Video: Tangents and Normals to the Graph of a Function

In this video, we will learn how to find the slope and equation of the tangent to a curve at a given point using derivatives.

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Video Transcript

In this video, we will see how to apply differentiation to find an equation of a tangent to a curve at a given point. We’ll also discuss what is meant by the normal to a curve at a given point and see examples of how to find the equations of both tangents and normals to curves.

We’ll recall, first of all, that a tangent to a curve at a particular point is a straight line, which touches the curve at that point but does not cross the curve. We will recall also that the gradient or slope of a curve at a given point is defined to be the slope of the tangent to the curve at that point. Therefore, it follows that if we can use differentiation to find the gradient function of a curve d𝑦 by d𝑥, then we can evaluate the slope of the curve and hence the slope of the tangent to the curve at a given point by substituting the 𝑥-value at that point into our gradient function d𝑦 by d𝑥.

We’ll recall also that the general equation of a straight line with slope 𝑚 passing through the point 𝑥 one, 𝑦 one is 𝑦 minus 𝑦 one equals 𝑚𝑥 minus 𝑥 one. So we can use the slope, which we’ve calculated, and the coordinates of the point at which we’re looking to find the tangent in order to find the equation of a tangent.

Let’s see how this works in an example.

Determine the equation of the line tangent to the curve 𝑦 equals four 𝑥 cubed minus two 𝑥 squared plus four at the point negative one, negative two.

So we’ve been given the equation of a curve. And we need to determine the equation of the line that is tangent to this curve at a particular point. We’re going to use the formula for the general equation of a straight line 𝑦 minus 𝑦 one equals 𝑚 𝑥 minus 𝑥 one. We already know the coordinates 𝑥 one, 𝑦 one. It’s the point negative one, negative two. But what about 𝑚, the slope of this line? Well, we recall that the slope of a curve is equal to the slope of the tangent to the curve at that point. So in order to find the slope of this tangent, we’re first going to find the gradient function of the curve d𝑦 by d𝑥.

We can do this by applying the power rule of differentiation, giving d𝑦 by d𝑥 equals four multiplied by three 𝑥 squared minus two multiplied by two 𝑥. Remember, a constant differentiates to zero. So that plus four just differentiates to zero in our derivative, which simplifies to 12𝑥 squared minus four 𝑥. Now, that’s the general gradient function of this curve. But we want to know the gradient at a particular point. So we need to evaluate d𝑦 by d𝑥 when 𝑥 is equal to negative one because that’s our 𝑥-coordinate at this point. This gives 12 multiplied by negative one squared minus four multiplied by negative one, which simplifies to 16.

We now know that the slope of this tangent is 16 and the coordinates of a point that it passes through are negative one, negative two. So we have all the information we need in order to use the formula for the general equation of a straight line. Substituting the values of 𝑚, 𝑥 one, and 𝑦 one gives 𝑦 minus negative two equals 16 𝑥 minus negative one. That’s 𝑦 plus two equals 16𝑥 plus 16. And, then, subtracting two from each side in order to collect the constants gives 𝑦 equals 16𝑥 plus 14. So this is the equation of the line tangent to the given curve at the point negative one, negative two. It passes through the point negative one, negative two and it has the same gradient as the curve at that point.

Now, let’s consider a second example.

Find the point on the curve 𝑦 equals negative 40𝑥 squared plus 40 at which the tangent to the curve is parallel to the 𝑥-axis.

Now, let’s think about what it means for a line to be parallel to the 𝑥-axis. The 𝑥-axis is a horizontal line. So if another line is parallel to the 𝑥-axis, then it must also be a horizontal line. And what do we know about the slopes of horizontal lines? Well, they’re equal to zero. So we know that the slope of the tangent we’re looking to find must be equal to zero. Remember also that the slope of a tangent is equal to the slope of the curve at that point. So we also know that the slope of the curve at this point must also be equal to zero.

The slope of the curve is its gradient function d𝑦 by d𝑥. So what we’re going to do is find the gradient function d𝑦 by d𝑥 by differentiating 𝑦 with respect to 𝑥 and then set it equal to zero. We’ll be able to solve the resulting equation to find the 𝑥-coordinates of the point on the curve where the gradient is equal to zero. Step one then is to find d𝑦 by d𝑥, which we can do by applying the power rule of differentiation. It gives negative 40 multiplied by two 𝑥, which is negative 80𝑥. And remember, the derivative of a constant is zero. So our gradient function d𝑦 by d𝑥 is just negative 80𝑥.

Then, we set d𝑦 by d𝑥 equal to zero and solve the resulting equation. We have negative 80𝑥 equals zero. And by dividing both sides of this equation by negative 80, we find that 𝑥 is equal to zero. So we know that the 𝑥-coordinate of the point on this curve at which the tangent is parallel to the 𝑥-axis is zero. We also need to find the 𝑦-coordinate, which we can do by substituting 𝑥 equals zero back into the equation of the curve. When 𝑥 is equal to zero, 𝑦 is equal to negative 40 multiplied by zero squared plus 40 which is equal to 40. So we find that the point on this curve at which the tangent to the curve is parallel to the 𝑥-axis is the point with coordinates zero, 40.

Now, we could also have seen this by considering what the graph of 𝑦 equals negative 40𝑥 squared plus 40 actually looks like. It’s a negative parabola as the coefficient of 𝑥 squared is negative 40 and it has a 𝑦-intercept of positive 40. We can see from our sketch that this function has a critical point at the point with coordinates zero, 40. In fact, it’s a local maximum. At the critical points for a function, the gradient of the curve and tangent are equal to zero. And so, we see that at the zero, 40 — the critical point of this curve — the tangent at this point will be parallel to the 𝑥-axis.

Let’s now consider another example.

The line 𝑥 minus 𝑦 minus three equals zero touches the curve 𝑦 equals 𝑎𝑥 cubed plus 𝑏𝑥 squared at one, negative two. Find 𝑎 and 𝑏.

The key information given in this question is that the line and the curve touch at this point with coordinates one, negative two. But the line does not cross the curve, which means that the line 𝑥 minus 𝑦 minus three equals zero is a tangent to the given curve at this point. We know that the gradient of a curve is equal to the gradient of the tangent to the curve at that point. The equation of our line is 𝑥 minus 𝑦 minus three equals zero. And when rearranging, we see that this is equivalent to 𝑦 equals 𝑥 minus three. Comparing with 𝑦 equals 𝑚𝑥 plus 𝑐, that’s the general form of the equation of a straight line in slope-intercept form, we see that the slope of our tangent is one. Can we find an expression for the slope of the curve? Well, we can do this by differentiation. By applying the power rule, we see that d𝑦 by d𝑥 is equal to three 𝑎𝑥 squared plus two 𝑏𝑥.

Next, we evaluate this gradient function at the point one, negative, two. So we substitute 𝑥 equals one into our gradient function, giving three 𝑎 plus two 𝑏. We can then equate the gradient of the curve at this point with the gradient of the tangent to the curve at this point. And it gives an equation involving 𝑎 and 𝑏: three 𝑎 plus two 𝑏 is equal to one. We can’t solve this equation because we have only one equation and two unknowns. So we’re going to need to find a second equation.

The point one, negative two lies on both the curve and the tangent. So if we substitute the values of one and negative two into the equation of the curve, we’ll get a second equation connecting 𝑎 and 𝑏. We have 𝑎 multiplied by one cubed plus 𝑏 multiplied by one squared is equal to negative two, simplifying to 𝑎 and 𝑏 equals negative two. We now have two linear equations in 𝑎 and 𝑏, which we need to solve simultaneously. We can multiply equation two by two, as this will make the coefficient of 𝑏 the same as it is in equation one.

We’ll then subtract the second equation from the first to eliminate the 𝑏 terms, giving 𝑎 equals five. Substituting this value for 𝑎 into our original equation two that’s 𝑎 and 𝑏 equals negative two gives five plus 𝑏 equals negative two. And by subtracting five, we see that 𝑏 is equal to negative seven. So we found the values of 𝑎 and 𝑏. 𝑎 is equal to five and 𝑏 is equal to negative seven.

A reminder is that the key point that we used in this question is that the slope of a curve is equal to the slope of the tangent to the curve at that point.

Let’s consider another type of example.

Find the equation of the tangent to the curve 𝑦 equals 𝑥 cubed plus nine 𝑥 squared plus 26𝑥 that makes an angle of 135 degrees with the positive 𝑥-axis.

So we’ve been asked to find the equation of a tangent to a particular curve, which we know we can do using differentiation and the general equation of a straight line. But what does it mean when it says this tangent makes an angle of 135 degrees with the positive 𝑥-axis? Let’s consider a sketch. Well, it will look something like this. The tangent here is shown in pink. And we can see that when intersects with the 𝑥-axis, the angle between the positive 𝑥-axis and the tangent is 135 degrees.

In order to apply the general equation of a straight line 𝑦 minus 𝑦 one equals 𝑚 𝑥 minus 𝑥 one, we either need to know the slope 𝑚 of our line or the coordinates of a point 𝑥 one, 𝑦 one which lies on the line. So how does knowing that our tangent makes an angle of 135 degrees with the positive 𝑥-axis help with determining either of those? Well, the angle on the other side of this line will be 45 degrees because we know that angles on a straight line sum to 180 degrees. We can sketch in a right-angled triangle below this line and recall that the slope of a line is change in 𝑦 over change in 𝑥. That’s the vertical height of this triangle divided by the horizontal distance. But in that right triangle, those sides are the opposite and adjacent in relation to the angle of 45 degrees. So we’re dividing the length of the opposite by the length of the adjacent.

As the line is sloping downwards though, that vertical change is actually the negative of the value of the opposite. So we have that the slope is equal to negative opposite over adjacent. Opposite divided by adjacent defines the tangent ratio. So in fact, this is equal to negative tan of 45 degrees. And tan of 45 degrees is just one. So by considering this right-angled triangle, we found that the slope of this line is negative one. So, we found the slope of our tangent. But we don’t yet know the coordinates of the point on the curve where this tangent is being drawn. To find this, we need to find the point on the curve where the gradient or slope is equal to negative one.

We begin by differentiating the equation of the curve with respect to 𝑥 and applying the power rule of differentiation, giving d𝑦 by d𝑥 equals three 𝑥 squared plus 18𝑥 plus 26. We then set this expression equal to negative one to find the 𝑥-coordinate of the point on the curve, where the gradient is negative one. This simplifies to three 𝑥 squared plus 18𝑥 plus 27 equals zero. And then dividing through by three gives 𝑥 squared plus six 𝑥 plus nine equals zero. We should notice that this is, in fact, a perfect square. We can write it as 𝑥 plus three all squared. Solving this equation then, this means that 𝑥 plus three must be equal to zero. And so, 𝑥 is equal to negative three.

Next, we need to find the value of 𝑦 when 𝑥 is equal to negative three, which we do by substituting negative three into the equation of the curve. And it gives negative 24. We now know that this tangent has a slope of negative one at the point negative three, negative 24. All that’s left is to substitute into the general equation of the straight line. 𝑦 minus negative 24 equals negative one multiplied by 𝑥 minus negative three. That all simplifies to 𝑦 plus 𝑥 plus 27 equals zero.

The key steps in this question then were to use some trigonometric reasoning to identify that if a line makes an angle of 135 degrees with the positive 𝑥-axis. Then, its gradient or slope is equal to negative tan 45 degrees, which is equal to negative one. We then use the gradient function of the curve to identify the 𝑥-value at which the gradient was equal to negative one. We found the corresponding 𝑦-value by substituting into the equation of the curve and then finally used the general equation of a straight line 𝑦 minus 𝑦 one equals 𝑚 𝑥 minus 𝑥 one to find the equation of this tangent.

Finally, in this video, we’re going to discuss what is meant by a normal to a curve. And we’ll do this in the context of an example.

List the equations of the normals to 𝑦 equals 𝑥 squared plus two 𝑥 at the points where the curve meets the line 𝑦 minus four 𝑥 equals zero.

What is meant by the term normal in this context? Well, we recall first of all that the tangent to a curve has the same gradient as the curve at that point. The normal, however, passes through that same point, but it is perpendicular to the tangent at that point. We can use properties of perpendicular lines to deduce the relationship that exists between the gradient of the tangent and the gradient of the normal to a curve at a given point. The product of the two gradients will be equal to negative one and they will be negative reciprocals of one another.

We must make sure that we’re clear whether we’ve been asked to find the equation of a tangent or a normal when we’re answering questions like this. So now that we know what normals are, let’s see how we can answer this question. We’ve been asked to list the equations of the normals to a given curve at the point where this curve meets another line. So our first step is going to be to find these points of intersection.

We can rearrange the equation of the line to give 𝑦 equals four 𝑥 and then set the two expressions for 𝑦 equal to one another to give an equation in 𝑥 only. We can subtract four 𝑥 from each side and then factor the resulting quadratic to give 𝑥 multiplied by 𝑥 minus two is equal to zero. The two roots of this equation are 𝑥 equals zero or 𝑥 equals two. So we know the 𝑥-coordinates of our points of intersection. To find the corresponding 𝑦-coordinates, we substitute each 𝑥-value back into the equation of the curve to give 𝑦 equals zero when 𝑥 equals zero and 𝑦 equals eight when 𝑥 equals two.

So we now know the two points of intersection. And we, therefore, know the coordinates of one point that lies on each normal. But we need to determine the gradient or slope of each normal. First, we can find the slope of each tangent by differentiating 𝑦 with respect to 𝑥, giving d𝑦 by d𝑥 equals two 𝑥 plus two. When 𝑥 equals zero, the slope will be two. And when 𝑥 equals two, the slope will be six. But remember, this is the slope of the tangent, not the slope of the normal. To find the slope of each normal, we need to take the negative reciprocal of the slope of each tangent. So the slope of our first normal is negative a half and the slope of our second is negative one-sixth.

Finally, we can apply the formula for the general equation of a straight line. For the first normal with a slope of negative a half passing through the point zero, zero, we get the equation two 𝑦 plus 𝑥 equals zero. And for the second with a slope of negative one-sixth passing through the point two, eight, we get the equation six 𝑦 plus 𝑥 minus 50 equals zero. So we found the equations of the two normals. We must be really careful on questions like this. Remember, the slope of the normal is not the same as the slope of the tangent. It’s equal to the negative reciprocal of the slope of the tangent because the two lines are perpendicular to one another.

Let’s summarize what we’ve seen in this video. Firstly, we reminded ourselves that the gradient of a curve is equal to the gradient of the tangent to the curve at that point. So by differentiating and then substituting the 𝑥-value at that point, we can find the slope of the tangent to a curve at any given point. We can then substitute the slope and the coordinates of the point into the general equation of a straight line 𝑦 minus 𝑦 one equals 𝑚𝑥 minus 𝑥 one in order to find the equation of the tangent to the curve at that point.

We also saw that the normal to a curve is perpendicular to the tangent to the curve at that point. And therefore, the product of their slopes is equal to negative one. We can apply all of these key results in order to find the equations of tangents and normals to a variety of different curves.

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