Tangent Line h(t) = sec t / t @ (Pi, 1/pi)
1 answer:
Take derivitive
note
the derivitive of sec(x)=sec(x)tan(x)
so
remember the quotient rule
the derivitive of
so
the derivitive of
so now evaluate when t=pi
we get
sec(pi)=-1
tan(pi)=0
we get
slope=1/pi
use slope point form
for
slope=m and point is (x1,y1)
equation is
y-y1=m(x-x1)
slope is 1/pi
point is (pi,1/pi)
y-1/π=1/π(x-π)
times both sides by π
πy-1=x-π
πy=x-π+1
y=(1/π)x-1+(1/π)
or, alternately
-(1/π)x+y=(1/π)-1
x-πy=π-1
note
the derivitive of sec(x)=sec(x)tan(x)
so
remember the quotient rule
the derivitive of
so
the derivitive of
so now evaluate when t=pi
we get
sec(pi)=-1
tan(pi)=0
we get
slope=1/pi
use slope point form
for
slope=m and point is (x1,y1)
equation is
y-y1=m(x-x1)
slope is 1/pi
point is (pi,1/pi)
y-1/π=1/π(x-π)
times both sides by π
πy-1=x-π
πy=x-π+1
y=(1/π)x-1+(1/π)
or, alternately
-(1/π)x+y=(1/π)-1
x-πy=π-1
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We have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.
We know that vertex of this quadratic function occurs at (5,2).
Further, we know that range of this function is .
If we restrict the domain of this function to either or
, it will become one to one function.
Let us know find its inverse.
Upon interchanging x and y, we get:
Let us now solve this function for y.
Hence, the inverse function would be if we restrict the domain of original function to
and the inverse function would be
if we restrict the domain to
.