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

Quote:

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:___

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- Sep 14th 2006, 07:33 PMhelpmeslope of tangent line
can someone explain to me on how to do this problem?

Quote:

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:___

- Sep 14th 2006, 09:53 PMCaptainBlack
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