Question

At what points on the given curve x = 2t3, y = 5 12t − 7t2 does the tangent line have slope 1?

Answer 1

To find the points on the curve x = 2t^3, y = 5/12t - 7t^2 where the tangent line has a slope of 1, calculate the derivatives of x and y with respect to t, set the ratio of dy/dt to dx/dt equal to 1, solve for t, and then find the respective x and y coordinates.

Step-by-step explanation:

How to Find the Points on a Curve Where the Tangent Line Has a Slope of 1

To find the points on the curve x = 2t3, y = ⅓t - 7t2 where the tangent line has a slope of 1, we need to follow these steps:

1. Determine the derivatives of x and y with respect to t to find the slopes of the tangent lines. The derivative of x with respect to t is dx/dt = 6t2, and the derivative of y with respect to t is dy/dt = ⅓ - 14t.
2. Find the slope of the tangent line at any point by taking the ratio dy/dx, which is equal to (dy/dt)/(dx/dt). Set this ratio equal to 1 and solve for t, since we are looking for a slope of 1.
3. Plug in the value of t found in step 2 into the original equations for x and y to find the coordinates of the point where the tangent has a slope of 1.

Here's an example applied to the given equations:

• First, compute the derivatives: dx/dt = 6t2, dy/dt = ⅓ - 14t.
• Next, find the slope of the tangent line by setting (⅓ - 14t) / (6t2) equal to 1 and solve for t.
• Finally, substitute the solved t values into the original equations to find the corresponding x and y coordinates.

Answer 2

The point on the curve 2t³ y = 4 + 12t - 7t², where the tangent line has a slope of 1 is (0, 4).

Step-by-step explanation:

To find the points on the given curve where the tangent line has a slope of 1, we first need to rewrite the equations in terms of the variable t:

Curve equation: x = 2t³ y = 4 + 12t - 7t²

Next, we will find the derivative of the curve with respect to t:

x’ = 6t² y’ = 12 - 14t

Now, we can set the slope of the tangent line equal to 1:

1 = x’/y’

Substitute the derivative equations into the slope equation:

1 = (6t²)/(12 - 14t)

Now, we need to solve the equation for t:

Step 1: Factor out the common factor of t² from the numerator and the denominator:

1 = (6t²)/(12 - 14t) = (6t²)/(t²(12 - 14t))

Step 2: Divide both sides by t²:

1 = (6t²)/(t²(12 - 14t)) = 6/(12 - 14t)

Step 3: Simplify the equation:

6 = 6t - 84t²

Step 4: Factor out t from the equation:

6t = 84t² - 84t²

Step 5: Simplify the equation:

6t = 0

Step 6: Solve for t:

t = 0

Now that we have found the value of t, we can plug it back into the original equations to find the points on the curve where the tangent line has a slope of 1:

x = 2t³ = 2(0)³ = 0

y = 4 + 12t - 7t² = 4 + 12(0) - 7(0)² = 4

So, the point on the curve is (0, 4).