Lesson Video: Tangents and Normals to the Graph of a Function | Nagwa Lesson Video: Tangents and Normals to the Graph of a Function | Nagwa

Lesson Video: Tangents and Normals to the Graph of a Function Mathematics

In this video, we will learn how to find the equations of tangents and normals to trigonometric, parametric, and implicitly defined curves using derivatives.

17:15

Video Transcript

In this video, we will learn how to find the equations of tangents and normals to trigonometric, parametric, and implicitly defined curves using derivatives. Let’s begin by recalling the basics of tangents and normals. We know that at any given point on a curve, the tangent to the curve has the same slope as the curve itself at that point. The slope of a curve is given by its first derivative, which may be d𝑦 by d𝑥 or 𝑓 prime of 𝑥 or something else, depending on how the equation of the curve is specified. We can obtain the equation of the tangent to a curve at a given point by substituting the coordinates of this point, which we’ll call 𝑥 one, 𝑦 one, and the slope of the tangent, which we’ll call 𝑚, into the point–slope form of the equation of the straight line, 𝑦 minus 𝑦 one equals 𝑚 𝑥 minus 𝑥 one.

The way in which we find an expression for the slope function, and hence evaluate the slope at the given point, will depend on how the curve is defined. In the course of this video, we’ll see examples of how we can do this for trigonometric, parametric, and implicitly defined curves. If instead we want to find the equation of the normal to a curve at a given point, we recall that the normal is perpendicular or orthogonal to the tangent at that point. We also recall that if two lines are perpendicular, then the product of their slopes is negative one. Or, put another way, their slopes are the negative reciprocals of one another.

We can therefore find the slope of the normal to a curve at a given point by finding the negative reciprocal of the slope of the tangent at that point. We then proceed in the same way by substituting this slope and the coordinates of the point at which we’re finding the normal into the point–slope form of the equation of a straight line. Let’s begin with an example in which we’ll find the equation of the normal to a curve which is defined implicitly.

Find the equation of the normal to the curve three 𝑦 squared minus nine 𝑦𝑥 plus seven 𝑥 squared equals one at the point negative one, negative one.

The curve we’ve been given has been defined implicitly. It is a function of both 𝑥 and 𝑦 and cannot be easily rearranged to a form in which we have 𝑦 as a function of 𝑥. To find the equation of the normal to this curve, we’re going to use the point–slope form of the equation of a straight line. That’s 𝑦 minus 𝑦 one equals 𝑚 𝑥 minus 𝑥 one. We know the coordinates of a point on the normal, the point negative one, negative one. But we need to calculate its slope. We recall that a normal to a curve is perpendicular to the tangent to the curve at that point. So first, we must calculate the slope of the tangent. To do this, we’re going to need to find the slope function of the curve. So we need to recall how we differentiate a function which is defined implicitly.

Implicit differentiation is an application of the chain rule. Recall that if 𝑦 is a function of 𝑢 and 𝑢 is a function of 𝑥, then the chain rule states that d𝑦 by d𝑥 is equal to d𝑦 by d𝑢 multiplied by d𝑢 by d𝑥. So the derivative of 𝑦 with respect to 𝑥 is found by taking the derivative of 𝑦 with respect to 𝑢 and then multiplying this by the derivative of 𝑢 with respect to 𝑥. Let’s consider then how we can apply this to the equation of our curve. We’ll do this term by term, starting with the first term three 𝑦 squared. As three 𝑦 squared is a function of 𝑦 and we can consider 𝑦 to be a function of 𝑥, by the chain rule, we have that the derivative with respect to 𝑥 of three 𝑦 squared is equal to the derivative with respect to 𝑦 of three 𝑦 squared multiplied by d𝑦 by d𝑥.

By the power rule of differentiation, the derivative with respect to 𝑦 of three 𝑦 squared is simply six 𝑦. So we find that the derivative with respect to 𝑥 of three 𝑦 squared is six 𝑦 d𝑦 by d𝑥. So, we’ve differentiated the first term in our equation implicitly. Differentiating the third term is straightforward because it is a function of 𝑥 only. By the power rule of differentiation, the derivative with respect to 𝑥 of seven 𝑥 squared is 14𝑥. And differentiating the term on the right-hand side of the equation is straightforward because it is a constant. And we know that the derivative of any constant with respect to 𝑥 is simply zero.

We have one term left to differentiate, negative nine 𝑦𝑥. And this is a little more complicated as it is a product involving both 𝑦 and 𝑥. We’ll need to use implicit differentiation. But we also need to recall the product rule. This states that for two differentiable functions 𝑢 and 𝑣, the derivative with respect to 𝑥 of their product 𝑢𝑣 is equal to 𝑢 times d𝑣 by d𝑥 plus 𝑣 times d𝑢 by d𝑥. We’ll therefore let 𝑢 equal negative nine 𝑦 and 𝑣 equal 𝑥. We then need to find their individual derivatives with respect to 𝑥. d𝑣 by d𝑥 is straightforward because 𝑣 is a function of 𝑥 only. If 𝑣 is equal to 𝑥, then d𝑣 by d𝑥 is equal to one.

For d𝑢 by d𝑥 though, as 𝑢 is a function of 𝑦, we’re going to need to use implicit differentiation again. By the chain rule, we have that the derivative with respect to 𝑥 of negative nine 𝑦 is equal to the derivative with respect to 𝑦 of negative nine 𝑦 multiplied by d𝑦 by d𝑥. The derivative with respect to 𝑦 of negative nine 𝑦 is simply negative nine. So we have that d𝑢 by d𝑥 is equal to negative nine d𝑦 by d𝑥. We then need to substitute each of these expressions into the product rule. We have that the derivative with respect to 𝑥 of negative nine 𝑦𝑥 is equal to 𝑢 times d𝑣 by d𝑥. That’s negative nine 𝑦 multiplied by one. Then we add 𝑣 times d𝑢 by d𝑥. That’s 𝑥 multiplied by negative nine d𝑦 by d𝑥. And this will simplify to negative nine 𝑦 minus nine 𝑥 d𝑦 by d𝑥.

So we’ve now differentiated the entire equation, giving six 𝑦 d𝑦 by d𝑥 minus nine 𝑦 minus nine 𝑥 d𝑦 by d𝑥 plus 14𝑥 is equal to zero. We want to evaluate the slope function at a particular point. So first, we need to rearrange this equation to give an explicit expression for d𝑦 by d𝑥. In other words, we need to make d𝑦 by d𝑥 the subject of this equation. We begin by collecting the terms that do involve d𝑦 by d𝑥 on one side of the equation and the terms that don’t on the other. So we have six 𝑦 d𝑦 by d𝑥 minus nine 𝑥 d𝑦 by d𝑥 is equal to nine 𝑦 minus 14𝑥. We can then factor the left-hand side of the equation to give six 𝑦 minus nine 𝑥 multiplied by d𝑦 by d𝑥 is equal to nine 𝑦 minus 14𝑥.

And to find an expression for d𝑦 by d𝑥 which is in terms of both 𝑥 and 𝑦, we divide both sides of the equation by six 𝑦 minus nine 𝑥. So we have d𝑦 by d𝑥 is equal to nine 𝑦 minus 14𝑥 over six 𝑦 minus nine 𝑥. We have an expression for the general slope function of the curve. But we need to evaluate the slope at a particular point, the point negative one, negative one. We therefore need to substitute the value negative one for both 𝑥 and 𝑦. We have negative nine plus 14 in the numerator and negative six plus nine in the denominator, which simplifies to the fraction five over three.

Remember though that this is the slope of the tangent to the curve at the point negative one, negative one, whereas we’re looking to find the equation of the normal. But we know that the slope of the normal is the negative reciprocal of the slope of the tangent. So we have that the slope of the normal is equal to negative three-fifths. We can invert the fraction and change the sign.

As we now know the coordinates of a point on the normal and its slope, we can find its equation using the point–slope form of the equation of a straight line. We have 𝑦 minus negative one is equal to negative three-fifths 𝑥 minus negative one. Of course, that’s just 𝑦 plus one equals negative three-fifths 𝑥 plus one. We can multiply through by five and distribute the parentheses on the right-hand side to give five 𝑦 plus five equals negative three 𝑥 minus three. And finally, we’ll group all of the terms on the left-hand side of the equation. Using implicit differentiation then, we found that the equation of the normal to the given curve at the point negative one, negative one is five 𝑦 plus three 𝑥 plus eight equals zero.

Let’s now consider an example in which we find the equation of the tangent to a curve which is defined parametrically.

Find the equation of the tangent to the curve 𝑥 equals 𝑡 cubed plus one, 𝑦 equals 𝑡 to the fourth power plus 𝑡 at the point corresponding to the value 𝑡 equals negative one.

The equation of this curve has been given in parametric form. 𝑥 and 𝑦 are both given as functions of a third variable 𝑡. We could write 𝑥 is equal to some function 𝑓 of 𝑡 and 𝑦 is equal to another function 𝑔 of 𝑡, if we wish. To find the equation of the tangent, we can use the point–slope form of the equation of a straight line. So we need to know two things: the coordinates of a point that lies on the tangent and its slope. Let’s begin by finding the coordinates for the point at which we’re finding the tangent. We’re told that this point corresponds to the value 𝑡 equals negative one. So we can find the coordinates of this point by substituting 𝑡 equals negative one into the parametric equations for 𝑥 and 𝑦.

When 𝑡 equals negative one, 𝑥 is equal to negative one cubed plus one, which is negative one plus one, which is zero. And 𝑦 is equal to negative one to the fourth power plus negative one. That’s one minus one, which is also equal to zero. So we now know that the tangent we’re looking for is the tangent at the origin, the point zero, zero.

Next, we need to find the slope of the tangent, which we can do by using parametric differentiation to find the slope of the curve at this point. We recall that if 𝑥 and 𝑦 are each functions of a third parameter 𝑡, then d𝑦 by d𝑥 is equal to d𝑦 by d𝑡 over d𝑥 by d𝑡. We need to find the individual derivatives of 𝑥 and 𝑦 with respect to 𝑡, which we can do using the power rule of differentiation. d𝑥 by d𝑡 will be equal to three 𝑡 squared, and d𝑦 by d𝑡 is equal to four 𝑡 cubed plus one. Substituting each of these expressions into our formula for d𝑦 by d𝑥, we have that d𝑦 by d𝑥 is equal to four 𝑡 cubed plus one over three 𝑡 squared. This is, of course, in terms of the parameter 𝑡.

To find the slope at the point where 𝑡 is equal to negative one, we substitute 𝑡 equals negative one into our expression for d𝑦 by d𝑥. This gives negative four plus one over three multiplied by one. That’s negative three over three, which is equal to negative one. And so we find that the slope of this tangent is negative one. We now know that this tangent passes through the origin and has a slope of negative one. We can go through the formal process of substituting these values into the point–slope form of the equation of a straight line if we wish. Or we may be able to recognize the equation of this tangent. It’s 𝑦 equals negative 𝑥.

In our final example, we’ll find the equation of the normal to a curve whose definition involves trigonometric and reciprocal trigonometric functions.

Find the equation of the normal to the curve 𝑦 equals eight cos 𝑥 minus three sec 𝑥 at 𝑥 equals 𝜋 by three.

To find the equation of the normal to any curve, we need to know the coordinates of a point that lies on the normal and its slope. We’re asked for the normal at the point where 𝑥 is equal to 𝜋 by three. We can find the 𝑦-value at this point by substituting 𝑥 equals 𝜋 by three into the equation of the curve. cos of 𝜋 by three is one-half, so sec of 𝜋 by three is two. We have eight multiplied by a half minus three multiplied by two. That’s four minus six, which is equal to negative two. So, the coordinates of the point at which we’re finding the normal are 𝜋 by three, negative two.

Next, we need to find the slope of the normal. Recall that the normal at any point on a curve is perpendicular or orthogonal to the tangent at that same point. And hence, their slopes are the negative reciprocals of one another. The slope of the tangent is the same as the slope of the curve itself. And we can find the slope function of the curve using differentiation. We need to recall two general rules for differentiating trigonometric functions. These rules only apply when the angle 𝑥 is measured in radians. Firstly, the derivative with respect to 𝑥 of cos 𝑥 is negative sin 𝑥. And secondly, the derivative with respect to 𝑥 of sec 𝑥 is sec 𝑥 tan 𝑥. We should be familiar with the derivatives of the three trigonometric functions sin, cos, and tan as well as their reciprocals csc, sec, and cot.

Applying these results then, we have that d𝑦 by d𝑥 is equal to eight multiplied by negative sin 𝑥 minus three multiplied by sec 𝑥 tan 𝑥. So we have the general slope function of the curve. We need to evaluate this at the point where 𝑥 is equal to 𝜋 by three. We have negative eight sin 𝜋 by three minus three sec 𝜋 by three tan 𝜋 by three. Evaluating on a calculator or recalling these results from memory, we have negative eight multiplied by root three over two minus three multiplied by two multiplied by root three. That’s negative four root three minus six root three, which simplifies to negative 10 root three.

Remember though that this is the slope of the tangent at the point where 𝑥 equals 𝜋 by three. The slope of the normal is the negative reciprocal of this. We can cancel the negatives and multiply both the numerator and denominator by root three in order to rationalize the denominator. So we find that the slope of the normal is root three over 30.

Finally, we’re able to calculate the equation of the normal by substituting the coordinates of the point and the slope we found into the point–slope form of the equation of a straight line. We have 𝑦 minus negative two equals root three over 30 multiplied by 𝑥 minus 𝜋 by three. Simplifying on the left-hand side and then distributing the parentheses on the right-hand side, we have 𝑦 plus two is equal to root three 𝑥 over 30 minus root three 𝜋 over 90. Finally, we’ll group all the terms on the same side of the equation. And we have that the equation of the normal to the given curve at the point where 𝑥 equals 𝜋 by three is 𝑦 minus root three 𝑥 over 30 plus root three 𝜋 over 90 plus two equals zero.

Let’s now summarize the key points from this video. Firstly, we recalled the basics of tangents and normals. The tangent to a curve at any given point has the same slope as the curve itself at that point. The normal is perpendicular to the tangent, and the product of their slopes are negative one, which means that the slopes are the negative reciprocals of one another. We can find the slope function of a curve using differentiation. And the method we use will depend on how the curve is defined. For a curve that has been defined implicitly, we use implicit differentiation, which is an application of the chain rule. If 𝑦 is a function of 𝑢 and 𝑢 is a function of 𝑥, then d𝑦 by d𝑥 is equal to d𝑦 by d𝑢 multiplied by d𝑢 by d𝑥.

For a curve defined parametrically in terms of a third parameter 𝑡, d𝑦 by d𝑥 is equal to d𝑦 by d𝑡 over d𝑥 by d𝑡. We also need to recall the derivatives of trigonometric functions where the argument is given in radians. The derivative with respect to 𝑥 of sin 𝑥 is cos 𝑥. The derivative with respect to 𝑥 of cos 𝑥 is negative sin 𝑥. And the derivative with respect to 𝑥 of tan 𝑥 is sec squared 𝑥. And finally, we should also recall the standard results for the derivatives of the three reciprocal trigonometric functions. These results and techniques enable us to find the equations of tangents and normals to the graphs of more advanced functions.

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