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Numerade Educator

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Problem 32 Hard Difficulty

Find the tangential and normal components of the acceleration vector.
$$\mathbf{r}(t)=t \mathbf{i}+\cos ^{2} t \mathbf{j}+\sin ^{2} t \mathbf{k}$$

Answer

$$
\frac{2 \sin 4 t}{\sqrt{1+2 \sin ^{2} 2 t}}
$$
$$
\frac{2 \sqrt{2\cos 2 t} }{\sqrt{1+2 \sin ^{2} 2 t}}
$$

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Video Transcript

So in this video are given off tea and were asked to determine the door bone tangential components of the exploration vector. So our team, you simply t I have Moscow sine squared Teoh a sine squared. So our priority way We just have to find the derivative of each component with respect so derivative of tea with respect to use one the derivative of course I square to use the changeable is just too close I t time. So this is a derivative of the outside derivative would be inside the derivative close scientists just negative sign. So in total we get negative to co scientist. Citing plus the derivative of science were Teoh perspective. You have to use a chain room. So it's to sigh Inti, that's a derivative of the outside times. The derivative of sine of t is closely so to scientific on 70 K Now we can use our trick identities so we know that well to sign beta co sign the lease just signed to theater So then these were going to get one. I have minus sign to t j have plus sign to Kiev I know are now the second derivative of our majestic, the derivative of our prime to teach. They're a bit of a former with respected t zero that inevitable negative cited. You respected t again. Train role begin Negative too. Uh, negative to sco sign t j a happ plus again they're gonna sign to feel respected. T general, We used to be willing to a close eye. Inti Chaos. Okay, so now we want to determine the magnitude of our privacy. Sort of take each component, square it and add it to bath. Together they take the square root. So swear Rose the square. So what was once we're plus signs where two teeth puts ice cream, I would take the square root of that. So that's just one plus two sides square two teeth. Okay. No. A sub t The tangential component of the explorations. Your spot prime of t dot r double o. T divided by our problems. So basically, what we're gonna do is we're gonna take each component and multiply it with each other. So one times 00 negative Sign, uh, to tee times Negative to co sign t will be positive to sign. T uh co signed 30 from Sai Inti and then, uh, that could multiply The cake combines together we get to sign to ti Ko sai. All right, so this is when we get right over here. So this is good zero. And then we have to sign to become some 20 plus to sign to t close I. And what we can do right now is that we can add the's together, so we're just gonna get So we add these two together, we just get for sign to t coastline, and that's pretty much it. So four signed. Two teeth close identity divided by the square root of one plus two have Science Square. All right now to determine that the normal component of the acceleration. First we determined the cross product. So have I. Had they had happened? We write our each component are far prime of being art double private t. So we get one negative scientist et sign to t zero negative to co sign into T two course on three teeth to return to find the I had component across these two. And then we find their determinant of once on the inside. This just negative scientific T times to close identity plus to co sign to t sign titty. So that's how we determine it. And then we realized that two components are the same. So this is simple zero now to find the jihad component across this across this point determining so one times to co sign to t buying his your attach signed to me, that's just too close. I duty. But since the jihad component, we have to foot the sign So it's negative to co sign to T. J. But the bill finally, where the key component across this across this we get across brother or we don't refine determinant. So it's one times negative Two cops trying to do is just negative two from sanctity buying a zero times negative scientist to you just get zero. So we have negative to t coastline. Do you have? So what survives are these? So we have zero. I have minus two coastline to t. J have my studio site T k A. And now, to determine the magnitude of what across product, we're gonna square each complaint and then take the square root of we have zero squared plus negative to co sign to t square loss. Negative to co sign to square which is just simply eight co sign square to t And now we take the square root of that So you can write that we can pull out a force are going to get to Times Square, vote off to t coastline square to t when we can, right two times the square root of two times the absolute value of coastline to because it's under the square roots which should always be positive. So that's where we put the absolute value, of course. All right now, Ace event is just the magnitude of our prime mufti cross with our double primal divides by the magnitude of our private T and we determined all the same to Times Square would have to I was the absolute by go side to t divided by one plus two sides square to t. And that is the normal component of the acceleration vector