Thread: slope of tangent line

can someone explain to me on how to do this problem?

the slope of the tangent line to the curve y = 4sqrt(x) at point (7,10.5831) is:___
The equation of this tangent line can be written in the form y = mx+b where m is:___ and b is:___

Originally Posted by helpme

can someone explain to me on how to do this problem?

Code:

the slope of the tangent line to the curve
y = 4sqrt(x)  at point (7,10.5831) is:___ 
 
The equation of this tangent line can be written in
the form  y = mx+b where m is:___ and b is:___

The slope of the tangent line at the point is the same thing as the
gradient of the curve at that point. You find the gradient by taking
the derivative of the equation of the curve, so:

dy/dx=4* 1/2 * 1/sqrt(x)

(note I am using "*" as the multiplication symbol here, which is a common
convention when writting maths in plain ASCII)

At the point in question x=7, so the gradient is:

dy/dx = 2/sqrt(7)~=0.75593.

Now we know the slope of the line we can write the equation of the
line in the form:

y=0.75593*x+b.

but we also know that this line is a tangent to the curve at (7,10.5831),
so we know that:

10.5831=0.75593*7+b.

Which can be solved for b to give the full equation of the desired line.

RonL