SOLUTION: please help im going insane!!!
{{{f(t)=1+tcos(pi*t),0<=t<=1}}} and {{{f(t)=1+cos(pi*t)+1/pi*sin(pi*t),t>1}}}
0 is < or equal to t < or equal to 1, and t>1 respectively.
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-> SOLUTION: please help im going insane!!!
{{{f(t)=1+tcos(pi*t),0<=t<=1}}} and {{{f(t)=1+cos(pi*t)+1/pi*sin(pi*t),t>1}}}
0 is < or equal to t < or equal to 1, and t>1 respectively.
I ne
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Question 190138: please help im going insane!!! and
0 is < or equal to t < or equal to 1, and t>1 respectively.
I need to show f is continuous and differentiable when t=1 and find whether (f')' is also continuous at t=1 Answer by jim_thompson5910(35256) (Show Source):
In order to show continuity, you simply need to plug in the given value into both expressions and see if they are equal. If they are equal, then the function is continuous at 1. If not, then the function has a jump discontinuity.
Set the first expression equal to g(t)
Plug in t=1
Multiply
Evaluate the cosine of to get -1
Combine like terms.
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Set the second expression equal to h(t)
Plug in t=1
Multiply
Evaluate the cosine of to get -1. Evaluate the sine of to get 0
Multiply and simplify
Combine like terms.
Since (ie they both equal 0), this means that when t=1. So this shows that f(t) is continuous at t=1
If you graphed this piecewise function, you'll find that at t=1 the graphs are joined since the f(t) (or y value) is equal at that point.
Now, derive each expression (separately) to find f'(t)
Now to show that f(t) is differentiable at t=1, simply follow the same technique shown above. In other words, plug in t=1 into both first derivatives of the 2 separate expressions and see if they are equal. Let me know if you need help with this.
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