Calculus Finding Slope and Equations to Tangent Line

Question

Find the slope, m, of the tangent to the curve y = 2√x at the point where x = a. Find equations of the tangent lines at the following points (25, 10) and (4, 4)

Answer 1

y = 2sqrx = 2x^(1/2)

take the derivative, y' = (1/2)(2)x^(1/2-1) = x^(-1/2) = 1/x^(1/2) = 1/sqrx

slope = m = y' = 1/sqrx

for (25,10) y' = 1/sqr25 = 1/5

tangent line at that point is y=mx + b where m= 1/5

10 = (1/5)(25) + b

b=5

y=x/5 + 5

or 5y = x + 25

5y-x = 25 is the tangent line at (25,10)

for (4,4)

y=x/5 + b

4=4/5 + b

b= 20/5 -4/5 = 16/5

y= x/5 + 16/5

5y = x + 16

5y-x = 16 is the tangent line at (4,4)

at the point where x=a, y= 2sqra = 2a^(1/2)

y=x/5 + b

2sqr(a) = a/5 + b

b = 2sqr(a) - a/5

y= x/5 + 2sq(a) - a/5

5y = x + 10sqr(a) - a

Answer 2

y = 2sqrx = 2x^(1/2)

take the derivative, y' = (1/2)(2)x^(1/2-1) = x^(-1/2) = 1/x^(1/2) = 1/sqrx

slope = m = y' = 1/sqrx

for (25,10) y' = 1/sqr25 = 1/5

tangent line at that point is y=mx + b where m= 1/5

10 = (1/5)(25) + b

b=5

y=x/5 + 5

or 5y = x + 25

5y-x = 25 is the tangent line at (25,10)

for (4,4)

y=x/5 + b

4=4/5 + b

b= 20/5 -4/5 = 16/5

y= x/5 + 16/5

5y = x + 16

5y-x = 16 is the tangent line at (4,4)