Unit #5 - Implicit Differentiation, Related Rates Section 3.7

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1 Unit #5 - Implicit Differentiation, Relate Rates Section 3.7 Some material from Calculus, Single an MultiVariable b Hughes-Hallett, Gleason, McCallum et. al. Copright 005 b John Wile & Sons, Inc. This material is use b permission of John Wile & Sons, Inc. TEST PREPARATION PROBLEMS 3. = so ) = ) + = 0 = = At the point, ), 0, 0 ) =, -) is = ) ) =, so the tangent line with slope / an point = ) The solution at the back of the tet epans the ) prouct, to obtain = 3, but this is unnecessar. 4. [ln)] = [] + ) = Epan: + = Solve for : = = At, e ), = e ) = e, so the tangent line is ) = e )+e This can be epane to = e e +e = e if esire but again, not necessar).

2 5. Group Epan: [ ] = [ 4 = 4) + 4) = 8 3 4) terms: + 3 4) = 8 4) 3 ) + 4) = 8 4) Finall, = 8 4) + 3 4) ] ) 6. At the point 4, ), = 0 all that work for zero?). Given the point 4,), the tangent line is = 0 4)+ or = Note: since we know we are onl going to be computing at a single point, we coul have replace an with 4 an respectivel earlier in the process. It is goo practice, though, to go through solving for completel with the variables. = [ ] +a ) +0 +a) = +a) = +a +a) +a) = +a +a) Rather than simplif completel, we know we onl want at the point 0,0), so we can sub those values in for an now: =,)=0,0) 0+a 0 0+a) =,)=0,0) a Using the point 0,0) an the slope, our tangent line is a = a 0)+0 = a

3 7. Cancel the 3 [ /3 + /3] = a/3 3 /3 + 3 /3 = 0 an simplif: Negative eponents are reciprocals: = /3 /3 = +/3 +/3 = 3 3 At the point,) = a,0), this gives a slope of = 0. Thus the tangent line is horizontal slope zero), an goes through the point a, 0), or = 0 a)+0 or simpl = 0 Yes, even mathematicians fin this a little anti-climactic.) 8. a) Separate ) = 5) = 0 terms from the others: +7 = +4 +7) = +4 = b) horizontal tangents: This curve has a horizontal tangent line when = 0 an,) lies on the curve. For = 0, +4 = 0 or =. Wherever the graph reaches =, the tangent lines will be horizontal. A follow-up question might be Fin all the coorinates of the points on the graph which have horizontal tangents, but that wasn t specificall in the question.) vertical tangents: This curve has a vertical tangent lines when has an infinite limit, which will occur when its enominator equals zero. 7 7, this means +7 = 0 or =. Wherever the graph reaches =, For our the tangent lines will be vertical. 3

4 30. a) Taking + = 5 of both sies, + = 0 = When = 4, there are two possible values: 4 + = 5 so = ± 5 6 = ± 9 = ±3. At the point 4,3), the slope of the tangent is = 4,)=4,3) 3. The tangent line will be = 4 3 4)+3 At the point 4, 3), the slope of the tangent is = 4,)=4, 3) 3 = 4 3. The tangent line will be = 4 3 4) 3 b) The normal line has slope / negative reciprocal slope is perpenicular). At the point 4,3), the slope of the normal line is,)=4,3) 4 = 3 4. The normal line will be = 3 4 4)+3 or simpl = 3 4 Similarl, the normal line at the point 4, 3) will be = 3 3 4) 3 or more simpl = 4 4 c) Note that both these normal lines have zero intercepts, which means the will both pass through 0,0). This makes sense for a circle, as an line outwars from the origin will intersect the circle at right-angles. 3. The tangent line to = can be compute without the nee for implicit ifferentiation. At the point =, the tangent line to = will be = )+ or = A circle centere at 8,0) will be of the form 8) + = R We fin the slope of its tangents b implicit ifferentiation: 8)+ = 0 = 8) = 8 3 4

5 We want to fin both a point,) an a value for R such that the tangent line to the circle at that point is also =. To have this tangent line, the circle must be at point where the slope is : = 8 = 8 Also, the full tangent line must equal =. To satisf both equations, the values on the tangent line an efine the circle point s slope must be equal, or ) 8 = 4 = 5 = 5 = an using = to fin, = ) = 3 Thus the point on the circle is,) =,3), so the raius of the circle satisfies 8) +3 = R R = 45 For reference, here are two graphs showing the relationship between the circle an the parabola, an the share tangent line a) Differentiating both sies with respect to P gives 4f ) P P f = K P = 0 5

6 Implicitl, f is a function of P, so using a combination of the quotient, prouct an chain rules, [ ] f f P 4 P +f f ) [ f P ] ) f f P f ) = 0 First multipl both sies b f ) to remove the enominator, an ivie b 4, [ fp f ] P +f f )+f 3 P f P = 0 fp f )+f 3 P ) f P = f f ) so f P = f f ) fp f )+f 3 P) The question is about the sign of this erivative value, an since f is a fraction an so positive an less than ) an P is a pressure in atmospheres an also positive), the numerator f f ) must be negative, an the enominator must be positive both terms positive, an ae together) so f P must alwas be negative. 6

Try It. Implicit and Explicit Functions. Video. Exploration A. Differentiating with Respect to x

Try It. Implicit and Explicit Functions. Video. Exploration A. Differentiating with Respect to x SECTION 5 Implicit Differentiation Section 5 Implicit Differentiation Distinguish between functions written in implicit form an eplicit form Use implicit ifferentiation to fin the erivative of a function