How do you find the equation of the tangent line to the curve #y=xsqrtx# that is parallel to the line #y=1+3x#?

Question

William Abdalla · Accepted Answer

The equation is #y = 3x- 4#.

A parallel line means equal slope. The slope of the curve at #x =a# is given by evaluating #f'(a)#, where #f'(x)# is the derivative. We need find the derivative, therefore. First of all, the function can be rewritten as #y= x(x)^(1/2) = x^(1 + 1/2) = x^(3/2)#. We can now use the power rule. #dy/dx = 3/2x^(1/2)# We know the slope of the line #y = 1 + 3x# is #3#, so we set #dy/dx# to #3# and solve for #x#. #3 = 3/2x^(1/2)# #3/(3/2) = x^(1/2)# #2 = x^(1/2)# #x = 4# We now use this point to determine the corresponding y-coordinate. #y= 4sqrt(4) = 4(2) = 8# We now know the slope and the point of contact . The equation is given by: #y- y_1 = m(x- x_1)# #y - 8 = 3(x- 4)# #y- 8 = 3x - 12# #y= 3x - 4# Hopefully this helps!

Adam Adkins · Answer

undefined To find the equation of the tangent line to the curve y = x√x that is parallel to the line y = 1 + 3x, we need to find the derivative of the curve and set it equal to the slope of the given line.

First, we find the derivative of y = x√x using the power rule and the chain rule:

dy/dx = (3x√x + 2x^(3/2))/2√x

Next, we set the derivative equal to the slope of the given line, which is 3:

(3x√x + 2x^(3/2))/2√x = 3

Simplifying the equation, we get:

3x√x + 2x^(3/2) = 6√x

Rearranging the terms, we have:

3x√x - 6√x = -2x^(3/2)

Factoring out √x, we get:

√x(3x - 6) = -2x^(3/2)

Dividing both sides by -2x^(3/2), we obtain:

√x(3x - 6)/-2x^(3/2) = 1

Simplifying further, we have:

(3x - 6)/-2x = 1/√x

Cross-multiplying, we get:

√x(3x - 6) = -2x

Expanding and rearranging the terms, we have:

3x√x - 6√x + 2x = 0

Combining like terms, we get:

3x√x - 6√x + 2x = 0

Factoring out x, we obtain:

x(3√x - 6 + 2) = 0

Simplifying further, we have:

x(3√x - 4) = 0

Setting each factor equal to zero, we get two possible values for x:

x = 0 or √x = 4/3

Solving for x, we find:

x = 0 or x = (4/3)^2 = 16/9

Now, we substitute these values of x back into the original equation y = x√x to find the corresponding y-values:

For x = 0, y = 0
For x = 16/9, y = (16/9)√(16/9) = 32/9

Therefore, the two points on the curve are (0, 0) and (16/9, 32/9).

To find the equation of the tangent line, we use the point-slope form:

y - y₁ = m(x - x₁)

Using the point (16/9, 32/9) and the slope 3, we have:

y - 32/9 = 3(x - 16/9)

Simplifying, we get:

y - 32/9 = 3x - 16/3

Multiplying through by 9 to eliminate the fractions, we obtain:

9y - 32 = 27x - 48

Finally, rearranging the terms, we have the equation of the tangent line:

27x - 9y = 16