Answer with Step-by-step explanation:
We are given that
[tex]T(x,y,z)=x^2+y^2+z^2[/tex]
[tex]\sigma(t)=(cost,sint,t)[/tex]
T(t) be the temperature at time t.
a.Substitute the value of x,y and z
Where [tex]x=cost,y=sint,z=t[/tex]
[tex]T(t)=cos^2t+sin^2t+t^2[/tex]
We know that
[tex]sin^2x+cos^2x=1[/tex]
By using the formula
[tex]T(t)=1+t^2[/tex]
Now, differentiate w.r.t t
[tex]T'(t)=2t[/tex]
b.[tex]t=\frac{\pi}{6}+0.01[/tex]
We have to find the approximate value of the temperature at given value of t.
Substitute [tex]x=\frac{\pi}{6}[/tex]
[tex]T(\frac{\pi}{6})=1+(\frac{\pi}{6})^2=1+\frac{\pi^2}{36}[/tex]
[tex]T'(\frac{\pi}{6})=2\times\frac{\pi}{6}=\frac{\pi}{3}[/tex]
Linear approximation formula:
[tex]f(x)\approx L(x)=f(a)+f'(a)(x-a)[/tex]
Where [tex]a=\frac{\pi}{6}[/tex]
and [tex]t=\frac{\pi}{6}+0.01[/tex]
Substitute the values
The approximate value of the temperature=[tex]T(\frac{\pi}{6})+T'(\frac{\pi}{6})(\frac{\pi}{6}+0.01-\frac{\pi}{6})[/tex]
The approximate value of the temperature=[tex]1+\frac{\pi^2}{36}+\frac{\pi}{3}(0.01)[/tex]
The approximate value of the temperature=[tex]1+\frac{(3.14)^2}{36}+\frac{3.14}{3}\times 0.01=1.284[/tex]
Where [tex]\pi=3.14[/tex]
