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Solve the given problems.The vertical displacement $y$ (in $\mathrm{cm}$ ) of the end of an industrial robot arm for each cycle is $y=2 t^{1.5}-\tan 0.1 t,$ where $t$ is the time (in s). Find its vertical velocity for $t=15 \mathrm{s}$.

Calculus 1 / AB

Chapter 27

Differentiation of Transcendental Functions

Section 2

Derivatives of the Other Trigonometric Functions

Derivatives

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University of Michigan - Ann Arbor

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Boston College

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for this problem we have the vertical displacement at the end of an industrial robot arm. It's going to be widely equals ju T. To the 1.5 mine ex candidate. 0.1 cheap. So when we evaluate this we know that we have different derivative rules that we know. So this the derivative of that would be Uh T to the 1.5 over the natural log of 1.5. And then the tangent of value when we take the derivative is going to be seeking squared of that value. And then because of the 0.1 inside we'd use chain rules. So we multiply the whole thing by 0.1. Um one way we can also evaluate this is by treating the function. Why as ffx and T as the value acts. I'm only here this we get F prime of X. Is this graph right here And we're trying to evaluate it at 15 seconds. So we see the f prime of 15 is going to end up giving us a negative 8.4 and that cm per second.

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