Slope of a Curve at a Point

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Question

Slope of a Curve at a Point

In Exercises 7–18, use the method in Example 3 to find (a) the slope of the curve at the given point P, and (b) an equation of the tangent line at P.

y=x³, P(2,8)

Accepted Answer

Welcome back, everyone. Calculate the slope of the curve at the specified points and find the equation of the tangent line at S. Y equals 3 x cubed, and S is a point with coordinates of X equals 3 and Y equals 81. So first of all, we're going to find the general equation of the slope at any point X. We're going to use the limit definition. We want to evaluate the limit as H approaches 0 of FX plus H minus F of X divided by H. We have to recall that this equation gives us the slope of the tangent line, let's call it M, right? So what we have to do is simply evaluate each part in the formula. We're going to take limit as H approaches 0. 1st of all, we want to take F of X plus H. This means that our X becomes X + H, so we can write that term as 3 multiplied by X plus H cub. Then we are subtracting F of X, which is 3 X cubed, and then we are dividing by H. From here, we have to calculate the result. Limit as each approaches 0 of 3, multiplied by. Now we have X plus H cube, so here we simply have to apply the formula. First of all, we are squaring the first term which gives us X cube. We're adding 3 multiplied by the square of the first term, so that's 3 X 2 multiplied by the second term, which is H. Then we are adding 3 multiplied by the first term, so 3 x multiplied by the square of the second term. So 3 X H squared, and finally we're adding the cube of the second term, so we're adding each cube. We are subtracting 3 x cubes and all of that is divided by H. From here, we're going to distribute limit as H 0 of 3 X cubed plus 9 X 2 H plus 9 XH 2+. We H cubed minus 3 x cubed, all of that divided by H. We can notice that we can cancel out 3x cubed and 3x cubed followed by fact realization, right, because each term hash, so we get limit as H approaches 0 of H incs we get 9 X2 + 9 X H plus 3 H2 divided by H. This finally allows us to cancel out the each term. And our limit becomes limit as H approaches 0 of 9 X2 + 9 X H + 3 H2. And when H approaches 0, all of the terms that now contain h are going to be equal to 0, right, because we're multiplying by H and H squared respectively. So our limit becomes 9 x 2, and this is the general expression of the slope. Now to evaluate the slope at the given point, we simply need to use the X coordinate, which is 3, right? So our slope M is equal to 9 multiplied by 3 squared. Which gives us 81. And now Since we have our slope, we are going to use the equation of a line Y minus Y1 equals M multiplied by X minus X1. The point slope form, right? We get Y minus Y1 is the Y coordinate of the given point, which is 81. We already know that our slope is 81 and it is multiplied by x minus X1, which is the X coordinate of points. So we're subtracting 3. And now we can just rewrite it, right? We're going to say that Y equals 81 X. Minus 243. And then we're simply adding 81 to both sides. Combining the like terms, we're going to get Y equals 81 X minus 162. Well done, so we have our final answers now. Our slope is 81 and the equation of the tangent line is Y equals 81 X minus 162. Thank you for watching.

Slope of a Curve at a Point

In Exercises 7–18, use the method in Example 3 to find (a) the slope of the curve at the given point P, and (b) an equation of the tangent line at P.

y=x³, P(2,8)

Key Concepts

Derivative

Tangent Line

Point-Slope Form

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