrite the denominator as a parenthesis to the minus 1 power in that way ends up in the numerator. Okay and remember: theta is my only variable. Everything else is a constant and i'm going to take the derivative just regard to theta. So when i take the derivative that m w in front that just goes along for the ride, that's just constant! Now i'm going to go ahead and do go ahead and do power rule first and then we're going to do chain rule so notice. We have a parenthesis to the minus 1 power, so our power rule will give us minus the parentheses. You sign a theta. You leave the inner argument the same at first here and that by power rule that parentheses is going to go to the minus 2 power, but we sootily by chain rule and chain rule is the derivative then at the inside. So, u goes along for the rye, but the derivative of sine of theta is cosine of theta, and the derivative of cosine of theta is minus sine of theta okay, so that is our derivative. Let'S go ahead and see if it can't clean it up a little bit, i'm going to go ahead and just go ahead and clean it up, because i have a minus sign here. I can switch the order of this expression and we'll just clean it up a little bit, so the ones that stay on top will be the sine of theta, because i distributed the minus so sine of theta minus mu, cosine of theta and then that power, the Minus 2, if i bring that parenthese down to the bottom, i can make it positive. 2 point. Okay, so that's the answer to part a we have here, circled in red. Okay, so are derivatives that are derivative of f with respects to theta. Is that lovely fraction with lots of terms, but there it is okay, next step, part b, so part b says: okay, what's that when our derivative equals? Let'S theta when our d f d theta equals 0? Well, we have d f d, theta from part a so we're going to set basically what we get equal to 0 and basically then solve for theta okay. So this is big. It has a lot that to make this equal to 0, then notice for it to be equal to 0. I need the numerator to be 0 and since these 2 terms are constants, i need this term to go to 0. So if we set that term equal to 0, that's going to help us solve it, so we're going to set sine of theta minus mu, cosine of theta equal to 0 point. Let'S go ahead and add mu cosine of theta to both sides and that will give us sine theta equals mu, cosine theta. If we divide both sides by cosine of theta, then we get tan at theta equal to mu. So if we want theta, it's going to be inverse tan of me, so that is our expression. Now, it's not till part c, we get numerical values, but that is our answer to part b, because i've just done it. Let'S also look in advance because we're going to need it also for part c, so in our calculator, if we find inverse tan notice in part c, we're gonna, let mu be .6. So if we do inverse tan of .6 notice, we get a value of 0.54042 point. Okay, so that's an advance. Canalis, look at part c in part c. What we're going to do is graph or function or original function f and see if having a derivative equals 0 makes sense. With the angle we got so it's pretty co now f has lots to repeat. As you know, a unit circle goes round and around and around from many thetas. So the part for us that is interesting of f will be the part that has this piece of. It looks something like that and if you use your graphing calculator to find the men, so you need to graph f right, we're graphing that functionary circled on top and if we use our graphing calculator to find the men the men were given is 0.54 o 42. As the x value and the y value is 25.725 okay, so when we find them, there notice that i should correspond to a horizontal tangent line, which is when the slope is 0 and notice that the calculator gives us the same value as when we do it. By hand the .54042 point so really call it's nice to see everything self consistent and it's a plan to a to do an application problem. All right, take care or se.

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James Stewart

Calculus: Early Transcendentals

Catherine Ross

Heather Zimmers

Kayleah Tsai

Samuel Hannah

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Catherine Ross

Heather Zimmers

Kayleah Tsai

Samuel Hannah

James StewartCalculus: Early Transcendentals

Catherine Ross

Heather Zimmers

Kayleah Tsai

Samuel Hannah

James Stewart

Calculus: Early Transcendentals