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In this video, we will see how to apply differentiation to find an equation of a tangent to a curve at a given point.
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Weβll 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.
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Weβll 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.
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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.
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Therefore, it follows that if we can use differentiation to find the gradient function of a curve dπ¦ by dπ₯, 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 π₯-value at that point into our gradient function dπ¦ by dπ₯.
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Weβll recall also that the general equation of a straight line with slope π passing through the point π₯ one, π¦ one is π¦ minus π¦ one equals ππ₯ minus π₯ one.
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So we can use the slope, which weβve calculated, and the coordinates of the point at which weβre looking to find the tangent in order to find the equation of a tangent.
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Letβs see how this works in an example.
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Determine the equation of the line tangent to the curve π¦ equals four π₯ cubed minus two π₯ squared plus four at the point negative one, negative two.
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So weβve been given the equation of a curve.
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And we need to determine the equation of the line that is tangent to this curve at a particular point.
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Weβre going to use the formula for the general equation of a straight line π¦ minus π¦ one equals π π₯ minus π₯ one.
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We already know the coordinates π₯ one, π¦ one.
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Itβs the point negative one, negative two.
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But what about π, the slope of this line?
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Well, we recall that the slope of a curve is equal to the slope of the tangent to the curve at that point.
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So in order to find the slope of this tangent, weβre first going to find the gradient function of the curve dπ¦ by dπ₯.
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We can do this by applying the power rule of differentiation, giving dπ¦ by dπ₯ equals four multiplied by three π₯ squared minus two multiplied by two π₯.
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Remember, a constant differentiates to zero.
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So that plus four just differentiates to zero in our derivative, which simplifies to 12π₯ squared minus four π₯.
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Now, thatβs the general gradient function of this curve.
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But we want to know the gradient at a particular point.
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So we need to evaluate dπ¦ by dπ₯ when π₯ is equal to negative one because thatβs our π₯-coordinate at this point.
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This gives 12 multiplied by negative one squared minus four multiplied by negative one, which simplifies to 16.
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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.
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So we have all the information we need in order to use the formula for the general equation of a straight line.
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Substituting the values of π, π₯ one, and π¦ one gives π¦ minus negative two equals 16 π₯ minus negative one.
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Thatβs π¦ plus two equals 16π₯ plus 16.
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And, then, subtracting two from each side in order to collect the constants gives π¦ equals 16π₯ plus 14.
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So this is the equation of the line tangent to the given curve at the point negative one, negative two.
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It passes through the point negative one, negative two and it has the same gradient as the curve at that point.
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Now, letβs consider a second example.
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Find the point on the curve π¦ equals negative 40π₯ squared plus 40 at which the tangent to the curve is parallel to the π₯-axis.
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Now, letβs think about what it means for a line to be parallel to the π₯-axis.
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The π₯-axis is a horizontal line.
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So if another line is parallel to the π₯-axis, then it must also be a horizontal line.
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And what do we know about the slopes of horizontal lines?
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Well, theyβre equal to zero.
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So we know that the slope of the tangent weβre looking to find must be equal to zero.
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Remember also that the slope of a tangent is equal to the slope of the curve at that point.
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So we also know that the slope of the curve at this point must also be equal to zero.
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The slope of the curve is its gradient function dπ¦ by dπ₯.
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So what weβre going to do is find the gradient function dπ¦ by dπ₯ by differentiating π¦ with respect to π₯ and then set it equal to zero.
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Weβll be able to solve the resulting equation to find the π₯-coordinates of the point on the curve where the gradient is equal to zero.
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Step one then is to find dπ¦ by dπ₯, which we can do by applying the power rule of differentiation.
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It gives negative 40 multiplied by two π₯, which is negative 80π₯.
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And remember, the derivative of a constant is zero.
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So our gradient function dπ¦ by dπ₯ is just negative 80π₯.
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Then, we set dπ¦ by dπ₯ equal to zero and solve the resulting equation.
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We have negative 80π₯ equals zero.
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And by dividing both sides of this equation by negative 80, we find that π₯ is equal to zero.
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So we know that the π₯-coordinate of the point on this curve at which the tangent is parallel to the π₯-axis is zero.
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We also need to find the π¦-coordinate, which we can do by substituting π₯ equals zero back into the equation of the curve.
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When π₯ is equal to zero, π¦ is equal to negative 40 multiplied by zero squared plus 40 which is equal to 40.
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So we find that the point on this curve at which the tangent to the curve is parallel to the π₯-axis is the point with coordinates zero, 40.
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Now, we could also have seen this by considering what the graph of π¦ equals negative 40π₯ squared plus 40 actually looks like.
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Itβs a negative parabola as the coefficient of π₯ squared is negative 40 and it has a π¦-intercept of positive 40.
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We can see from our sketch that this function has a critical point at the point with coordinates zero, 40.
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In fact, itβs a local maximum.
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At the critical points for a function, the gradient of the curve and tangent are equal to zero.
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And so, we see that at the zero, 40 β the critical point of this curve β the tangent at this point will be parallel to the π₯-axis.
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Letβs now consider another example.
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The line π₯ minus π¦ minus three equals zero touches the curve π¦ equals ππ₯ cubed plus ππ₯ squared at one, negative two.
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Find π and π.
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The key information given in this question is that the line and the curve touch at this point with coordinates one, negative two.
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But the line does not cross the curve, which means that the line π₯ minus π¦ minus three equals zero is a tangent to the given curve at this point.
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We know that the gradient of a curve is equal to the gradient of the tangent to the curve at that point.
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The equation of our line is π₯ minus π¦ minus three equals zero.
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And when rearranging, we see that this is equivalent to π¦ equals π₯ minus three.
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Comparing with π¦ equals ππ₯ plus π, thatβs the general form of the equation of a straight line in slope-intercept form, we see that the slope of our tangent is one.
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Can we find an expression for the slope of the curve?
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Well, we can do this by differentiation.
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By applying the power rule, we see that dπ¦ by dπ₯ is equal to three ππ₯ squared plus two ππ₯.
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Next, we evaluate this gradient function at the point one, negative, two.
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So we substitute π₯ equals one into our gradient function, giving three π plus two π.
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We can then equate the gradient of the curve at this point with the gradient of the tangent to the curve at this point.
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And it gives an equation involving π and π: three π plus two π is equal to one.
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We canβt solve this equation because we have only one equation and two unknowns.
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So weβre going to need to find a second equation.
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The point one, negative two lies on both the curve and the tangent.
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So if we substitute the values of one and negative two into the equation of the curve, weβll get a second equation connecting π and π.
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We have π multiplied by one cubed plus π multiplied by one squared is equal to negative two, simplifying to π and π equals negative two.
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We now have two linear equations in π and π, which we need to solve simultaneously.
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We can multiply equation two by two, as this will make the coefficient of π the same as it is in equation one.
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Weβll then subtract the second equation from the first to eliminate the π terms, giving π equals five.
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Substituting this value for π into our original equation two thatβs π and π equals negative two gives five plus π equals negative two.
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And by subtracting five, we see that π is equal to negative seven.
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So we found the values of π and π.
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π is equal to five and π is equal to negative seven.
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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.
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Letβs consider another type of example.
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Find the equation of the tangent to the curve π¦ equals π₯ cubed plus nine π₯ squared plus 26π₯ that makes an angle of 135 degrees with the positive π₯-axis.
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So weβve 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.
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But what does it mean when it says this tangent makes an angle of 135 degrees with the positive π₯-axis?
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Letβs consider a sketch.
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Well, it will look something like this.
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The tangent here is shown in pink.
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And we can see that when intersects with the π₯-axis, the angle between the positive π₯-axis and the tangent is 135 degrees.
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In order to apply the general equation of a straight line π¦ minus π¦ one equals π π₯ minus π₯ one, we either need to know the slope π of our line or the coordinates of a point π₯ one, π¦ one which lies on the line.
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So how does knowing that our tangent makes an angle of 135 degrees with the positive π₯-axis help with determining either of those?
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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.
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We can sketch in a right-angled triangle below this line and recall that the slope of a line is change in π¦ over change in π₯.
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Thatβs the vertical height of this triangle divided by the horizontal distance.
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But in that right triangle, those sides are the opposite and adjacent in relation to the angle of 45 degrees.
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So weβre dividing the length of the opposite by the length of the adjacent.
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As the line is sloping downwards though, that vertical change is actually the negative of the value of the opposite.
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So we have that the slope is equal to negative opposite over adjacent.
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Opposite divided by adjacent defines the tangent ratio.
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So in fact, this is equal to negative tan of 45 degrees.
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And tan of 45 degrees is just one.
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So by considering this right-angled triangle, we found that the slope of this line is negative one.
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So, we found the slope of our tangent.
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But we donβt yet know the coordinates of the point on the curve where this tangent is being drawn.
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To find this, we need to find the point on the curve where the gradient or slope is equal to negative one.
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We begin by differentiating the equation of the curve with respect to π₯ and applying the power rule of differentiation, giving dπ¦ by dπ₯ equals three π₯ squared plus 18π₯ plus 26.
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We then set this expression equal to negative one to find the π₯-coordinate of the point on the curve, where the gradient is negative one.
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This simplifies to three π₯ squared plus 18π₯ plus 27 equals zero.
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And then dividing through by three gives π₯ squared plus six π₯ plus nine equals zero.
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We should notice that this is, in fact, a perfect square.
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We can write it as π₯ plus three all squared.
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Solving this equation then, this means that π₯ plus three must be equal to zero.
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And so, π₯ is equal to negative three.
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Next, we need to find the value of π¦ when π₯ is equal to negative three, which we do by substituting negative three into the equation of the curve.
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And it gives negative 24.
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We now know that this tangent has a slope of negative one at the point negative three, negative 24.
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All thatβs left is to substitute into the general equation of the straight line.
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π¦ minus negative 24 equals negative one multiplied by π₯ minus negative three.
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That all simplifies to π¦ plus π₯ plus 27 equals zero.
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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 π₯-axis.
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Then, its gradient or slope is equal to negative tan 45 degrees, which is equal to negative one.
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We then use the gradient function of the curve to identify the π₯-value at which the gradient was equal to negative one.
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We found the corresponding π¦-value by substituting into the equation of the curve and then finally used the general equation of a straight line π¦ minus π¦ one equals π π₯ minus π₯ one to find the equation of this tangent.
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Finally, in this video, weβre going to discuss what is meant by a normal to a curve.
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And weβll do this in the context of an example.
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List the equations of the normals to π¦ equals π₯ squared plus two π₯ at the points where the curve meets the line π¦ minus four π₯ equals zero.
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What is meant by the term normal in this context?
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Well, we recall first of all that the tangent to a curve has the same gradient as the curve at that point.
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The normal, however, passes through that same point, but it is perpendicular to the tangent at that point.
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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.
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The product of the two gradients will be equal to negative one and they will be negative reciprocals of one another.
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We must make sure that weβre clear whether weβve been asked to find the equation of a tangent or a normal when weβre answering questions like this.
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So now that we know what normals are, letβs see how we can answer this question.
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Weβve been asked to list the equations of the normals to a given curve at the point where this curve meets another line.
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So our first step is going to be to find these points of intersection.
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We can rearrange the equation of the line to give π¦ equals four π₯ and then set the two expressions for π¦ equal to one another to give an equation in π₯ only.
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We can subtract four π₯ from each side and then factor the resulting quadratic to give π₯ multiplied by π₯ minus two is equal to zero.
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The two roots of this equation are π₯ equals zero or π₯ equals two.
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So we know the π₯-coordinates of our points of intersection.
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To find the corresponding π¦-coordinates, we substitute each π₯-value back into the equation of the curve to give π¦ equals zero when π₯ equals zero and π¦ equals eight when π₯ equals two.
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So we now know the two points of intersection.
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And we, therefore, know the coordinates of one point that lies on each normal.
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But we need to determine the gradient or slope of each normal.
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First, we can find the slope of each tangent by differentiating π¦ with respect to π₯, giving dπ¦ by dπ₯ equals two π₯ plus two.
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When π₯ equals zero, the slope will be two.
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And when π₯ equals two, the slope will be six.
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But remember, this is the slope of the tangent, not the slope of the normal.
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To find the slope of each normal, we need to take the negative reciprocal of the slope of each tangent.
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So the slope of our first normal is negative a half and the slope of our second is negative one-sixth.
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Finally, we can apply the formula for the general equation of a straight line.
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For the first normal with a slope of negative a half passing through the point zero, zero, we get the equation two π¦ plus π₯ equals zero.
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And for the second with a slope of negative one-sixth passing through the point two, eight, we get the equation six π¦ plus π₯ minus 50 equals zero.
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So we found the equations of the two normals.
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We must be really careful on questions like this.
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Remember, the slope of the normal is not the same as the slope of the tangent.
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Itβs equal to the negative reciprocal of the slope of the tangent because the two lines are perpendicular to one another.
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Letβs summarize what weβve seen in this video.
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Firstly, we reminded ourselves that the gradient of a curve is equal to the gradient of the tangent to the curve at that point.
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So by differentiating and then substituting the π₯-value at that point, we can find the slope of the tangent to a curve at any given point.
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We can then substitute the slope and the coordinates of the point into the general equation of a straight line π¦ minus π¦ one equals ππ₯ minus π₯ one in order to find the equation of the tangent to the curve at that point.
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We also saw that the normal to a curve is perpendicular to the tangent to the curve at that point.
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And therefore, the product of their slopes is equal to negative one.
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We can apply all of these key results in order to find the equations of tangents and normals to a variety of different curves.