Suppose that the temperature at the point (x, y, z) in space is T(x, y, z) = x2 + y2 + z2. Let a particle follow the right-circular helix σ(

Question

Suppose that the temperature at the point (x, y, z) in space is T(x, y, z) = x2 + y2 + z2. Let a particle follow the right-circular helix σ(t) = (cos(t), sin(t), t) and let T(t) be its temperature at time t. (a) What is T ‘(t)? T ‘(t) = (b) Find an approximate value for the temperature at t = π 6 + 0.01.

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Arya 4 weeks 2021-12-29T00:11:41+00:00 1 Answer 0 views 0

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    2021-12-29T00:13:02+00:00

    Answer with Step-by-step explanation:

    We are given that

    T(x,y,z)=x^2+y^2+z^2

    \sigma(t)=(cost,sint,t)

    T(t) be the temperature at time t.

    a.Substitute the value of x,y and z

    Where x=cost,y=sint,z=t

    T(t)=cos^2t+sin^2t+t^2

    We know that

    sin^2x+cos^2x=1

    By using the formula

    T(t)=1+t^2

    Now, differentiate w.r.t t

    T'(t)=2t

    b.t=\frac{\pi}{6}+0.01

    We have to find the approximate value of the temperature at given value of t.

    Substitute x=\frac{\pi}{6}

    T(\frac{\pi}{6})=1+(\frac{\pi}{6})^2=1+\frac{\pi^2}{36}

    T'(\frac{\pi}{6})=2\times\frac{\pi}{6}=\frac{\pi}{3}

    Linear approximation formula:

    f(x)\approx L(x)=f(a)+f'(a)(x-a)

    Where a=\frac{\pi}{6}

    and t=\frac{\pi}{6}+0.01

    Substitute the values

    The approximate value of the temperature=T(\frac{\pi}{6})+T'(\frac{\pi}{6})(\frac{\pi}{6}+0.01-\frac{\pi}{6})

    The approximate value of the temperature=1+\frac{\pi^2}{36}+\frac{\pi}{3}(0.01)

    The approximate value of the temperature=1+\frac{(3.14)^2}{36}+\frac{3.14}{3}\times 0.01=1.284

    Where \pi=3.14

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